Translatory motion

Class 11 Physics, unit 3, book pages 32–52

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This unit describes how bodies move along straight lines and curved paths, and what happens when they interact. It derives the three equations of uniformly accelerated motion and uses them for free fall and for projectile motion, where horizontal and vertical motions are treated independently. It then introduces the law of conservation of momentum, which explains explosions, recoil and collisions, and compares elastic and inelastic collisions.

Learning outcomes (7)

Derive the equations of motion [for uniform acceleration cases only. Derive from the definitions of velocity and acceleration as well as graphically].
Solve problems using the equations of motion [for the cases of uniformly accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance. This also includes situations where the equations of motion need to be resolved into vertical and horizontal components for 2-D motion].
Evaluate and analyse projectile motion in the absence of air resistance [this includes solving problems making use of the facts that (i) the horizontal component ( $v_{H}$ ) of velocity is constant, (ii) the acceleration is in the vertical direction and is the same as that of a vertically free falling object, (iii) the horizontal motion and vertical motion are independent of each other]. Situations may require students to determine for projectiles: how high does it go, how far would it go along the level land, where would it be after a given time, how long will it remain in flight. Situations may also require students to calculate for a projectile launched from ground height the launch angle that results in the maximum range and the relation between the launch angles that result in the same range.
Predict qualitatively how air resistance affects projectile motion [this includes analysis of both the horizontal component and vertical component of velocity and hence predicting qualitatively the range of the projectile].
Apply the principle of conservation of momentum to solve simple problems [including elastic and inelastic interactions between objects in both one and two dimensions. Knowledge of the concept of coefficient of restitution is not required]. Examples of applications include: karate chops to break a pile of bricks, car crashes, ball and bat, and the motion under thrust of a rocket in a straight line considering short thrusts during which the mass remains constant.
Predict and analyse motion for elastic collisions [this includes making use of the fact that for an elastic collision, total kinetic energy is conserved and the relative speed of approach is equal to the relative speed of separation].
Justify why though the momentum of a closed system is always conserved, some change in kinetic energy may take place.

Introduction

A body can show three types of motion: translational motion, rotational motion and vibrational motion. In this unit we study only translational motion, such as motion along a straight line and projectile motion.

We begin with kinematics, the branch of mechanics that deals with the study of motion without considering the causes of motion. It uses the terms displacement, velocity and acceleration. First we derive the equations of motion for constant acceleration and use them to solve problems of straight-line motion, including bodies falling in a uniform gravitational field without air resistance. The same ideas are then extended to objects moving along curved paths, such as projectile motion under the action of gravity only.

Next we study dynamics, the branch of mechanics that deals with the forces that cause objects and systems to move. This includes momentum and the associated law of conservation of momentum.

Kinematics: The branch of mechanics that deals with the study of motion without considering the causes of motion.
Dynamics: The branch of mechanics that deals with the study of the forces that cause objects and systems to move.

3.1Equations of uniformly accelerated motionOften examined

There are three equations of motion. They show the relation between the initial velocity, final velocity, acceleration, displacement and time, and they describe and predict the motion of objects under constant acceleration. To keep the derivations simple, the following assumptions are made.

The object is moving along a straight line.
Acceleration is constant.
Only the magnitudes of vectors such as displacement, velocity and acceleration are considered.
The direction of the initial velocity, and of all quantities in that direction, is taken as positive.

The three equations of motion are:

v_{f} = v_{i} + a t

S = v_{i} t + \frac{1}{2} a t^{2}

2 a S = v_{f}^{2} - v_{i}^{2}

Symbols used in the equations of motion.
Symbol	Meaning
$v_{i}$	initial velocity
$v_{f}$	final velocity after time $t$
$a$	constant acceleration
$S$	distance (displacement) travelled in time $t$
$t$	time

3.1.1First equation of motionOften examined

Derivation First equation of motion

Aim. Derive $v_{f} = v_{i} + a t$ from the velocity-time graph.

Assumptions.

A body moves in a straight line with constant acceleration $a$ .
Its velocity changes from $v_{i}$ to $v_{f}$ in time $t$ .

t

v_{i}

v_{f}

time

velocity

O

A

B

C

D

Area = v_{i} t

\frac{1}{2} t (a t)

Figure 3.1: Velocity-time graph for a body moving with constant acceleration. The slope of AB is the acceleration

a

, and the area under AB is the distance travelled.

The slope of a velocity-time graph gives the acceleration. In Figure 3.1 the graph is the straight line AB, so the acceleration is the slope of AB.
$a = slope of line AB = \frac{B C}{A C}$
Draw BC vertical and AC horizontal. BC is the rise BD $-$ CD, and AC is equal to OD.
$a = \frac{B D - C D}{O D} (3.1)$
Read the lengths from the graph: BD is the final velocity $v_{f}$ , CD is the initial velocity $v_{i}$ (CD equals OA) and OD is the time $t$ .
$a = \frac{v _{f} - v _{i}}{t}$
Multiply both sides by $t$ .
$a t = v_{f} - v_{i}$
Rearrange for $v_{f}$ .
$v_{f} = v_{i} + a t (3.2)$

Result

v_{f} = v_{i} + a t

Equation (3.2) is the first equation of motion. It relates initial velocity, final velocity, acceleration and time. The gain in speed $v_{f} - v_{i} = a t$ is the height BC of the triangle in the graph.

3.1.2Second equation of motionOften examined

Derivation Second equation of motion

Aim. Derive $S = v_{i} t + \frac{1}{2} a t^{2}$ from the velocity-time graph.

Assumptions.

A body moves in a straight line with initial velocity $v_{i}$ and constant acceleration $a$ .
Its velocity becomes $v_{f}$ after time $t$ , as in Figure 3.1.

t

v_{i}

v_{f}

time

velocity

O

A

B

C

D

Area = v_{i} t

\frac{1}{2} t (a t)

Figure 3.1: Velocity-time graph for a body moving with constant acceleration. The slope of AB is the acceleration

a

, and the area under AB is the distance travelled.

The distance $S$ travelled is equal to the total area under the velocity-time graph. Here that area is the trapezium OABD.
$S = area of trapezium OABD = \frac{1}{2} (sum of parallel sides) \times (distance between the parallel sides)$
OA and BD are the parallel sides, and OD is the distance between them.
$S = \frac{1}{2} (O A + B D) \times O D$
Put $O A = v_{i}$ , $B D = v_{f}$ and $O D = t$ .
$S = \frac{1}{2} (v_{i} + v_{f}) \times t$
Replace $v_{f}$ using the first equation of motion, $v_{f} = v_{i} + a t$ .
$S = \frac{1}{2} (v_{i} + v_{i} + a t) \times t$
Add the two $v_{i}$ terms.
$S = \frac{1}{2} (2 v_{i} + a t) \times t$
Multiply out the bracket.
$S = \frac{1}{2} (2 v_{i} t + a t^{2})$
Divide each term by 2.
$S = v_{i} t + \frac{1}{2} a t^{2} (3.3)$

Result

S = v_{i} t + \frac{1}{2} a t^{2}

Equation (3.3) is the second equation of motion. It relates initial velocity, acceleration, distance and time. The two terms are the two pieces of area in Figure 3.1: the rectangle $v_{i} t$ and the triangle $\frac{1}{2} t (a t)$ .

3.1.3Third equation of motionOften examined

Derivation Third equation of motion

Aim. Derive $2 a S = v_{f}^{2} - v_{i}^{2}$ from the velocity-time graph.

Assumptions.

A body moves in a straight line with initial velocity $v_{i}$ and constant acceleration $a$ .
Its velocity becomes $v_{f}$ after time $t$ , as in Figure 3.1.

t

v_{i}

v_{f}

time

velocity

O

A

B

C

D

Area = v_{i} t

\frac{1}{2} t (a t)

Figure 3.1: Velocity-time graph for a body moving with constant acceleration. The slope of AB is the acceleration

a

, and the area under AB is the distance travelled.

The distance $S$ is the area of the trapezium OABD under the graph.
$S = \frac{1}{2} (O A + B D) \times O D$
Multiply both sides by 2.
$2 S = (O A + B D) \times O D$
Multiply both sides by $\frac{B C}{O D}$ . This brings in the acceleration without bringing in the time.
$2 S \times \frac{B C}{O D} = (O A + B D) \times O D \times \frac{B C}{O D}$
The slope of AB is $\frac{B C}{O D} = a$ , and $O D$ cancels on the right.
$2 S \times a = (O A + B D) \times B C$
Put $O A = v_{i}$ , $B D = v_{f}$ and $B C = v_{f} - v_{i}$ .
$2 a S = (v_{f} + v_{i}) (v_{f} - v_{i})$
Use $(x + y) (x - y) = x^{2} - y^{2}$ .
$2 a S = v_{f}^{2} - v_{i}^{2} (3.4)$

Result

2 a S = v_{f}^{2} - v_{i}^{2}

Equation (3.4) is the third equation of motion. It relates initial velocity, final velocity, acceleration and distance, with no time in it, so it is the one to use when $t$ is not given.

Worked examples

Example 3.1

A bus is travelling at

10 m s^{- 1}

and accelerates uniformly at

2 m s^{- 2}

. Calculate the velocity of the bus after

5 s

Given

$v_{i} = 10 m s^{- 1}$
$a = 2 m s^{- 2}$
$t = 5 s$

To find

$v_{f}$

Time and acceleration are known and $S$ is not involved, so use the first equation of motion.
$v_{f} = v_{i} + a t$
Substitute the values.
$v_{f} = 10 + 2 (5)$
Evaluate.
$v_{f} = 20 m s^{- 1}$

v_{f} = 20 m s^{- 1}

Example 3.2

A car is travelling initially with a velocity of

4 m s^{- 1}

then it accelerates at

1 m s^{- 2}

for

10 s

. Find the distance travelled by the car during this time.

Given

$v_{i} = 4 m s^{- 1}$
$a = 1 m s^{- 2}$
$t = 10 s$

To find

The distance is wanted and $v_{f}$ is not given, so use the second equation of motion.
$S = v_{i} t + \frac{1}{2} a t^{2}$
Substitute the values.
$S = 4 (10) + \frac{1}{2} (1) (10)^{2}$
Evaluate each term.
$S = 40 + 50$
Add.
$S = 90 m$

S = 90 m

Assignment 3.1 (1)

A train slows down from

80 km h^{- 1}

with a uniform retardation of

2 m s^{- 2}

. How long will it take to attain a speed of

20 km h^{- 1}

Given

$v_{i} = 80 km h^{- 1}$
$v_{f} = 20 km h^{- 1}$
$a = - 2 m s^{- 2}$

To find

Convert the speeds to $m s^{- 1}$ by multiplying by $\frac{1000}{3600}$ .
$v_{i} = 80 \times \frac{1000}{3600} = 22.2 m s^{- 1}, v_{f} = 20 \times \frac{1000}{3600} = 5.56 m s^{- 1}$
Retardation is negative acceleration. Use the first equation of motion and solve for $t$ .
$t = \frac{v _{f} - v _{i}}{a}$
Substitute.
$t = \frac{5.56 - 22.2}{- 2} = 8.33 s$

t \approx 8.3 s

Assignment 3.1 (2)

A car travels with a velocity of

5 m s^{- 1}

. It then accelerates uniformly and travels a distance of

50 m

. If the velocity reached is

15 m s^{- 1}

, find the acceleration and the time to travel this distance.

Given

$v_{i} = 5 m s^{- 1}$
$v_{f} = 15 m s^{- 1}$
$S = 50 m$

To find

Time is not given, so find $a$ from the third equation of motion.
$a = \frac{v _{f}^{2} - v _{i}^{2}}{2 S}$
Substitute.
$a = \frac{1 5 ^{2} - 5 ^{2}}{2 ( 50 )} = \frac{225 - 25}{100} = 2 m s^{- 2}$
Now use the first equation of motion to find $t$ .
$t = \frac{v _{f} - v _{i}}{a} = \frac{15 - 5}{2} = 5 s$
Check with the second equation: $S = 5 (5) + \frac{1}{2} (2) (5)^{2} = 25 + 25 = 50 m$ , as given.

a = 2 m s^{- 2}, t = 5 s

Worked example: stopping distance when the speed doubles

A car moving at

20 m s^{- 1}

brakes with a uniform retardation of

5 m s^{- 2}

. (a) Find the time and the distance to stop. (b) Find the stopping distance if it was moving at

40 m s^{- 1}

and explain why it is not just double.

Given

$v_{i} = 20 m s^{- 1}$
$v_{f} = 0$
$a = - 5 m s^{- 2}$

To find

(a) $t$ and $S$
(b) $S$ for $v_{i} = 40 m s^{- 1}$

(a) Retardation means the acceleration is opposite to the motion, so $a$ is negative. The car stops, so $v_{f} = 0$ . The first equation has no $S$ , so use it for the time.
$t = \frac{v _{f} - v _{i}}{a} = \frac{0 - 20}{- 5} = 4 s$
The third equation has no $t$ , so use it for the distance.
$S = \frac{v _{f}^{2} - v _{i}^{2}}{2 a} = \frac{0 - 400}{2 ( - 5 )} = 40 m$
Check with the second equation.
$S = v_{i} t + \frac{1}{2} a t^{2} = 20 (4) - \frac{1}{2} (5) (16) = 80 - 40 = 40 m$
(b) Same retardation, new initial speed.
$S = \frac{0 - 4 0 ^{2}}{2 ( - 5 )} = 160 m$
Why is the stopping distance not just double? With $v_{f} = 0$ and the same retardation, the third equation gives $S = \frac{v _{i}^{2}}{2∣ a ∣}$ , so $S$ is proportional to the square of the initial speed. Doubling the speed makes the distance four times as large.
$S \propto v_{i}^{2} \Rightarrow S = 4 \times 40 = 160 m$

t (s)

v (m s^{- 1})

40 m s^{- 1}

20 m s^{- 1}

Velocity against time for the braking car. The area under each line is the stopping distance:

40 m

from

20 m s^{- 1}

and

160 m

from

40 m s^{- 1}

t = 4 s, S = 40 m (a); S = 160 m (b)

3.1.4Free-fall motionOften examined

An example of uniformly accelerated motion is the motion of a free-falling body. In the absence of air resistance, all objects (lighter or heavier) fall freely near the surface of the Earth with the same acceleration, independent of their masses. This acceleration is called the acceleration due to gravity, denoted by g. It has a value of 9.81\,\mathrm{m\,s^{-2}} and is directed towards the centre of the Earth.

Substituting a=g and S=h in equations (3.2), (3.3) and (3.4) gives the equations of motion for free fall.

v_f=v_i+gt

h=v_it+\tfrac12gt^{2}

2gh=v_f^{2}-v_i^{2}

These equations let us solve problems on bodies falling in a uniform gravitational field without air resistance.

Example 3.3

A kangaroo can jump over an object 2.50\,\mathrm{m} high. (a) Considering just its vertical motion, calculate its vertical speed when it leaves the ground. (b) How long a time is it in the air?

Given

h = 2.50\,\mathrm{m}
v_f = 0\,\mathrm{m\,s^{-1}}\ \text{(at the highest point)}
g = -9.8\,\mathrm{m\,s^{-2}}\ \text{(moving up)}

To find

(a) v_i
(b) total time in the air T

(a) Initial velocity

No time is given, so use the third equation of motion.
2gh=v_f^{2}-v_i^{2}
Substitute with g=-9.8 because the kangaroo moves upwards.
2(-9.8)(2.5)=0-v_i^{2}
Simplify.
v_i^{2}=49\,\mathrm{m^{2}\,s^{-2}}
Take the square root.
v_i=7\,\mathrm{m\,s^{-1}}

(b) Time in the air

Find the time t to reach the highest point with the first equation of motion.
v_f=v_i+gt
Substitute (g is negative for upward motion).
0=7-9.8\,t
Solve.
t=0.71\,\mathrm{s}
The fall takes as long as the rise, so the total time in the air is twice this.
T=2(0.71)=1.42\,\mathrm{s}

v_i=7\,\mathrm{m\,s^{-1}},\quad T=1.42\,\mathrm{s}

Assignment 3.2

A 1\,\mathrm{kg} ball is dropped from the top of the leaning tower of Pisa. The ball reaches the ground in 3.34\,\mathrm{s}. Find (a) the height of the tower, (b) the velocity of the ball when it strikes the ground.

Given

v_i = 0
t = 3.34\,\mathrm{s}
g = 9.8\,\mathrm{m\,s^{-2}}\ \text{(downwards is positive)}

To find

(a) h
(b) v_f

Dropped means v_i=0. Height from the second equation of motion.
h=v_it+\tfrac12gt^{2}=0+\tfrac12(9.8)(3.34)^{2}
Evaluate.
h=4.9\times11.16=54.7\,\mathrm{m}
Final velocity from the first equation of motion.
v_f=v_i+gt=0+9.8(3.34)=32.7\,\mathrm{m\,s^{-1}}

h\approx54.7\,\mathrm{m},\quad v_f\approx32.7\,\mathrm{m\,s^{-1}}

Worked example: a ball thrown straight up

A ball is thrown vertically upwards at 19.6\,\mathrm{m\,s^{-1}}. Find (a) the time to reach the top, (b) the maximum height, (c) the total time in the air and the speed on returning to the thrower's hand.

Given

v_i = 19.6\,\mathrm{m\,s^{-1}}
g = -9.8\,\mathrm{m\,s^{-2}}
v_f\ \text{(top)} = 0

To find

(a) t
(b) h
(c) total time and landing speed

Take upward as positive, so g=-9.8\,\mathrm{m\,s^{-2}}. (a) At the top the velocity is zero.
t=\dfrac{v_f-v_i}{g}=\dfrac{0-19.6}{-9.8}=2\,\mathrm{s}
(b) Use the equation without t.
h=\dfrac{v_f^{2}-v_i^{2}}{2g}=\dfrac{0-384.16}{2(-9.8)}=19.6\,\mathrm{m}
(c) The fall from the top is free fall from rest through 19.6\,\mathrm{m}, which takes the same time as the rise. So the total time is double.
T=2\times2=4\,\mathrm{s}
Check with h=v_it+\tfrac12gt^{2} at t=4\,\mathrm{s}: the displacement must be zero (back at the hand).
h=19.6(4)-\tfrac12(9.8)(16)=78.4-78.4=0
Speed on return.
v_f=v_i+gt=19.6-9.8(4)=-19.6\,\mathrm{m\,s^{-1}}

The minus sign only means downwards: the ball returns with the same speed 19.6\,\mathrm{m\,s^{-1}} with which it left.

Velocity against time for the ball thrown upwards. The velocity falls by 9.8\,\mathrm{m\,s^{-1}} every second, is zero at the top (t=2\,\mathrm{s}) and the area above the axis (going up) equals the area below it (coming down).

t=2\,\mathrm{s},\quad h=19.6\,\mathrm{m},\quad T=4\,\mathrm{s},\quad |v|=19.6\,\mathrm{m\,s^{-1}}\ \text{downwards}

3.2Projectile motionOften examined

So far we have studied motion of bodies in a straight line, either horizontal or vertical. There are many situations in which a body moves along a curved path in a plane, having both vertical and horizontal components. For example, when a player kicks a football to another player it moves in a curved path (Figure 3.2). Similarly, a stone thrown horizontally from the top of a building, or a long jumper leaving the ground at an angle, are examples of projectile motion.

Projectile motion: A two-dimensional motion of an object thrown in the air under the action of gravitational force only.

In our discussion of projectile motion, the effects of other forces, such as air friction and the rotation of the Earth, are neglected. The object that is thrown is called the projectile and its path is called its trajectory. A football, a cricket ball or a baseball when thrown in the air are examples of projectiles.

Experimentally, it can be proved that the horizontal and vertical motions are completely independent of each other. Consider two balls, blue and green, as in Figure 3.3. The blue ball is dropped vertically downward while the green ball is thrown horizontally at the same time from a certain cliff. Neglecting air resistance, both balls hit the ground at the same time. The key to analysing two-dimensional projectile motion is to break the motion into two components, one along the horizontal axis and the other along the vertical axis.

Figure 3.3: The two balls, one thrown horizontally and the other dropped vertically, hit the ground at the same time. The horizontal component of velocity stays constant, while the vertical component grows.

3.2.1Projectile motion for an object launched horizontallyOften examined

In many situations a projectile is thrown horizontally from a certain height with some initial velocity. It travels forward as well as falling downward until it strikes the ground. Neglecting air resistance, the motion follows projectile motion.

Consider the green ball of Figure 3.3 thrown horizontally from a cliff with velocity v_i. There is no vertical component of the initial velocity, that is v_{iy}=0. Hence:

v_{ix}=v_i

initial horizontal component

v_{iy}=0

initial vertical component

Thus there is no horizontal acceleration, while the vertical acceleration is the acceleration due to gravity:

a_x=0,\qquad a_y=g

The acceleration due to gravity g is taken positive when the ball is falling downward and negative when the ball is moving upward.

Derivation Velocity and displacement of a horizontally launched projectile

Aim. Find the velocity components, the speed and direction, and the displacements of a projectile launched horizontally, at any instant t.

Assumptions.

Air resistance is neglected.
The projectile is launched horizontally with speed v_i from a height, so v_{ix}=v_i, v_{iy}=0, a_x=0 and a_y=g (downward positive).

The horizontal component of velocity remains constant because there is no horizontal acceleration.
v_{fx}=v_{ix}=v_i
The vertical component increases, so apply the first equation of motion in the y-direction.
v_{fy}=v_{iy}+a_yt=0+gt
Simplify.
v_{fy}=gt
The speed at time t is the magnitude of the resultant of the two perpendicular components.
v_f=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}\qquad(3.9)
The direction \theta of the velocity, measured from the horizontal, comes from the ratio of the components.
\theta=\tan^{-1}\!\left(\dfrac{v_{fy}}{v_{fx}}\right)\qquad(3.10)
Horizontal displacement: the horizontal velocity is constant, so distance equals velocity times time.
X=v_{fx}\,t\qquad(3.11)
Vertical displacement: apply the second equation of motion in the y-direction with v_{iy}=0.
Y=v_{iy}t+\tfrac12a_yt^{2}=\tfrac12gt^{2}\qquad(3.12)

Resultv_{fx}=v_i,\quad v_{fy}=gt,\quad X=v_it,\quad Y=\tfrac12gt^{2}

The horizontal distance grows steadily with time while the vertical distance grows with t^{2}. Eliminating t shows that the path is a parabola.

Worked example: a ball rolling off a table

A ball rolls off the edge of a table 1.25\,\mathrm{m} high with a horizontal speed of 2.0\,\mathrm{m\,s^{-1}}. Find (a) the time to reach the floor, (b) where it lands, (c) its speed and direction on landing, (d) what changes if the table speed is doubled.

Given

v_{ix} = 2.0\,\mathrm{m\,s^{-1}}
v_{iy} = 0
Y = 1.25\,\mathrm{m}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(a) t
(b) X
(c) v_f and \theta
(d) effect of doubling v_i

(a) The vertical motion is free fall from rest, independent of the horizontal speed.
Y=\tfrac12gt^{2}\ \Rightarrow\ t=\sqrt{\dfrac{2Y}{g}}=\sqrt{\dfrac{2(1.25)}{9.8}}=0.505\,\mathrm{s}
(b) Horizontally there is no acceleration, so the speed stays constant.
X=v_it=2.0(0.505)=1.01\,\mathrm{m}
(c) Find the two components on landing.
v_{fx}=2.0\,\mathrm{m\,s^{-1}},\qquad v_{fy}=gt=9.8(0.505)=4.95\,\mathrm{m\,s^{-1}}
Combine them with Pythagoras and take the angle below the horizontal.
v_f=\sqrt{2.0^{2}+4.95^{2}}=5.34\,\mathrm{m\,s^{-1}},\qquad\theta=\tan^{-1}\dfrac{4.95}{2.0}=68.0^\circ
(d) The time depends only on the height, so t=0.505\,\mathrm{s} and v_{fy}=4.95\,\mathrm{m\,s^{-1}} do not change. Only the horizontal quantities double.
\begin{aligned}X&=4.0(0.505)=2.02\,\mathrm{m}\\ v_f&=\sqrt{4.0^{2}+4.95^{2}}=6.36\,\mathrm{m\,s^{-1}}\\ \theta&=\tan^{-1}\dfrac{4.95}{4.0}=51.1^\circ\end{aligned}

Ball leaving the table with v_i and landing a distance X away. At landing the velocity v has a constant horizontal component v_x and a vertical component v_y that has grown during the fall.

t=0.505\,\mathrm{s},\ X=1.01\,\mathrm{m},\ v_f=5.34\,\mathrm{m\,s^{-1}}\ \text{at }68.0^\circ\text{ below the horizontal}

3.2.2Projectile motion for an object launched at some angle with horizontalOften examined

There are many situations in which a projectile is thrown with velocity v_i at an angle \theta with the horizontal. After the initial push, it moves under the action of the force of gravity only (air resistance neglected), as in Figure 3.4.

There is no horizontal acceleration, while the vertical acceleration is the acceleration due to gravity, which is downward and so negative when the upward direction is positive:

a_x=0,\qquad a_y=-g

Resolve the initial velocity into its horizontal and vertical components:

v_{ix}=v_i\cos\theta

v_{iy}=v_i\sin\theta

Figure 3.4: Rectangular components of the initial velocity v_i at angle \theta to the horizontal.

Derivation Velocity and displacement at any instant

Aim. Find the velocity components, speed, direction and displacement of a projectile launched at angle \theta, at any instant t.

Assumptions.

Air resistance is neglected.
Upward is positive, so a_x=0 and a_y=-g.
The projectile starts from the origin.

The horizontal component of velocity stays constant.
v_{fx}=v_{ix}=v_i\cos\theta
The vertical component changes. Apply the first equation of motion in the y-direction.
v_{fy}=v_{iy}+a_yt
Substitute v_{iy}=v_i\sin\theta and a_y=-g.
v_{fy}=v_i\sin\theta-gt
The magnitude of the instantaneous velocity is the resultant of the two components.
v_f=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}
Its direction is found from the ratio of the components.
\theta=\tan^{-1}\!\left(\dfrac{v_{fy}}{v_{fx}}\right)
Horizontal displacement: constant horizontal velocity, so x=v_{fx}t.
x=v_i\cos\theta\times t
Vertical displacement: second equation of motion in the y-direction.
y=v_{iy}t+\tfrac12a_yt^{2}
Substitute v_{iy} and a_y.
y=(v_i\sin\theta)\,t-\tfrac12gt^{2}\qquad(3.13)

Resultx=(v_i\cos\theta)\,t,\qquad y=(v_i\sin\theta)\,t-\tfrac12gt^{2}

These two equations give the position of the projectile at any time. The next three derivations give the height, the time of flight and the range.

The maximum height, the time of flight and the range of a projectile are described below (Figure 3.5).

Figure 3.5: Height and range of a projectile.

Height of the projectile H

The height of the projectile is the maximum vertical distance attained during the projectile motion. It is denoted by H.

Derivation Maximum height of a projectile

Aim. Derive H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}.

Assumptions.

Projection speed v_i at angle \theta to the horizontal.
No air resistance, so a_y=-g.
At the highest point the body is momentarily at rest vertically.

Use the third equation of motion in the y-direction, with the vertical displacement equal to H.
2a_yH=v_{fy}^{2}-v_{iy}^{2}\qquad(3.14)
At the maximum height v_{fy}=0, since the body momentarily comes to rest vertically. For upward motion a_y=-g.
2(-g)H=0-v_{iy}^{2}
Put v_{iy}=v_i\sin\theta.
-2gH=0-v_i^{2}\sin^{2}\theta
Cancel the minus signs and divide by 2g.
H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}\qquad(3.15)

ResultH=\dfrac{v_i^{2}\sin^{2}\theta}{2g}

If the magnitude and direction of the initial velocity are known, this equation gives the height of the projectile. Only the vertical component v_i\sin\theta decides the height.

Time of flight T

The time taken by the projectile from the point of projection to the point where it hits the ground at the same level is called the time of flight, denoted by T.

Derivation Time of flight of a projectile

Aim. Derive T=\dfrac{2v_i\sin\theta}{g}.

Assumptions.

The projectile lands at the same level from which it was launched.
No air resistance.

Use the second equation of motion, s=v_it+\tfrac12at^{2} (3.16), in the y-direction.
H=v_{iy}t+\tfrac12a_yt^{2}
The projectile returns to the ground at the same height, so the vertical displacement is zero: H=0. Put a_y=-g, v_{iy}=v_i\sin\theta and t=T.
0=(v_i\sin\theta)\,T-\tfrac12gT^{2}
Take T common. The solution T=0 is the instant of launch, so the other factor must be zero.
T\left(v_i\sin\theta-\tfrac12gT\right)=0
Solve for T.
T=\dfrac{2v_i\sin\theta}{g}\qquad(3.17)

ResultT=\dfrac{2v_i\sin\theta}{g}

The time of flight depends only on the vertical component of the launch velocity.

The time taken by the projectile to reach the highest point is called the time of summit, denoted by T'. The rise and the fall take equal times, so it is half the time of flight:

T'=\dfrac{T}{2}=\dfrac{v_i\sin\theta}{g}\qquad(3.18)

Range of the projectile R

The horizontal distance travelled by a projectile is called its range, denoted by R.

Derivation Range of a projectile

Aim. Derive R=\dfrac{v_i^{2}\sin2\theta}{g}.

Assumptions.

Level ground: the projectile lands at the height of launch.
No air resistance, so the horizontal velocity stays at v_i\cos\theta.

The range is a horizontal distance, so use X=v_xt.
X=v_xt
Substitute X=R, v_x=v_i\cos\theta and t=T=\dfrac{2v_i\sin\theta}{g}.
R=v_i\cos\theta\times\dfrac{2v_i\sin\theta}{g}
Rearrange and use the identity 2\sin\theta\cos\theta=\sin2\theta.
R=\dfrac{v_i^{2}\,(2\sin\theta\cos\theta)}{g}=\dfrac{v_i^{2}\sin2\theta}{g}\qquad(3.19)

ResultR=\dfrac{v_i^{2}\sin2\theta}{g}

Knowing the magnitude and direction of the initial velocity, we can find the height, range and time of flight of the projectile.

Maximum range R_{max}

The greater the initial speed, the greater the range. For a given value of the initial speed v_i, the range is maximum when \sin2\theta=1, which happens when 2\theta=90^\circ, that is \theta=45^\circ. Putting \theta=45^\circ in equation (3.19) gives the maximum range:

R_{max}=\dfrac{v_i^{2}}{g}\qquad(3.20)

Same range

If the initial speed v_i and the acceleration due to gravity g remain constant, there are always two angles for which the projectile has the same range (Figure 3.6 (a)). These angles are complementary to each other, that is \theta and 90^\circ-\theta. Hence a projectile has the same range for pairs of angles such as (75^\circ,15^\circ), (60^\circ,30^\circ) and (70^\circ,20^\circ).

Figure 3.6 (a): Height against range for projectiles launched at v_i=50\,\mathrm{m\,s^{-1}}. The angles 15^\circ and 75^\circ (complementary angles) give the same range, and 45^\circ gives the greatest range. The book also shows 30^\circ and 60^\circ.

Why complementary angles give the same range

The range depends on \sin2\theta.
R=\dfrac{v_i^{2}\sin2\theta}{g}
Replace \theta by its complement 90^\circ-\theta.
R'=\dfrac{v_i^{2}\sin2(90^\circ-\theta)}{g}=\dfrac{v_i^{2}\sin(180^\circ-2\theta)}{g}
Since \sin(180^\circ-A)=\sin A, the two ranges are equal.
R'=\dfrac{v_i^{2}\sin2\theta}{g}=R

Worked example: complementary launch angles

A ball is kicked at 20\,\mathrm{m\,s^{-1}} once at 30^\circ and once at 60^\circ to the horizontal. For each kick find the range, the maximum height and the time of flight, and compare.

Given

v_i = 20\,\mathrm{m\,s^{-1}}
\theta = 30^\circ\ \text{and}\ 60^\circ
g = 9.8\,\mathrm{m\,s^{-2}}

To find

R, H, T for each angle

Range for 30^\circ (with \sin60^\circ=0.866).
R=\dfrac{v_i^{2}\sin2\theta}{g}=\dfrac{400\sin60^\circ}{9.8}=35.3\,\mathrm{m}
Range for 60^\circ. Here \sin120^\circ=\sin60^\circ, so the range is the same.
R=\dfrac{400\sin120^\circ}{9.8}=35.3\,\mathrm{m}
Maximum height.
H_{30}=\dfrac{400\sin^{2}30^\circ}{2(9.8)}=5.10\,\mathrm{m},\qquad H_{60}=\dfrac{400\sin^{2}60^\circ}{2(9.8)}=15.3\,\mathrm{m}
Time of flight.
T_{30}=\dfrac{2(20)\sin30^\circ}{9.8}=2.04\,\mathrm{s},\qquad T_{60}=\dfrac{2(20)\sin60^\circ}{9.8}=3.53\,\mathrm{s}
Compare the two kicks. The ranges are equal because 30^\circ and 60^\circ are complementary angles. The steeper kick rises three times as high and stays in the air \sqrt3 times as long.
\dfrac{H_{60}}{H_{30}}=\dfrac{\sin^{2}60^\circ}{\sin^{2}30^\circ}=3,\qquad\dfrac{T_{60}}{T_{30}}=\dfrac{\sin60^\circ}{\sin30^\circ}=1.73

Paths of a ball launched at 20\,\mathrm{m\,s^{-1}} at 30^\circ and 60^\circ. Both land 35.3\,\mathrm{m} away, but the 60^\circ path is higher and takes longer.

R=35.3\,\mathrm{m}\ \text{(both)};\ H=5.10\,\mathrm{m},\,15.3\,\mathrm{m};\ T=2.04\,\mathrm{s},\,3.53\,\mathrm{s}

Effect of air resistance on projectile motion

Air resistance affects the motion just like friction. Due to air resistance the particle's energy decreases. Generally, air resistance decreases the velocity of the projectile. As a result, both the horizontal and vertical components of velocity decrease.

Air resistance affects the parabolic motion of a projectile by reducing its range and maximum height. Hence air resistance can significantly alter the trajectory of the motion, as shown in Figure 3.6 (b). Air resistance also increases the time of flight of the projectile.

Figure 3.6 (b): Trajectory of a projectile with and without air resistance.

Worked examples

Example 3.4

If a projectile is launched horizontally with a speed of 12.0\,\mathrm{m\,s^{-1}} from the top of a 24.6\,\mathrm{m} high building, determine the horizontal displacement of the projectile.

Given

v_{ix} = 12.0\,\mathrm{m\,s^{-1}}
y = 24.6\,\mathrm{m}
v_{iy} = 0
a_y=g = 9.8\,\mathrm{m\,s^{-2}}
a_x = 0

To find

First find the time of fall. Use the second equation of motion in the y-direction.
y=v_{iy}t+\tfrac12a_yt^{2}
For a horizontal launch v_{iy}=0 and a_y=g=9.8\,\mathrm{m\,s^{-2}}.
24.6=(0)t+\tfrac12(9.8)t^{2}
Solve for t.
t^{2}=\dfrac{2(24.6)}{9.8}=5.02\quad\Rightarrow\quad t=2.24\,\mathrm{s}
Now use the second equation of motion in the x-direction, with a_x=0.
x=v_{ix}t+\tfrac12a_xt^{2}=(12)(2.24)+0
Evaluate.
x=26.9\,\mathrm{m}

x=26.9\,\mathrm{m}

Example 3.5

A projectile is launched with an initial speed of 200.0\,\mathrm{m\,s^{-1}} at an angle of 30^\circ with the horizontal. (a) Determine the time of flight of the projectile. (b) Determine the peak height of the projectile. (c) Determine the horizontal displacement of the projectile.

Given

v_i = 200\,\mathrm{m\,s^{-1}}
\theta = 30^\circ

To find

(a) T
(b) H
(c) R

(a) Time of flight

Use the time-of-flight formula.
T=\dfrac{2v_i\sin\theta}{g}
Substitute.
T=\dfrac{2(200)\sin30^\circ}{9.8}=\dfrac{2(200)(0.5)}{9.8}
Evaluate.
T=20.41\,\mathrm{s}

(b) Maximum height

Use the height formula.
H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}
Substitute.
H=\dfrac{(200)^{2}\sin^{2}30^\circ}{2(9.8)}=\dfrac{40000\times0.25}{19.6}
Evaluate.
H=510.2\,\mathrm{m}

(c) Range

Use the range formula.
R=\dfrac{v_i^{2}\sin2\theta}{g}
Substitute.
R=\dfrac{(200)^{2}\sin2(30^\circ)}{9.8}=\dfrac{40000\times0.866}{9.8}
Evaluate.
R=3534.7\,\mathrm{m}

T=20.41\,\mathrm{s},\quad H=510.2\,\mathrm{m},\quad R=3534.7\,\mathrm{m}

Assignment 3.3

A ball is thrown with a speed of 30\,\mathrm{m\,s^{-1}} at an angle of 30^\circ with the horizontal. Determine: (a) the height to which it rises, (b) the time of flight and (c) the horizontal range.

Given

v_i = 30\,\mathrm{m\,s^{-1}}
\theta = 30^\circ
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(a) H
(b) T
(c) R

Height.
H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}=\dfrac{(30)^{2}(0.5)^{2}}{2(9.8)}=\dfrac{225}{19.6}=11.5\,\mathrm{m}
Time of flight.
T=\dfrac{2v_i\sin\theta}{g}=\dfrac{2(30)(0.5)}{9.8}=3.06\,\mathrm{s}
Range.
R=\dfrac{v_i^{2}\sin2\theta}{g}=\dfrac{(30)^{2}\sin60^\circ}{9.8}=\dfrac{900\times0.866}{9.8}=79.5\,\mathrm{m}

H=11.5\,\mathrm{m},\quad T=3.06\,\mathrm{s},\quad R=79.5\,\mathrm{m}

3.3Law of conservation of momentumOften examined

If a system consists of particles, each moving with a different velocity, then the total momentum of the system is the sum of the momenta of the individual particles. A system is said to be isolated if and only if the net external force, such as the gravitational force or friction, acting on the system is zero. There is no ideally isolated system in the universe, but we consider an isolated system as one that does not interact with its environment.

Law of conservation of momentum: The total momentum of an isolated system of interacting particles is conserved.

If p_i and p_f are the initial and final momenta of an isolated system, then according to the law of conservation of momentum:

p_f=p_i

p_f-p_i=0

\Delta p=0\qquad(3.21)

The law of conservation of momentum is useful in collision problems. When a collision occurs in an isolated system, the forces involved are internal forces. They change the momentum of each particle, but they do not contribute to the net force, which remains zero. Hence the total momentum of the system does not change with the passage of time. The law is also applicable in explosions.

3.3.1Applications of conservation of momentum

Explosions

In an explosion, chemical energy (stored in the bonds of the atoms) is transformed into the kinetic energy of the fragments. Using the principle of conservation of momentum, numerical values can be predicted for the fragments after the explosion.

All objects before an explosion are generally considered at rest. After and during the explosion, the objects fly away in different directions with different speeds. Momentum is always conserved, so in this case the initial momentum of everything is zero. Knowing this, the final momentum should be zero too.

Consider the explosion of a bomb into two fragments A and B, as shown in Figure 3.7 (a). The initial momentum of the system before the explosion is zero. As the momentum must be conserved, the sum of the final momenta of the two fragments must be zero. Therefore fragments of the same mass must move in opposite directions with equal speed for their momentum to be conserved.

Figure 3.7 (a): Explosion of a bomb into two fragments A and B. Before the explosion the momentum is zero, so after it m_1v_1+m_2v_2=0.

An explosion in a cannon also follows conservation of momentum. For example, consider a cannon on frictionless ground shooting a cannon ball, as shown in Figure 3.7 (b). The sum of the initial momentum is zero, because nothing is moving. After the explosion inside the cannon, the cannon ball is shot forward at very high speed, while the cannon recoils at a much slower speed because of its heavy mass. The sum of the final momentum is zero, so the initial momentum and the final momentum are equal.

Figure 3.7 (b): Momentum conservation in a cannon. The ball goes forward with a large velocity v_b; the heavy cannon recoils with a small velocity v_c.

Karate chop

A karate player can break a pile of tiles with a single blow because he strikes the pile with his hand very fast (Figure 3.8 (a)). In doing so, the large momentum of his hand is reduced to zero in a very short time interval. This exerts a large force on the pile of tiles, which is sufficient to break them apart.

Ball and bat

Conservation of momentum is also applied to a ball and a bat. When a ball hits the bat, the total momentum before and after the collision must be equal (you have to add up the momentum of ball and bat). So the momentum of ball and bat before and after the collision must be equal, as in Figure 3.8 (b).

Rockets and jet engines

Rockets and jet engines also work on the law of conservation of momentum. In these machines, hot gases produced by burning of fuel rush out with large momentum. As a result, the machines gain an equal and opposite momentum. This enables them to move with very high velocity.

Car crashes

Conservation of momentum is also applied when two cars hit each other. The net momentum of the cars before and after the collision must be equal.

Example 3.6

A 46\,\mathrm{g} tennis ball is launched from a 1.35\,\mathrm{kg} homemade cannon. If the cannon recoils with a speed of 2.1\,\mathrm{m\,s^{-1}}, determine the muzzle speed of the tennis ball.

Given

m_b = 46\,\mathrm{g}=0.046\,\mathrm{kg}
m_c = 1.35\,\mathrm{kg}
v_c = -2.1\,\mathrm{m\,s^{-1}}\ \text{(recoil)}

To find

Initially the ball is in the cannon and both objects are at rest, so the total momentum of the system is zero.
p_i=0
After the explosion the total momentum must also be zero. The cannon's backward momentum is equal and opposite to the ball's forward momentum.
p_f=0\;\Rightarrow\;m_cv_c+m_bv_b=0
Substitute the values.
(1.35)(-2.1)+(0.046)\,v_b=0
Simplify.
-2.8+0.046\,v_b=0
Solve for v_b.
v_b=\dfrac{2.8}{0.046}=60.87\,\mathrm{m\,s^{-1}}

v_b=60.87\,\mathrm{m\,s^{-1}}

Assignment 3.4

A bullet of mass 20\,\mathrm{g} is fired from a gun with a muzzle velocity of 100\,\mathrm{m\,s^{-1}}. Find the recoil of the gun if its mass is 5\,\mathrm{kg}.

Given

m_b = 20\,\mathrm{g}=0.020\,\mathrm{kg}
v_b = 100\,\mathrm{m\,s^{-1}}
m_g = 5\,\mathrm{kg}

To find

Gun and bullet are at rest before firing, so the total momentum is zero and stays zero.
m_gv_g+m_bv_b=0
Solve for the gun's velocity.
v_g=-\dfrac{m_bv_b}{m_g}=-\dfrac{0.020\times100}{5}
Evaluate. The minus sign means the gun moves opposite to the bullet.
v_g=-0.4\,\mathrm{m\,s^{-1}}

v_g=-0.4\,\mathrm{m\,s^{-1}}\ \text{(recoil of }0.4\,\mathrm{m\,s^{-1}}\text{)}

3.4Elastic and inelastic collisionsOften examined

A collision occurs when two bodies come in physical contact with each other for a short interval of time and then separate. Examples are the collision of a ball with a bat and the collision of two cars. There are two types of collisions: elastic collision and inelastic collision. Momentum remains conserved in all types of collisions, but kinetic energy may change.

Elastic collision

Elastic collision: A collision in which both the momentum and the kinetic energy of the system are conserved.

For example, the collisions between atomic and subatomic particles are elastic in nature. In such collisions the two objects collide and return to their original shapes with no loss of total kinetic energy, that is, the kinetic energy does not change into other types of energy.

Inelastic collision

Inelastic collision: A collision in which the momentum of the system is conserved but the kinetic energy is not conserved.

For example, a meteorite falls on the Earth. During inelastic collisions, the kinetic energy is transformed into other forms of energy, such as heat energy, sound energy and material deformation.

Comparison of the two types of collision (Short question 3.8).
	Elastic collision	Inelastic collision
Momentum	conserved	conserved
Kinetic energy	conserved	not conserved (changed into heat, sound, deformation)
Example	atomic and subatomic particles	a meteorite falling on the Earth

3.4.1Elastic collisionOften examined

Consider an isolated system of two spherical bodies, such as billiard balls, of masses m_1 and m_2 moving to the right with velocities u_1 and u_2 along a straight line without rotation. The speed of the body of mass m_1 is greater than the speed of the body of mass m_2 and it is approaching it, so the two bodies collide head-on elastically. After the collision they move with velocities v_1 and v_2 along the same straight line (Figure 3.9).

Figure 3.9: Two balls (a) before collision, (b) during collision, (c) after collision.

Take the velocity as positive if a body moves to the right and negative if it moves to the left. Since the collision is elastic, we have two conservation laws.

Derivation Velocities after an elastic head-on collision

Aim. Find v_1 and v_2 after the collision in terms of m_1, m_2, u_1 and u_2, and prove that the relative speed of approach equals the relative speed of separation.

Assumptions.

Isolated system of two bodies in a straight line (head-on, one dimension).
The collision is elastic: both momentum and kinetic energy are conserved.
Right is positive.

Conservation of momentum.
m_1u_1+m_2u_2=m_1v_1+m_2v_2\qquad(3.22)
Collect the terms of body 1 on the left and of body 2 on the right.
m_1u_1-m_1v_1=m_2v_2-m_2u_2
Take the masses out as common factors.
m_1(u_1-v_1)=m_2(v_2-u_2)\qquad(3.23)
Conservation of kinetic energy.
\tfrac12m_1u_1^{2}+\tfrac12m_2u_2^{2}=\tfrac12m_1v_1^{2}+\tfrac12m_2v_2^{2}
Cancel the factor \tfrac12 and rearrange as in step 2.
m_1(u_1^{2}-v_1^{2})=m_2(v_2^{2}-u_2^{2})
Factorise each difference of squares.
m_1(u_1-v_1)(u_1+v_1)=m_2(v_2-u_2)(v_2+u_2)\qquad(3.24)
Divide equation (3.24) by equation (3.23). The factors m_1(u_1-v_1) and m_2(v_2-u_2) are equal and cancel.
u_1+v_1=v_2+u_2
Rearrange to bring the initial velocities to one side.
u_1-u_2=-(v_1-v_2)\qquad(3.25)
Equation (3.25) says that the relative velocity before the collision is equal and opposite to the relative velocity after it. So the relative speed of approach before collision equals the relative speed of separation after collision. From it we can write v_2 in terms of v_1.
v_2=u_1+v_1-u_2\qquad(3.26)
Put this v_2 in the momentum equation (3.22).
m_1u_1+m_2u_2=m_1v_1+m_2(u_1+v_1-u_2)
Expand the bracket.
m_1u_1+m_2u_2=m_1v_1+m_2u_1+m_2v_1-m_2u_2
Collect the v_1 terms on the left and the rest on the right.
m_1v_1+m_2v_1=m_1u_1+m_2u_2-m_2u_1+m_2u_2
Factorise both sides.
(m_1+m_2)v_1=(m_1-m_2)u_1+2m_2u_2
Divide by (m_1+m_2).
v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1+\dfrac{2m_2}{m_1+m_2}\,u_2\qquad(3.27)
Similarly, from (3.25) write v_1 in terms of v_2.
v_1=u_2+v_2-u_1\qquad(3.28)
Put this v_1 in equation (3.22).
m_1u_1+m_2u_2=m_1(u_2+v_2-u_1)+m_2v_2
Expand.
m_1u_1+m_2u_2=m_1u_2+m_1v_2-m_1u_1+m_2v_2
Collect the v_2 terms on the left.
m_1v_2+m_2v_2=m_1u_1+m_2u_2-m_1u_2+m_1u_1
Factorise both sides.
(m_1+m_2)v_2=2m_1u_1-(m_1-m_2)u_2
Divide by (m_1+m_2).
v_2=\dfrac{2m_1}{m_1+m_2}\,u_1-\dfrac{m_1-m_2}{m_1+m_2}\,u_2\qquad(3.29)

Resultv_1=\dfrac{m_1-m_2}{m_1+m_2}u_1+\dfrac{2m_2}{m_1+m_2}u_2,\qquad v_2=\dfrac{2m_1}{m_1+m_2}u_1-\dfrac{m_1-m_2}{m_1+m_2}u_2

Equations (3.27) and (3.29) give the velocities of the two bodies after the collision. Equation (3.25) is the relative-speed result: for an elastic collision the relative speed of approach equals the relative speed of separation.

Special cases

**Case 1: bodies of equal mass, m_1=m_2.** From equations (3.27) and (3.29):

v_1=u_2\qquad\text{and}\qquad v_2=u_1

The velocities of the bodies are exchanged during an elastic collision of two bodies of equal mass.

**Case 2: equal masses, m_1=m_2, and the second body (target) at rest, u_2=0.** From (3.27) and (3.29):

v_1=0\qquad\text{and}\qquad v_2=u_1

When the first body comes to rest, the second body moves with the initial velocity of the first body.

**Case 3: a lighter body (m_1) collides with a massive body (m_2\gg m_1) at rest, u_2=0.** Then m_1 can be neglected, m_1\approx0, and from (3.27) and (3.29):

v_1=-u_1\qquad\text{and}\qquad v_2=0

The first body (the lighter one) rebounds with the same initial speed but in the opposite direction. The second body (the heavier one) remains at rest even after the collision. For example, if a ball is thrown at a fixed wall, the ball bounces back with the same speed it had initially, but in the opposite direction.

**Case 4: a massive body (m_1\gg m_2) collides with a lighter body at rest, u_2=0.** Then m_2 can be neglected, m_2\approx0, and from (3.27) and (3.29):

v_1=u_1\qquad\text{and}\qquad v_2=2u_1

The first body (the heavier one) continues to move with the same initial velocity, while the second body (the lighter one) moves with twice the initial velocity of the first body.

Summary of the four special cases of a head-on elastic collision.
Case	Masses	Before	After
1	m_1=m_2	u_1,\ u_2	v_1=u_2,\ v_2=u_1 (velocities exchange)
2	m_1=m_2, u_2=0	u_1,\ 0	v_1=0,\ v_2=u_1
3	m_2\gg m_1, u_2=0	u_1,\ 0	v_1=-u_1,\ v_2=0 (rebound)
4	m_1\gg m_2, u_2=0	u_1,\ 0	v_1=u_1,\ v_2=2u_1

Example 3.7

An object of mass 5\,\mathrm{kg} moving at 10\,\mathrm{m\,s^{-1}} collides with another object of mass 10\,\mathrm{kg} moving in the same direction at 5\,\mathrm{m\,s^{-1}}. Assume that the collision is a one-dimensional elastic collision. What will be the speed of both objects after the collision?

Given

m_1 = 5\,\mathrm{kg}
u_1 = 10\,\mathrm{m\,s^{-1}}
m_2 = 10\,\mathrm{kg}
u_2 = 5\,\mathrm{m\,s^{-1}}

To find

Use the relation (3.27) for v_1.
v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1+\dfrac{2m_2}{m_1+m_2}\,u_2
Substitute the values.
v_1=\dfrac{5-10}{5+10}(10)+\dfrac{2(10)}{5+10}(5)
Evaluate each term.
v_1=-3.33+6.67=\dfrac{50}{15}=3.3\,\mathrm{m\,s^{-1}}
Use the relation (3.29) for v_2.
v_2=\dfrac{2m_1}{m_1+m_2}\,u_1-\dfrac{m_1-m_2}{m_1+m_2}\,u_2
Substitute the values.
v_2=\dfrac{2(5)}{5+10}(10)-\dfrac{5-10}{5+10}(5)
Evaluate.
v_2=6.67+1.67=8.33\,\mathrm{m\,s^{-1}}
Check with equation (3.25): the relative speed of approach is 10-5=5\,\mathrm{m\,s^{-1}} and the relative speed of separation is 8.33-3.33=5\,\mathrm{m\,s^{-1}}, equal as they must be.

The velocity of the lighter object decreases to 3.3\,\mathrm{m\,s^{-1}} and the heavier object moves with a velocity of 8.33\,\mathrm{m\,s^{-1}}.

v_1=3.3\,\mathrm{m\,s^{-1}},\quad v_2=8.33\,\mathrm{m\,s^{-1}}

Worked example: elastic collision with a heavier body at rest

A 2\,\mathrm{kg} trolley moving at 6\,\mathrm{m\,s^{-1}} makes a head-on elastic collision with a 4\,\mathrm{kg} trolley at rest. Find both velocities after the collision and verify the conservation laws.

Given

m_1 = 2\,\mathrm{kg}
u_1 = 6\,\mathrm{m\,s^{-1}}
m_2 = 4\,\mathrm{kg}
u_2 = 0

To find

v_1, v_2
check of momentum and kinetic energy

Use the book's formula for body 1 with u_2=0.
v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1=\dfrac{2-4}{6}(6)=-2\,\mathrm{m\,s^{-1}}
Body 2.
v_2=\dfrac{2m_1}{m_1+m_2}u_1=\dfrac{2(2)}{6}(6)=4\,\mathrm{m\,s^{-1}}
The minus sign shows that the lighter trolley rebounds. Momentum check.
m_1u_1=12\,\mathrm{kg\,m\,s^{-1}},\qquad m_1v_1+m_2v_2=-4+16=12\,\mathrm{kg\,m\,s^{-1}}
Kinetic energy check.
\tfrac12(2)(36)=36\,\mathrm{J},\qquad \tfrac12(2)(4)+\tfrac12(4)(16)=4+32=36\,\mathrm{J}
Relative speed check (Eq. 3.25): the relative speed of approach equals the relative speed of separation.
u_1-u_2=6-0=6\,\mathrm{m\,s^{-1}},\qquad v_2-v_1=4-(-2)=6\,\mathrm{m\,s^{-1}}

The lighter trolley rebounds at 2\,\mathrm{m\,s^{-1}} and the heavier one moves off at 4\,\mathrm{m\,s^{-1}}.

v_1=-2\,\mathrm{m\,s^{-1}},\qquad v_2=4\,\mathrm{m\,s^{-1}}

Formula sheet

First equation of motion

v_f=v_i+at

v_i, v_f: initial and final velocity; a: constant acceleration; t: time

Unit: \mathrm{m\,s^{-1}}

Second equation of motion

S=v_it+\tfrac12at^{2}

S: distance travelled in time t

Unit: \mathrm{m}

Third equation of motion

2aS=v_f^{2}-v_i^{2}

S: distance; no time involved

Unit: \mathrm{m^{2}\,s^{-2}}

Equations of free fall

v_f=v_i+gt,\quad h=v_it+\tfrac12gt^{2},\quad 2gh=v_f^{2}-v_i^{2}

g: acceleration due to gravity (9.81\,\mathrm{m\,s^{-2}}, taken -9.8 for upward motion); h: height

Unit: \mathrm{m\,s^{-2}} for g

Velocity components (horizontal launch)

v_{fx}=v_i,\qquad v_{fy}=gt

v_i: horizontal launch speed; t: time

Unit: \mathrm{m\,s^{-1}}

Speed and direction of a projectile

v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}},\qquad\theta=\tan^{-1}\dfrac{v_{fy}}{v_{fx}}

v_{fx}, v_{fy}: velocity components at time t

Units: \mathrm{m\,s^{-1}}; \theta in degrees

Displacements (horizontal launch)

X=v_it,\qquad Y=\tfrac12gt^{2}

X: horizontal displacement; Y: vertical drop

Unit: \mathrm{m}

Components of initial velocity at angle \theta

v_{ix}=v_i\cos\theta,\qquad v_{iy}=v_i\sin\theta

\theta: launch angle above the horizontal

Unit: \mathrm{m\,s^{-1}}

Velocity at time t (angle launch)

v_{fx}=v_i\cos\theta,\qquad v_{fy}=v_i\sin\theta-gt

Unit: \mathrm{m\,s^{-1}}

Position at time t (angle launch)

x=(v_i\cos\theta)\,t,\qquad y=(v_i\sin\theta)\,t-\tfrac12gt^{2}

x, y: horizontal and vertical positions

Unit: \mathrm{m}

Maximum height

H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}

H: maximum height

Unit: \mathrm{m}

Time of flight

T=\dfrac{2v_i\sin\theta}{g}

T: time of flight on level ground

Unit: \mathrm{s}

Time of summit

T'=\dfrac{T}{2}=\dfrac{v_i\sin\theta}{g}

T': time to reach the highest point

Unit: \mathrm{s}

Range

R=\dfrac{v_i^{2}\sin2\theta}{g}

R: horizontal range on level ground

Unit: \mathrm{m}

Maximum range

R_{max}=\dfrac{v_i^{2}}{g}\quad(\theta=45^\circ)

R_{max}: greatest range for a given v_i

Unit: \mathrm{m}

Conservation of momentum

p_f=p_i\quad\text{or}\quad\Delta p=0

p_i, p_f: total momentum of the isolated system before and after

Unit: \mathrm{kg\,m\,s^{-1}}

Momentum in a two-body interaction

m_1u_1+m_2u_2=m_1v_1+m_2v_2

u: velocities before; v: velocities after

Unit: \mathrm{kg\,m\,s^{-1}}

Relative speed in an elastic collision

u_1-u_2=-(v_1-v_2)

relative speed of approach equals relative speed of separation

Unit: \mathrm{m\,s^{-1}}

Velocity of body 1 after elastic collision

v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1+\dfrac{2m_2}{m_1+m_2}u_2

m_1, m_2: masses; u_1, u_2: initial velocities

Unit: \mathrm{m\,s^{-1}}

Velocity of body 2 after elastic collision

v_2=\dfrac{2m_1}{m_1+m_2}u_1-\dfrac{m_1-m_2}{m_1+m_2}u_2

same symbols as above

Unit: \mathrm{m\,s^{-1}}

Key terms

Kinematics: The branch of mechanics that deals with the study of motion without considering the causes of motion.
Dynamics: The branch of mechanics that deals with the study of the forces that cause objects and systems to move.
Acceleration due to gravity (g): The constant acceleration with which all objects fall freely near the Earth's surface in the absence of air resistance, independent of their masses. Its value is 9.81\,\mathrm{m\,s^{-2}}, directed towards the centre of the Earth.
Projectile motion: A two-dimensional motion of an object thrown in the air under the action of gravitational force only.
Projectile: The object that is thrown in projectile motion.
Trajectory: The path of a projectile.
Height of the projectile: The maximum vertical distance attained by a projectile during its motion.
Time of flight: The time taken by a projectile from the point of projection to the point where it hits the ground at the same level.
Time of summit: The time taken by a projectile to reach its highest point; it is half the time of flight.
Range: The horizontal distance travelled by a projectile.
Isolated system: A system on which the net external force (such as gravitational force or friction) is zero.
Law of conservation of momentum: The total momentum of an isolated system of interacting particles is conserved.
Collision: The coming of two bodies into physical contact with each other for a short interval of time, after which they separate.
Elastic collision: A collision in which both the momentum and the kinetic energy of the system are conserved.
Inelastic collision: A collision in which the momentum of the system is conserved but the kinetic energy is not conserved.

Multiple choice questionsp. 49–50

Each answer is the definition or one line of working that the question tests.

A projectile thrown upward moves in its parabolic path, the velocity and acceleration vectors for the projectile are perpendicular to each other at:

Anowhere
Bthe highest pointAnswer
Cthe launch point
Dthe landing point

Why

At the highest point the vertical component of velocity is zero, so the velocity is horizontal. The acceleration is g, always vertically downwards. A horizontal velocity and a vertical acceleration are at right angles.

At the launch and landing points the velocity makes an angle \theta with the horizontal, so its angle with the downward g is 90^\circ+\theta or 90^\circ-\theta, which is not 90^\circ for a projectile launched at an angle that is not zero.

A truck driving along a highway road has a large amount of momentum. If it moves at the same speed but has twice as much mass, its momentum is ______.

Azero
Bquadrupled
CdoubledAnswer
Dunchanged

Why

Momentum is p=mv. The speed v is unchanged and the mass is doubled, so p_{\text{new}}=2m\,v=2p. Momentum is doubled.

Option D is wrong because momentum depends on mass as well as speed. Option B would need the speed to double too.

A 5\,\mathrm{N} force is applied to a 3\,\mathrm{kg} ball to change its velocity from 9\,\mathrm{m\,s^{-1}} to 3\,\mathrm{m\,s^{-1}}. This impulse causes the momentum change of the ball to be ______ \mathrm{kg\,m\,s^{-1}}.

A-2.5
B-10
C-18Answer
D-45

Why

The impulse equals the change in momentum, \Delta p = m(v-u). The force is not needed, because the momentum change is found from the speeds alone.

\Delta p = 3\,(3-9)=-18\,\mathrm{kg\,m\,s^{-1}}

Which of the following statements is true about the projectile motion?

AProjectile motion is the motion of an object projected vertically upward into the air and moving under the influence of gravity.
BProjectile motion is the motion of an object projected into the air and moving independently of gravity.
CProjectile motion is the motion of an object projected into the air and moving under the influence of gravity.Answer
DProjectile motion is the motion of an object projected horizontally into the air and moving independently of gravity.

Why

The book defines projectile motion as the two-dimensional motion of an object thrown in the air under the action of gravitational force only. Option C states this.

Option A is too narrow, because a projectile need not be thrown vertically. Options B and D are wrong, because gravity is the only force acting and the motion is not independent of it.

The vertical component of velocity of a projectile is smallest at:

AThe instant it is thrown.
BHalfway to the top.
CThe top.Answer
DThe landing point.

Why

The vertical component is v_y=v_i\sin\theta-gt. At launch it is v_i\sin\theta, at halfway up it is smaller, and at the top it is zero because the projectile momentarily stops rising.

At the landing point the vertical component has the same size as at launch but points downwards. Read as a size (magnitude), it is largest there, so 'smallest' means the top.

Note. The book prints no key for this question. It is read as the magnitude of the vertical component. If the sign is kept, the landing point has the most negative vertical component, so the question is ambiguous; the intended answer is the top.

A 4\,\mathrm{kg} object has a momentum of 12\,\mathrm{kg\,m\,s^{-1}}. The object's speed is:

A3\,\mathrm{m\,s^{-1}}Answer
B4\,\mathrm{m\,s^{-1}}
C12\,\mathrm{m\,s^{-1}}
D48\,\mathrm{m\,s^{-1}}

Why

From p=mv, the speed is v=\dfrac{p}{m}=\dfrac{12}{4}=3\,\mathrm{m\,s^{-1}}. Option C is the momentum itself, not the speed.

A bomb of mass 9\,\mathrm{kg} explodes into 2 pieces of masses 3\,\mathrm{kg} and 6\,\mathrm{kg}. The velocity of mass 3\,\mathrm{kg} is 1.6\,\mathrm{m\,s^{-1}}, the kinetic energy of mass 6\,\mathrm{kg} is:

A3.84\,\mathrm{J}
B9.6\,\mathrm{J}
C1.92\,\mathrm{J}Answer
D2.92\,\mathrm{J}

Why

The bomb is at rest, so the total momentum is zero before and after. The two pieces must have equal and opposite momenta.

p_3=3\times1.6=4.8\,\mathrm{kg\,m\,s^{-1}},\qquad v_6=\dfrac{p_6}{6}=\dfrac{4.8}{6}=0.8\,\mathrm{m\,s^{-1}}

E_{k6}=\tfrac12\,m_6v_6^2=\tfrac12\times6\times0.8^{2}=1.92\,\mathrm{J}

What is the force experienced by a projectile after the initial force that launched it into the air, in the absence of air resistance?

AThe gravitational forceAnswer
BThe nuclear force
CThe contact force
DThe electromagnetic force

Why

After launch the projectile is no longer pushed or touched by anything, and air resistance is neglected. The only force left is its weight, which is gravitational.

The contact force needs something touching the projectile, and the nuclear force acts only at distances of the size of a nucleus, so both are wrong.

Note. The book prints no key for this question. Air resistance, which would be electromagnetic in origin, is excluded by the question.

If a projectile is launched on level ground, what launch angle maximizes the range of the projectile?

A0^\circ
B30^\circ
C45^\circAnswer
D90^\circ

Why

On level ground the range is R=\dfrac{v_i^2\sin2\theta}{g}. It is largest when \sin2\theta=1, that is 2\theta=90^\circ, so \theta=45^\circ.

At 0^\circ and 90^\circ the range is zero, and 30^\circ gives less than the maximum.

Note. The book prints no key for this question. Worked out from the range formula.

Short questionsp. 50–51

3.1

What are the conditions for using the equations of motion?

Solution

The equations of motion v_f=v_i+at, s=v_it+\tfrac12at^2 and v_f^2=v_i^2+2as apply when:

the motion is along a straight line (or the component being analysed is along a straight line);
the acceleration is constant (uniform);
only magnitudes of displacement, velocity and acceleration are used, with the direction of the initial velocity taken as positive and other quantities signed against it.

For a projectile the equations are applied separately to the horizontal and vertical components. Each has constant acceleration: zero horizontally and g vertically.

3.2

You throw a small ball vertically up in the air. How are the velocity and acceleration of the ball oriented with respect to one another (a) when the ball is moving upward (b) when the ball is moving downward?

Solution

The only force on the ball in the air is its weight, so its acceleration is always g directed vertically downwards, at every point including the top.

(a) Moving upward: the velocity is upward and the acceleration is downward. They are opposite (antiparallel), so the ball slows down.

(b) Moving downward: the velocity and the acceleration are both downward. They are in the same direction, so the ball speeds up.

3.3

For a projectile motion, is the velocity ever zero? Is the acceleration ever zero?

Solution

Velocity. At the highest point the vertical component of velocity is zero, but the horizontal component v_i\cos\theta is not, because it stays constant. The velocity is therefore zero only when \cos\theta=0, that is when the projectile is thrown vertically (\theta=90^\circ), and then it is zero at the top. For any other launch angle the velocity is never zero.

Acceleration. The acceleration is never zero. It is g directed downwards throughout the flight, because gravity is the only force and it does not change with time or position near the surface.

3.4

Draw a diagram showing the velocity and acceleration of a projectile at several points along its path, assuming (a) the projectile is launched horizontally and (b) the projectile is launched at an angle \theta with the horizontal.

Solution

The velocity is always along the tangent to the path. The acceleration is g vertically downwards at every point, with the same length in each diagram. The horizontal component of velocity is constant, while the vertical component changes.

(a) The projectile is launched horizontally.

The horizontal velocity arrows are the same length at every point. The vertical component grows downwards as the projectile falls, so the velocity arrow tilts further down along the path.

Figure 3.4(a): Velocity (accent) and acceleration (ink) of a projectile launched horizontally. The horizontal component of velocity stays the same, the vertical component grows, and the acceleration is g downwards at every point.

(b) The projectile is launched at an angle \theta with the horizontal.

At launch the velocity is inclined at \theta above the horizontal. Its vertical component falls to zero at the top, where the velocity is horizontal, and then it points downwards. The acceleration is still g downwards at every point, including the top.

Figure 3.4(b): Velocity (accent) and acceleration (ink) of a projectile launched at angle \theta. The acceleration is g downwards everywhere; the velocity is horizontal at the top, where v_y=0.

3.5

An aeroplane while flying horizontally drops a bomb when it reaches exactly above the target, but misses it. Explain why?

Solution

When the bomb is released it has the horizontal velocity of the aeroplane. Air resistance is neglected, so no horizontal force acts on it and its horizontal velocity stays constant while it falls under gravity.

The bomb therefore keeps moving forward with the aeroplane while it falls. In the time t=\sqrt{2h/g} that it takes to reach the ground from height h, it moves a horizontal distance x=v_x t, so it lands ahead of the target. The pilot must release the bomb earlier, while still behind the target by this distance.

3.62026 Annual I, Set D, Q.2 (vii)

How air resistance affects projectile motion?

Solution

Air resistance acts opposite to the velocity, like friction. It removes energy from the projectile, so its velocity decreases:

both the horizontal and the vertical components of velocity decrease, so the horizontal component is no longer constant;
the range and the maximum height are reduced;
the path is no longer a parabola, and it is not symmetrical about the top (Figure 3.6(b) in the book).

Note. The book (p. 42) also states that air resistance increases the time of flight, and an examiner marking from the book may look for that point. Strictly, drag makes the rise quicker and the fall slower, so the change in total time is small.

3.7

Why do a slow-moving loaded truck and a speeding rifle bullet each have a large momentum?

Solution

Momentum is p=mv, the product of mass and velocity. A loaded truck has a very large mass, so even its small speed gives a large momentum. A rifle bullet has a very high speed, so even its small mass gives a large momentum. Each has a large product, though for different reasons.

3.8

What is the difference between elastic and inelastic collision?

Solution

Elastic collision: both the momentum and the kinetic energy of the system are conserved. The bodies return to their original shapes with no loss of kinetic energy.
Inelastic collision: the momentum of the system is conserved but the kinetic energy is not. Some kinetic energy becomes heat, sound or deformation of the bodies.

3.9

An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy?

Solution

The object with the small mass has the largest kinetic energy. Kinetic energy can be written in terms of momentum:

E_k=\tfrac12mv^{2}=\dfrac{(mv)^{2}}{2m}=\dfrac{p^{2}}{2m}

For the same p, E_k is inversely proportional to m. For example, with p=10\,\mathrm{kg\,m\,s^{-1}}, a 1\,\mathrm{kg} object has E_k=50\,\mathrm{J}, while a 10\,\mathrm{kg} object has E_k=5\,\mathrm{J}.

3.10

Can objects in a system have momentum while the momentum of the system is zero? Explain your answer.

Solution

Yes. Momentum is a vector, so the momenta of the objects can be equal and opposite and cancel in the sum. The total momentum of the system is then zero while each object has momentum.

For example, when a bomb at rest explodes into two fragments of equal mass, the fragments fly apart with equal and opposite momenta. Each fragment has momentum, but the total momentum of the system is still zero, as the law of conservation of momentum requires.

3.11

For any specific velocity of projection, prove that the maximum range is equal to four times the corresponding height.

Solution

The maximum range is obtained when the angle is 45^\circ, since R=\dfrac{v_i^2\sin2\theta}{g} is largest when \sin2\theta=1. The corresponding height is the height for the same angle, \theta=45^\circ.

Write the height for \theta=45^\circ.
H=\dfrac{v_i^{2}\sin^{2}45^\circ}{2g}=\dfrac{v_i^{2}\times\tfrac12}{2g}=\dfrac{v_i^{2}}{4g}
Write the maximum range.
R_{\max}=\dfrac{v_i^{2}\sin90^\circ}{g}=\dfrac{v_i^{2}}{g}
Take the ratio.
\dfrac{R_{\max}}{H}=\dfrac{v_i^{2}/g}{v_i^{2}/(4g)}=4

So R_{\max}=4H for any speed of projection, which is what was to be proved.

R_{\max}=4H

3.12

Is momentum conserved when a bat hits a ball? How?

Solution

Yes. The momentum of the ball and the bat together is conserved in the collision. The total momentum of the two just before impact is equal to their total momentum just after it, even though the momentum of each changes.

The forces between the ball and the bat during the impact are much larger than any external force (such as the weight of the ball or the grip of the hands) for the very short time of contact, so the ball and bat act as a system with no net external force. The momentum of the ball reverses and increases, while the bat slows down by a corresponding amount.

Comprehensive questionsp. 51

3.1

Derive the equations of motion for uniformly accelerated objects.

Solution

Three equations relate initial velocity v_i, final velocity v_f, acceleration a, time t and displacement s. They are derived for a body moving along a straight line with constant acceleration, using magnitudes and taking the direction of v_i as positive. The velocity-time graph (Figure 3.1) shows the motion.

Derivation First equation of motion

Aim. Find v_f after time t in terms of v_i, a and t.

Assumptions.

Acceleration is constant, so the velocity-time graph is a straight line.

Figure 3.1: Velocity against time for a body with constant acceleration. The slope is a, and the area under the line is the displacement s.

The acceleration is the slope of the velocity-time graph, which is the change in velocity divided by the time.
a=\dfrac{v_f-v_i}{t}
Multiply both sides by t to clear the fraction.
at=v_f-v_i
Add v_i to both sides.
v_f=v_i+at

Resultv_f=v_i+at

The final velocity is the initial velocity plus the gain in speed at (the rectangle and triangle of Figure 3.1).

Derivation Second equation of motion

Aim. Find the displacement s after time t.

Assumptions.

The displacement is the area under the velocity-time graph, which is a trapezium.

The trapezium has parallel sides v_i and v_f and height t.
s=\tfrac12(v_i+v_f)\,t
Substitute v_f=v_i+at from the first equation.
s=\tfrac12(v_i+v_i+at)\,t
Combine the terms in v_i.
s=\tfrac12(2v_i+at)\,t
Expand.
s=v_it+\tfrac12at^{2}

Results=v_it+\tfrac12at^{2}

The displacement depends on the starting speed and on the extra distance from the acceleration.

Derivation Third equation of motion

Aim. Find a relation between v_f, v_i, a and s without t.

Assumptions.

The same assumptions as above.

Start from the trapezium form and multiply both sides by 2.
2s=(v_i+v_f)\,t
Multiply both sides by a.
2as=(v_i+v_f)\,at
Replace at by v_f-v_i from the first equation.
2as=(v_i+v_f)(v_f-v_i)
Expand the product of a sum and a difference.
2as=v_f^{2}-v_i^{2}

Result2as=v_f^{2}-v_i^{2}

This equation is used when time is not given.

These three equations hold only under the assumptions above. For free fall, a is replaced by g and s by h.

3.2

Define the law of conservation of momentum. Explain how it applies to a handball bouncing off a wall.

Solution

Law of conservation of momentum

The total momentum of an isolated system of interacting particles is conserved. If p_i and p_f are the total momenta before and after, then p_i=p_f, or \Delta p=0. An isolated system has no net external force.

During a collision the forces between the particles are internal. They change each particle's momentum but cancel in pairs (Newton's third law), so the total momentum does not change.

Handball bouncing off a wall

Derivation Momentum of the handball and the wall

Aim. Show that the total momentum of the ball, the wall and the Earth is the same before and after a ball of mass m hits the wall with speed u and bounces back with speed v.

Assumptions.

Take the direction towards the wall as positive. The ball is the only body whose momentum changes noticeably; the wall and the Earth are far more massive and hardly move.

Before impact the ball moves at u towards the wall, and the wall and Earth are at rest.
p_{i}=mu+0=mu
After the bounce the ball moves away from the wall with speed v, which is negative in this convention.
p_{\text{ball}}=-mv
The change in the ball's momentum is the impulse it receives from the wall.
\Delta p_{\text{ball}}=-mv-mu=-m(u+v)
By Newton's third law the wall and the Earth gain the equal and opposite momentum, +m(u+v). Their velocity change is negligible.
\Delta p_{\text{wall+Earth}}=+m(u+v)
Add the momenta after the bounce: the ball's -mv and the wall and Earth's +m(u+v).
p_{f}=-mv+m(u+v)=mu

Resultp_{f}=p_{i}=mu

The ball's momentum reverses, so the ball alone does not conserve momentum. The change is taken up by the wall and the Earth, and the total is conserved. With v=u the ball's momentum changes by -2mu.

3.32026 Annual I, Set D, Q.4

Explain elastic collision and prove that for two bodies colliding elastically, relative speed of approach before collision is equal to relative speed of separation after collision.

Solution

Elastic collision

In an elastic collision both the momentum and the kinetic energy of the system are conserved. The bodies do not lose kinetic energy to heat, sound or deformation, so they separate with the same total kinetic energy they had before.

Relative speed of approach and separation

Derivation Proof for a head-on elastic collision

Aim. Show that u_1-u_2=v_2-v_1, where u_1 and u_2 are the velocities before and v_1 and v_2 after the collision, all along one straight line.

Assumptions.

An isolated system of two bodies of masses m_1 and m_2 with no rotation, and a collision along the line of motion.
Velocity to the right is positive.

Figure 3.9: Two balls before (left) and after (right) a head-on elastic collision. The faster ball on the left approaches the slower one, and the two separate with the same relative speed.

Conservation of momentum gives one equation.
m_1u_1+m_2u_2=m_1v_1+m_2v_2
Rearrange so that the terms of each body are on one side.
m_1(u_1-v_1)=m_2(v_2-u_2)
Conservation of kinetic energy gives a second equation. Multiply it by 2 and rearrange in the same way.
m_1(u_1^{2}-v_1^{2})=m_2(v_2^{2}-u_2^{2})
Factorise both sides as differences of squares.
m_1(u_1-v_1)(u_1+v_1)=m_2(v_2-u_2)(v_2+u_2)
Divide the second equation by the first. Neither bracket is zero for a real collision, so the division is valid.
u_1+v_1=v_2+u_2
Rearrange so that the velocities before are on one side.
u_1-u_2=v_2-v_1

Resultu_1-u_2=v_2-v_1

The relative speed of approach before the collision equals the relative speed of separation after it. This holds only for elastic collisions, because it uses conservation of kinetic energy.

3.4

Discuss the motion of a projectile in the absence of air resistance.

Solution

A projectile is an object thrown into the air, moving under gravity only. With air resistance neglected, its motion is two-dimensional and is found by treating the horizontal and vertical motions separately. The two motions are independent of each other.

Components of the motion

Resolve the launch velocity v_i at angle \theta into its components (Figure 3.4). The horizontal acceleration is zero and the vertical acceleration is -g, so:

Figure 3.4: Rectangular components of the initial velocity v_i at angle \theta with the horizontal.

v_{fx}=v_i\cos\theta\ \text{(constant)},\qquad v_{fy}=v_i\sin\theta-gt

x=v_i\cos\theta\,t,\qquad y=v_i\sin\theta\,t-\tfrac12gt^{2}

The path is a parabola

Derivation Equation of the path

Aim. Eliminate t to show that y is a quadratic function of x.

Assumptions.

Launch from the origin on level ground.

From the horizontal equation, t=\dfrac{x}{v_i\cos\theta}.
t=\dfrac{x}{v_i\cos\theta}
Substitute into the vertical equation.
y=v_i\sin\theta\cdot\dfrac{x}{v_i\cos\theta}-\tfrac12g\left(\dfrac{x}{v_i\cos\theta}\right)^{2}
Simplify using \tan\theta=\sin\theta/\cos\theta.
y=x\tan\theta-\dfrac{gx^{2}}{2v_i^{2}\cos^{2}\theta}

Resulty=x\tan\theta-\dfrac{gx^{2}}{2v_i^{2}\cos^{2}\theta}

This has the form y=ax-bx^2, a parabola, which opens downwards.

Height, time of flight and range

From the book's results (equations 3.15, 3.17, 3.19 and 3.20):

H=\dfrac{v_i^{2}\sin^{2}\theta}{2g},\qquad T=\dfrac{2v_i\sin\theta}{g},\qquad R=\dfrac{v_i^{2}\sin2\theta}{g}

The top of the path is reached at half the time of flight, and the path is symmetrical about the vertical line through the top.

Same range and maximum range

Since \sin2\theta=\sin(180^\circ-2\theta), two launch angles that add up to 90^\circ give the same range, for example 75^\circ and 15^\circ. The range is greatest at \theta=45^\circ, where R_{\max}=v_i^{2}/g.

3.52026 Annual I, Set B, Q.32025 Annual I, Set B, Q.3 OR

Derive mathematical equations for (a) maximum height attained, (b) time of flight, (c) range of a projectile.

Solution

Take the launch point as the origin and the launch velocity v_i at angle \theta above the horizontal (Figure 3.5). The components are v_{ix}=v_i\cos\theta and v_{iy}=v_i\sin\theta. Horizontally a_x=0; vertically a_y=-g for the motion upwards.

Figure 3.5: Height H and range R of a projectile launched at angle \theta from level ground.

Derivation (a) Maximum height

Aim. Find the greatest vertical distance H above the launch point.

Assumptions.

Air resistance is neglected and the landing point is at the same level as the launch point.

At the highest point the vertical velocity is zero.
v_{fy}=0
Use the third equation of motion in the vertical direction.
2a_yH=v_{fy}^{2}-v_{iy}^{2}
Substitute a_y=-g, v_{fy}=0 and v_{iy}=v_i\sin\theta.
-2gH=0-v_i^{2}\sin^{2}\theta
Divide both sides by -2g.
H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}

ResultH=\dfrac{v_i^{2}\sin^{2}\theta}{2g}

The height depends on the vertical component of the launch velocity only.

Derivation (b) Time of flight

Aim. Find the total time T from launch until the projectile returns to the launch level.

Assumptions.

The landing height is y=0.

Use the second equation of motion in the vertical direction with y=0 at landing.
0=v_{iy}T-\tfrac12gT^{2}
Factorise out T. The root T=0 is the launch, so the other root is the flight time.
T\left(v_i\sin\theta-\tfrac12gT\right)=0
Solve for T.
T=\dfrac{2v_i\sin\theta}{g}

ResultT=\dfrac{2v_i\sin\theta}{g}

The time to reach the top is half of this, T'=\dfrac{v_i\sin\theta}{g}.

Derivation (c) Range

Aim. Find the horizontal distance R to the landing point.

Assumptions.

Horizontal velocity is constant, so x=v_{ix}t.

Substitute the horizontal velocity and the flight time.
R=v_i\cos\theta\cdot\dfrac{2v_i\sin\theta}{g}
Simplify, then use \sin2\theta=2\sin\theta\cos\theta.
R=\dfrac{2v_i^{2}\sin\theta\cos\theta}{g}=\dfrac{v_i^{2}\sin2\theta}{g}

ResultR=\dfrac{v_i^{2}\sin2\theta}{g}

For a given speed the range is greatest at \theta=45^\circ, where R_{\max}=\dfrac{v_i^{2}}{g}=4H at that angle.

3.6

Discuss that why though the momentum of a closed system is always conserved, but some change in kinetic energy may take place.

Solution

Momentum is conserved in a closed system because the net external force on it is zero. Internal forces in a collision or explosion change the momenta of the individual parts, but they are equal and opposite, so the total does not change.

Kinetic energy is not conserved in general. It depends on the way the mass moves, and internal forces can convert kinetic energy into other forms (heat, sound, deformation) or convert stored chemical energy into kinetic energy. Momentum is a conserved quantity; kinetic energy is not, unless the collision is elastic.

Example: two carts that stick together

Take two carts of equal mass m, one moving at +2\,\mathrm{m\,s^{-1}} and the other at -1\,\mathrm{m\,s^{-1}}. They collide and couple, so they move together with speed v:

Conservation of momentum gives the speed after the collision.
m(2)+m(-1)=(2m)\,v\quad\Rightarrow\quad v=0.5\,\mathrm{m\,s^{-1}}
Kinetic energy before the collision.
E_{k,\text{before}}=\tfrac12m(2)^{2}+\tfrac12m(1)^{2}=2.5\,m\ \mathrm{J}
Kinetic energy after the collision.
E_{k,\text{after}}=\tfrac12(2m)(0.5)^{2}=0.25\,m\ \mathrm{J}
Kinetic energy lost.
\Delta E_k=2.5m-0.25m=2.25\,m\ \mathrm{J}

The momentum is the same before and after (m in \mathrm{kg\,m\,s^{-1}}), but about 90\% of the kinetic energy is lost to deformation, heat and sound. This is an inelastic collision.

Example: explosion

In the bomb explosion of Figure 3.7(a) the total momentum stays zero, but the fragments have kinetic energy that came from the chemical energy of the explosive. Kinetic energy increases while momentum is still conserved.

Numerical problemsp. 51–52

3.1

On dry concrete, a car can decelerate at a rate of 7.00\,\mathrm{m\,s^{-2}}, whereas on wet concrete it can decelerate at only 5.00\,\mathrm{m\,s^{-2}}. Compare the distances necessary to stop the car moving at 30.0\,\mathrm{m\,s^{-1}} (about 110\,\mathrm{km\,h^{-1}}) (a) on dry concrete and (b) on wet concrete.

Solution

Given

v_i = 30.0\,\mathrm{m\,s^{-1}}
v_f = 0
a_{\text{dry}} = 7.00\,\mathrm{m\,s^{-2}}\text{ (deceleration)}
a_{\text{wet}} = 5.00\,\mathrm{m\,s^{-2}}\text{ (deceleration)}

To find

s_{\text{dry}}
s_{\text{wet}}

The car stops when its final velocity is zero. The third equation of motion, with a=-|a|, gives the stopping distance:

Write the third equation with v_f=0.
0=v_i^{2}-2|a|s\quad\Rightarrow\quad s=\dfrac{v_i^{2}}{2|a|}
Dry concrete.
s_{\text{dry}}=\dfrac{30.0^{2}}{2\times7.00}=\dfrac{900}{14.0}=64.3\,\mathrm{m}
Wet concrete.
s_{\text{wet}}=\dfrac{30.0^{2}}{2\times5.00}=\dfrac{900}{10.0}=90.0\,\mathrm{m}

(a)\ s_{\text{dry}}\approx64.3\,\mathrm{m};\qquad (b)\ s_{\text{wet}}=90.0\,\mathrm{m}

Book answer. 64.3\,\mathrm{m} / 90.0\,\mathrm{m}

3.2

Find the angle of projection of a projectile for which the maximum height and corresponding range are equal.

Solution

Given

H = R
v_i = \text{not given; the result does not depend on it}

To find

\theta

Write the maximum height and the range.
H=\dfrac{v_i^{2}\sin^{2}\theta}{2g},\qquad R=\dfrac{v_i^{2}\sin2\theta}{g}=\dfrac{2v_i^{2}\sin\theta\cos\theta}{g}
Set H=R and cancel v_i^{2}/g.
\dfrac{\sin^{2}\theta}{2}=2\sin\theta\cos\theta
Divide by \sin\theta/2 (the angle is not zero).
\sin\theta=4\cos\theta
Divide by \cos\theta.
\tan\theta=4
Take the inverse tangent.
\theta=\tan^{-1}4\approx76^\circ

\theta\approx76^\circ

Book answer. 76^\circ

3.3

A jet plane comes in for a landing with a speed of 100\,\mathrm{m\,s^{-1}}, and its acceleration can have a maximum magnitude of 5.00\,\mathrm{m\,s^{-2}} as it comes to rest.

Solution

(a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest?

Given

v_i = 100\,\mathrm{m\,s^{-1}}
v_f = 0
a = -5.00\,\mathrm{m\,s^{-2}}\text{ (maximum braking)}

To find

Use the first equation of motion.
v_f=v_i+at
Substitute.
0=100+(-5.00)\,t
Solve for t.
t=\dfrac{100}{5.00}=20.0\,\mathrm{s}

t=20\,\mathrm{s}

(b) Can this plane land on a small tropical island airport where the runway is 0.8\,\mathrm{km} long? Explain your answer.

Find the stopping distance with the second equation, or the third equation with v_f=0.
s=v_it+\tfrac12at^{2}=100\times20-\tfrac12\times5.00\times20^{2}=2000-1000
Compare with the runway length.
s=1000\,\mathrm{m}=1.0\,\mathrm{km}>0.8\,\mathrm{km}

The plane needs at least 1.0\,\mathrm{km} of runway, so it cannot land on the 0.8\,\mathrm{km} runway.

\text{No: a runway of at least } 1.0\,\mathrm{km} \text{ is needed.}

Book answer. 20 s, it cannot, it would require a runway of minimum length 1 km

3.4

A projectile is launched with an initial speed of 21.8\,\mathrm{m\,s^{-1}} at an angle of 35^\circ above the horizontal. Determine: (a) the time of flight of the projectile, (b) the peak height of the projectile, (c) the horizontal displacement of the projectile.

Solution

Given

v_i = 21.8\,\mathrm{m\,s^{-1}}
\theta = 35^\circ
g = 9.8\,\mathrm{m\,s^{-2}}

To find

Resolve the launch velocity.
v_{iy}=21.8\sin35^\circ=12.50\,\mathrm{m\,s^{-1}},\qquad v_{ix}=21.8\cos35^\circ=17.86\,\mathrm{m\,s^{-1}}

(a) Time of flight.

Use T=\dfrac{2v_i\sin\theta}{g}.
T=\dfrac{2\times12.503}{9.8}=2.55\,\mathrm{s}

T=2.55\,\mathrm{s}

(b) Peak height.

Use H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}, or H=\dfrac{v_{iy}^{2}}{2g}.
H=\dfrac{12.503^{2}}{2\times9.8}=7.98\,\mathrm{m}

H=7.98\,\mathrm{m}

The horizontal velocity is constant, so R=v_{ix}T. Use the unrounded flight time, T=2.5516\,\mathrm{s}.
R=17.86\times2.5516=45.6\,\mathrm{m}

R=45.6\,\mathrm{m}

Book answer. 2.55 s, 7.98 m, 45.6 m

Note. Use unrounded values: 17.86\times2.55 gives 45.5\,\mathrm{m}, but the unrounded flight time gives 45.6\,\mathrm{m}, which matches the book.

3.5

A projectile is launched horizontally from the top of a 45.2\,\mathrm{m} high cliff and lands a distance of 17.6\,\mathrm{m} from the base of the cliff. Determine the magnitude of the launch velocity.

Solution

Given

h = 45.2\,\mathrm{m}
x = 17.6\,\mathrm{m}
g = 9.8\,\mathrm{m\,s^{-2}}
v_{iy} = 0

To find

The vertical motion is free fall from rest. Use h=\tfrac12gt^{2} to find the time of fall.
t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times45.2}{9.8}}=3.04\,\mathrm{s}
The horizontal velocity is constant, so v_i=\dfrac{x}{t}.
v_i=\dfrac{17.6}{3.037}=5.79\,\mathrm{m\,s^{-1}}

v_i\approx5.79\,\mathrm{m\,s^{-1}}

Book answer. 5.79 m s^-1

3.6

Two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward at 2\,\mathrm{m\,s^{-1}} and the other cart rolls leftward at 1\,\mathrm{m\,s^{-1}}. After the carts collide, they couple and roll together. Ignoring resistive forces, find their combined speed.

Solution

Given

m_1=m_2 = m
u_1 = +2\,\mathrm{m\,s^{-1}}
u_2 = -1\,\mathrm{m\,s^{-1}}

To find

Conservation of momentum, with rightward positive.
m(2)+m(-1)=(m+m)\,v
Solve for v.
v=\dfrac{m}{2m}=0.5\,\mathrm{m\,s^{-1}}

v=0.5\,\mathrm{m\,s^{-1}}\text{, rightward}

Book answer. 0.5 m s^-1

3.7

A 0.5\,\mathrm{kg} ball travelling at a speed of 4\,\mathrm{m\,s^{-1}} to the right collides elastically with another ball of 3.5\,\mathrm{kg} which is initially at rest. Find the velocities of both the balls after collision.

Solution

Given

m_1 = 0.5\,\mathrm{kg}
u_1 = +4\,\mathrm{m\,s^{-1}}
m_2 = 3.5\,\mathrm{kg}
u_2 = 0

To find

For an elastic collision in one dimension use the two results derived from momentum and kinetic energy conservation.
v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1+\dfrac{2m_2}{m_1+m_2}\,u_2,\qquad v_2=\dfrac{2m_1}{m_1+m_2}\,u_1+\dfrac{m_2-m_1}{m_1+m_2}\,u_2
Substitute m_1+m_2=4.0\,\mathrm{kg} and u_2=0 for the first ball.
v_1=\dfrac{0.5-3.5}{4.0}\times4=-3.0\,\mathrm{m\,s^{-1}}
Second ball.
v_2=\dfrac{2\times0.5}{4.0}\times4=1.0\,\mathrm{m\,s^{-1}}
Check: momentum 0.5(4)=0.5(-3.0)+3.5(1.0)=2.0\,\mathrm{kg\,m\,s^{-1}}, and kinetic energy 4.0\,\mathrm{J} both before and after.

v_1=-3.0\,\mathrm{m\,s^{-1}}\ (\text{bounces back}),\qquad v_2=1.0\,\mathrm{m\,s^{-1}}

Book answer. -3.0 m s^-1, 1.0 m s^-1

3.8

A 17.5\,\mathrm{g} bullet is fired at a muzzle velocity of 582\,\mathrm{m\,s^{-1}} from a gun with a mass of 8.0\,\mathrm{kg} and a barrel length of 75.0\,\mathrm{cm}.

Solution

Given

m = 17.5\,\mathrm{g}=0.0175\,\mathrm{kg}
v = 582\,\mathrm{m\,s^{-1}}\text{ (at the muzzle)}
L = 75.0\,\mathrm{cm}=0.750\,\mathrm{m}
u = 0

To find

The bullet starts from rest and accelerates uniformly along the barrel. The mass of the gun is not needed for these parts.

(a) How long is the bullet travelled in the barrel?

Uniform acceleration from rest, so the average speed is \tfrac12v.
L=\dfrac{v}{2}\,t\quad\Rightarrow\quad t=\dfrac{2L}{v}
Substitute.
t=\dfrac{2\times0.750}{582}=2.58\times10^{-3}\,\mathrm{s}

t=2.58\times10^{-3}\,\mathrm{s}

(b) What is the force on the bullet while it is in the barrel?

Find the acceleration from v^{2}=2aL.
a=\dfrac{v^{2}}{2L}=\dfrac{582^{2}}{2\times0.750}=2.26\times10^{5}\,\mathrm{m\,s^{-2}}
Newton's second law.
F=ma=0.0175\times2.26\times10^{5}=3.95\times10^{3}\,\mathrm{N}

F=3.95\times10^{3}\,\mathrm{N}

Impulse equals the change in momentum.
J=\Delta p=m(v-u)=0.0175\times(582-0)=10.2\,\mathrm{N\,s}

J=10.2\,\mathrm{N\,s}

(d) Find the bullet's momentum as it leaves the barrel.

The momentum at the muzzle is mv, which is the same as the impulse because the bullet started from rest.
p=mv=0.0175\times582=10.2\,\mathrm{kg\,m\,s^{-1}}
Check: F\,t=3.95\times10^{3}\times2.58\times10^{-3}=10.2\,\mathrm{N\,s}.

p=10.2\,\mathrm{kg\,m\,s^{-1}}

Book answer. 0.00258 s, 3950 N, 10.2 kg m s^-1, 10.2 kg m s^-1

Note. The book's answers agree with these values. 3.95\times10^{3}\,\mathrm{N} is the book's 3950\,\mathrm{N} to three significant figures.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 31 mark

Time of flight of a projectile depends on:

AHorizontal velocity
BMass
CVertical component of velocityAnswer
DAcceleration only

Why

Time of flight is T=\dfrac{2v_i\sin\theta}{g}=\dfrac{2v_{iy}}{g}, so it is fixed by the vertical component of the launch velocity (the projectile stays up for as long as the vertical motion lasts). So C is correct.

Horizontal velocity (A) only decides how far the projectile travels in that time, not how long it stays in the air. Mass (B) does not appear because every body has the same acceleration g. Option D is incomplete: the acceleration g matters, but the time of flight also needs v_{iy}.

2026 Annual I, Set B, MCQ 41 mark

A car accelerates uniformly from rest to 60\,\mathrm{ms^{-1}} in 5 seconds. What distance does it cover during this time?

A150mAnswer
B450m
C300m
D600m

Why

The acceleration is uniform and the car starts from rest, so the average velocity is \dfrac{0+60}{2}=30\,\mathrm{m\,s^{-1}}. Then S=\bar v\,t=30\times5=150\,\mathrm m. Using the equations of motion: a=\dfrac{60-0}{5}=12\,\mathrm{m\,s^{-2}} and S=\tfrac12at^{2}=\tfrac12(12)(5)^{2}=150\,\mathrm m. So A is correct.

Option C (300\,\mathrm m) is 60\times5, which would be the distance if the car moved at the final speed for the whole 5 s. The car's speed is less than 60\,\mathrm{m\,s^{-1}} during the whole interval until the last instant.

2025 Annual I, Set B, Q.2 (ii)3 marks

The velocity of a missile traveling at 1500\,\mathrm{m\,s^{-1}} increases to 5000\,\mathrm{m\,s^{-1}} after moving through a distance of 70\,\mathrm{km}. Calculate the time for the missile to reach this velocity.

Solution

Given

u = 1500\,\mathrm{m\,s^{-1}}
v = 5000\,\mathrm{m\,s^{-1}}
S = 70\,\mathrm{km}=7.0\times10^{4}\,\mathrm{m}

To find

Acceleration (1.5 marks). The acceleration is uniform and the time is not given, so use the third equation of motion 2aS=v_f^{2}-v_i^{2}.
a=\dfrac{v^{2}-u^{2}}{2S}=\dfrac{(5000)^{2}-(1500)^{2}}{2(7.0\times10^{4})}=\dfrac{2.275\times10^{7}}{1.4\times10^{5}}=162.5\,\mathrm{m\,s^{-2}}
Time (1.5 marks). Now use the first equation of motion v_f=v_i+at.
t=\dfrac{v-u}{a}=\dfrac{5000-1500}{162.5}=21.5\,\mathrm{s}
Check. The average velocity is \dfrac{u+v}{2}=3250\,\mathrm{m\,s^{-1}}, so the distance is 3250\times21.54\approx7.0\times10^{4}\,\mathrm{m}, as given.

t\approx21.5\,\mathrm{s}

2026 Annual I, Set B, Q.2 (vii)3 marks

A 5\,\mathrm{kg} shell is fired with a muzzle velocity of 700\,\mathrm{ms^{-1}} from the gun barrel 2.4\,\mathrm{m} long. Calculate the average value of: i) The acceleration ii) The time to come out of the barrel

Solution

Given

m = 5\,\mathrm{kg}
v_i = 0
v_f = 700\,\mathrm{m\,s^{-1}}
S = 2.4\,\mathrm m

To find

average acceleration a
time t to leave the barrel

Acceleration (1.5 marks). The shell starts from rest and the time is not given, so use the third equation of motion 2aS=v_f^{2}-v_i^{2}.
a=\dfrac{v_f^{2}-v_i^{2}}{2S}=\dfrac{(700)^{2}-0}{2(2.4)}=\dfrac{4.9\times10^{5}}{4.8}=1.02\times10^{5}\,\mathrm{m\,s^{-2}}
Time (1.5 marks). Use the first equation of motion v_f=v_i+at.
t=\dfrac{v_f-v_i}{a}=\dfrac{700}{1.02\times10^{5}}=6.9\times10^{-3}\,\mathrm s
Check. The average velocity is 350\,\mathrm{m\,s^{-1}}, so t=S/\bar v=2.4/350=6.9\times10^{-3}\,\mathrm s.

The mass of the shell is not needed because the equations of motion describe the motion only.

a\approx1.02\times10^{5}\,\mathrm{m\,s^{-2}},\qquad t\approx6.9\times10^{-3}\,\mathrm s

2026 Annual I, Set B, Q.33+3 = 6 marks

Derive expressions for: a. Range of projectile and b. Height of projectile

Solution

A projectile is launched from level ground with speed v_i at angle \theta to the horizontal. Air resistance is neglected, so horizontally a_x=0 (the horizontal velocity stays constant) and vertically a_y=-g. The initial components are

v_{ix}=v_i\cos\theta,\qquad v_{iy}=v_i\sin\theta

Figure 3.5: A projectile launched from level ground with speed v_i at angle \theta. It rises to the maximum height H at half the time of flight and returns to the ground after travelling the range R.

(a) Range (3 marks)

Derivation Range of a projectile

Aim. Derive R=\dfrac{v_i^{2}\sin2\theta}{g}, the horizontal distance covered before the projectile returns to the level of projection.

Assumptions.

The projectile lands at the height of launch.
No air resistance.

First find the time of flight. Apply the second equation of motion in the vertical direction, with y=0 at the end of the flight and a_y=-g.
0=(v_i\sin\theta)\,T-\tfrac12gT^{2}
Take T common. The root T=0 is the instant of launch, so the other factor is zero.
T=\dfrac{2v_i\sin\theta}{g}
The horizontal velocity is constant, so the range is the horizontal velocity times the time of flight.
R=v_{ix}\,T=v_i\cos\theta\times\dfrac{2v_i\sin\theta}{g}
Use the identity 2\sin\theta\cos\theta=\sin2\theta.
R=\dfrac{v_i^{2}\sin2\theta}{g}

ResultR=\dfrac{v_i^{2}\sin2\theta}{g}

For a given speed the range is greatest when \sin2\theta=1, that is \theta=45^\circ, and then R_{max}=v_i^{2}/g.

(b) Height (3 marks)

Derivation Maximum height of a projectile

Aim. Derive H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}, the greatest height above the level of launch.

Assumptions.

At the highest point the vertical component of velocity is zero (v_{fy}=0).
No air resistance, so a_y=-g.

Apply the third equation of motion in the vertical direction between the launch point and the top.
v_{fy}^{2}-v_{iy}^{2}=2a_yH
Put v_{fy}=0, v_{iy}=v_i\sin\theta and a_y=-g.
0-(v_i\sin\theta)^{2}=2(-g)H
Solve for H.
H=\dfrac{v_i^{2}\sin^{2}\theta}{2g}

ResultH=\dfrac{v_i^{2}\sin^{2}\theta}{2g}

The height depends on the vertical component v_i\sin\theta only, and is greatest for a vertical launch (\theta=90^\circ), when H=v_i^{2}/2g.

2025 Annual I, Set B, Q.3 OR1+2+4 = 7 marks

What is meant by projectile? Derive the mathematical expressions of following for projectile motion: i) Time of flight ii) Range and maximum range of projectile.

Solution

Projectile (1 mark)

A projectile is an object thrown into the air that then moves under the action of gravity only. Its motion is two-dimensional, along a curved path called its trajectory. Air resistance and the rotation of the Earth are neglected. The horizontal and vertical motions are independent: horizontally a_x=0, so the horizontal velocity stays constant, and vertically a_y=-g. Examples are a football that is kicked, a cricket ball and a long jumper.

Let the projectile be launched from level ground with speed v_i at angle \theta to the horizontal. The initial components are

v_{ix}=v_i\cos\theta,\qquad v_{iy}=v_i\sin\theta

Figure 3.5: A projectile launched from level ground with speed v_i at angle \theta. It rises to the maximum height H at half the time of flight and returns to the ground after travelling the range R.

Time of flight (2 marks)

Derivation Time of flight

Aim. Derive T=\dfrac{2v_i\sin\theta}{g}, the time taken to return to the level of projection.

Assumptions.

The projectile lands at the same level from which it was launched.
No air resistance, so a_y=-g (upward positive).

Apply the second equation of motion in the vertical direction.
y=v_{iy}t+\tfrac12a_yt^{2}
At the end of the flight y=0 and t=T. Put v_{iy}=v_i\sin\theta and a_y=-g.
0=(v_i\sin\theta)\,T-\tfrac12gT^{2}
Take T common. The root T=0 is the instant of launch, so the other factor must be zero.
T\left(v_i\sin\theta-\tfrac12gT\right)=0
Solve for T.
T=\dfrac{2v_i\sin\theta}{g}

ResultT=\dfrac{2v_i\sin\theta}{g}

The time of flight depends only on the vertical component of the launch velocity. The rise and the fall take equal times, so the top is reached at T/2.

Range and maximum range (4 marks)

Derivation Range of a projectile

Aim. Derive R=\dfrac{v_i^{2}\sin2\theta}{g} and find the angle for the maximum range.

Assumptions.

Level ground: the projectile lands at the height of launch.
No air resistance, so the horizontal velocity stays at v_i\cos\theta.

The range is a horizontal distance covered at constant horizontal velocity, so distance equals velocity times time.
R=v_{x}\,T
Put v_x=v_i\cos\theta and the time of flight T=\dfrac{2v_i\sin\theta}{g}.
R=v_i\cos\theta\times\dfrac{2v_i\sin\theta}{g}
Rearrange and use the identity 2\sin\theta\cos\theta=\sin2\theta.
R=\dfrac{v_i^{2}\,(2\sin\theta\cos\theta)}{g}=\dfrac{v_i^{2}\sin2\theta}{g}
For a given speed v_i the range depends only on \sin2\theta, which has its largest value 1 when 2\theta=90^\circ. So the range is greatest for \theta=45^\circ.
R_{max}=\dfrac{v_i^{2}}{g}\quad(\theta=45^\circ)

ResultR=\dfrac{v_i^{2}\sin2\theta}{g},\qquad R_{max}=\dfrac{v_i^{2}}{g}\ \text{at}\ \theta=45^\circ

The range depends on the speed and the angle of projection. Two complementary angles, \theta and 90^\circ-\theta, give the same range because \sin2(90^\circ-\theta)=\sin(180^\circ-2\theta)=\sin2\theta.

2025 Annual I, Set D, MCQ 31 mark

When a massive body of mass M_1 is moving with velocity V_1 collides with lighter stationary body of mass m_2, the velocity of massive body after collision will be:

AV_1' = 2V_1
BV_2' = V_1
CV_1' = V_1Answer
DV_2' = 2V_2

Why

For a head-on elastic collision with the second body at rest (u_2=0), the book gives v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1. If m_1\gg m_2, then m_2 can be neglected, so v_1\approx u_1 and v_2=\dfrac{2m_1}{m_1+m_2}u_1\approx2u_1. The massive body keeps moving with its original velocity, so V_1'=V_1 and C is correct.

Option A (2V_1) is the velocity of the lighter body after the collision, not of the massive one. Option B would be true only for bodies of equal mass, where the first body stops and the second takes over its velocity. Option D, V_2'=2V_2, is zero because V_2=0.

2026 Annual I, Set D, MCQ 41 mark

Horizontal velocity in projectile motion:

Ais zero at max height
Bincreases
Cdecreases
Dremains constantAnswer

Why

In projectile motion with no air resistance there is no horizontal force, so a_x=0 and the horizontal component of velocity v_x=v_i\cos\theta stays constant throughout the flight. So D is correct.

Option A mixes up the two components: it is the vertical component v_y that is zero at the highest point, while v_x is still v_i\cos\theta there. The horizontal velocity neither increases (B) nor decreases (C), because gravity acts only vertically.

2026 Annual I, Set D, MCQ 51 mark

A bus travelling at 10\,\mathrm{ms^{-1}} accelerates uniformly at 2\,\mathrm{ms^{-2}}. Calculate the velocity of the bus after 5 s.

A0.02\,\mathrm{ms^{-1}}
B20\,\mathrm{ms^{-1}}Answer
C2.0\,\mathrm{ms^{-1}}
D0.2\,\mathrm{ms^{-1}}

Why

Use the first equation of motion with v_i=10\,\mathrm{m\,s^{-1}}, a=2\,\mathrm{m\,s^{-2}} and t=5\,\mathrm s: v_f=v_i+at=10+2\times5=20\,\mathrm{m\,s^{-1}}. So B is correct.

The bus is speeding up, so the final velocity must be greater than 10\,\mathrm{m\,s^{-1}}. Options A, C and D are all smaller than 10\,\mathrm{m\,s^{-1}} and come from dividing instead of multiplying.

2025 Annual I, Set D, Q.2 (iii) OR3 marks

Discuss the motion of two bodies after collision when m_1 \ll m_2 and V_2 = 0.

Solution

Here the first body is much lighter than the second (m_1\ll m_2) and the second is at rest. In the book's notation the velocities before the collision are u_1 and u_2=0 (the paper's V_1 and V_2=0), and after the collision v_1 and v_2. The collision is elastic and in one dimension.

Equations for an elastic collision (1 mark). Use the book's results for the final velocities.
v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1+\dfrac{2m_2}{m_1+m_2}\,u_2,\qquad v_2=\dfrac{2m_1}{m_1+m_2}\,u_1-\dfrac{m_1-m_2}{m_1+m_2}\,u_2
**Put u_2=0 and m_1\ll m_2 (1 mark).** Then m_1 can be neglected beside m_2 (m_1\approx0).
v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1\approx\dfrac{-m_2}{m_2}\,u_1=-u_1,\qquad v_2=\dfrac{2m_1}{m_1+m_2}\,u_1\approx0
Check. The relative speed of approach is u_1-0=u_1 and of separation is 0-(-u_1)=u_1, equal as they must be for an elastic collision.

Discussion (1 mark). The lighter body rebounds with the same speed it had, in the opposite direction (v_1=-u_1). The heavier body remains at rest even after the collision (v_2\approx0). A ball thrown at a fixed wall bounces back with the same speed it had before hitting the wall; the wall is the massive body.

v_1\approx-u_1,\qquad v_2\approx0

2026 Annual I, Set D, Q.2 (vii)2+1 = 3 marks

Evaluate effect of air resistance on projectile motion. Show graphically.

Solution

Effect (2 marks). Air resistance is a force that acts opposite to the velocity of the projectile, like friction. It removes energy from the projectile, so:

both the horizontal and the vertical components of the velocity decrease, so the horizontal component is no longer constant;
the maximum height and the range are reduced;
the path is no longer a parabola, and it is not symmetrical about the highest point: the descent is steeper than the ascent;
the book also states that the time of flight is increased.

Graph (1 mark). The figure compares the path without air resistance (a parabola) with the path when air resistance acts.

Figure 3.6: Path of a projectile launched at 45^\circ without air resistance (black, a symmetrical parabola) and with air resistance (coloured). With air resistance the maximum height and the range are smaller, and the path is no longer symmetrical: it falls more steeply than it rises.

Note. The book (p. 42) states that air resistance increases the time of flight. Strictly, drag makes the rise quicker and the fall slower, so the change in total time is small.

2025 Annual I, Set D, Q.31+4+2 = 7 marks

What is meant by projectile? Determine the expressions for velocity and displacement of a projectile thrown horizontally from a certain height.

Solution

Projectile (1 mark)

A projectile is an object thrown into the air that then moves under gravity only, neglecting air resistance. The path it follows is its trajectory. Its horizontal and vertical motions are independent of each other: horizontally there is no acceleration, and vertically the acceleration is g. A stone thrown horizontally from the top of a building is an example.

Velocity and displacement of a projectile thrown horizontally (4 marks)

Let the projectile be thrown horizontally from a height with speed v_i (Figure). Take the launch point as the origin and downward as positive for the vertical motion. Then

v_{ix}=v_i,\qquad v_{iy}=0,\qquad a_x=0,\qquad a_y=g

Derivation Velocity and displacement at time t

Aim. Find the velocity components, the speed and direction, and the displacements X and Y of the projectile at any time t.

Assumptions.

Air resistance is neglected.
v_{ix}=v_i, v_{iy}=0, a_x=0 and a_y=g (downward positive).

The projectile leaves O horizontally with speed v_i. At time t it has moved X horizontally and fallen Y; the velocity v is the resultant of the constant v_x and the growing v_y.

The horizontal component of velocity remains constant because there is no horizontal acceleration.
v_{fx}=v_{ix}=v_i
The vertical component grows. Apply the first equation of motion in the y-direction.
v_{fy}=v_{iy}+a_yt=0+gt=gt
The speed at time t is the magnitude of the resultant of the two perpendicular components.
v=\sqrt{v_{fx}^{2}+v_{fy}^{2}}=\sqrt{v_i^{2}+g^{2}t^{2}}
The direction \theta of the velocity below the horizontal comes from the ratio of the components.
\theta=\tan^{-1}\!\left(\dfrac{v_{fy}}{v_{fx}}\right)=\tan^{-1}\!\left(\dfrac{gt}{v_i}\right)
Horizontal displacement: the horizontal velocity is constant, so distance equals velocity times time.
X=v_{fx}\,t=v_it
Vertical displacement: apply the second equation of motion in the y-direction with v_{iy}=0.
Y=v_{iy}t+\tfrac12a_yt^{2}=\tfrac12gt^{2}

Resultv_{fx}=v_i,\quad v_{fy}=gt,\quad X=v_it,\quad Y=\tfrac12gt^{2}

The horizontal distance grows steadily with time while the vertical distance grows with t^{2}.

Path and time of fall (2 marks)

Eliminate t between X=v_it and Y=\tfrac12gt^{2}. Put t=X/v_i in the second equation:

Y=\dfrac{g}{2v_i^{2}}\,X^{2}

This is the equation of a parabola, so the trajectory of a horizontally thrown projectile is a parabola. If it is thrown from a height h, it reaches the ground when Y=h, so the time of fall is t=\sqrt{2h/g}, which does not depend on v_i; the horizontal distance covered is then v_i\sqrt{2h/g}.

2026 Annual I, Set D, Q.41+5 = 6 marks

Describe elastics collision in one dimension. Derive relation to show the relative velocities before and after collision.

Solution

Elastic collision (1 mark)

A collision in which both the momentum and the kinetic energy of the system are conserved is an elastic collision. The bodies do not lose kinetic energy as heat, sound or deformation, so they separate with the same total kinetic energy they had before.

Relative velocities before and after the collision (5 marks)

Two bodies of masses m_1 and m_2 move along the same straight line with velocities u_1 and u_2 before the collision and v_1 and v_2 after it (Figure). Velocity to the right is positive.

Derivation Relative speed of approach equals relative speed of separation

Aim. Show that u_1-u_2=v_2-v_1 for a head-on elastic collision.

Assumptions.

An isolated system of two bodies with no rotation, colliding along the line of motion.

Figure 3.9: Two balls before (left) and after (right) a head-on elastic collision. The faster ball on the left approaches the slower one, and the two separate with the same relative speed.

Conservation of momentum gives the first equation.
m_1u_1+m_2u_2=m_1v_1+m_2v_2
Rearrange so that the terms of each body are on one side.
m_1(u_1-v_1)=m_2(v_2-u_2)\qquad(1)
Conservation of kinetic energy gives the second equation. Multiply by 2 and rearrange in the same way.
m_1(u_1^{2}-v_1^{2})=m_2(v_2^{2}-u_2^{2})
Factorise both sides as differences of squares.
m_1(u_1-v_1)(u_1+v_1)=m_2(v_2-u_2)(v_2+u_2)\qquad(2)
Divide (2) by (1). The common factors cancel, which is valid because neither bracket is zero in a real collision.
u_1+v_1=v_2+u_2
Rearrange so that the velocities before the collision are on one side.
u_1-u_2=v_2-v_1

Resultu_1-u_2=v_2-v_1

The relative speed of approach before the collision equals the relative speed of separation after it. This holds only for elastic collisions, because it uses conservation of kinetic energy.

Exam focus

Derive the range and the maximum height of a projectile launched at an angle, with a figure; also the time of flight. (long)
2026 Set B Q.3 (6 marks, split 3+3: range and height); 2025 Set B Q.3 OR (7 marks: definition, time of flight, range and maximum range). Comprehensive question 3.5, worked examples 3.5 and Assignment 3.3.
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Derive the velocity and displacement of a projectile thrown horizontally from a height, and show the path is a parabola. (long)
2025 Set D Q.3 (7 marks, split 1+4+2). Section 3.4 derivation; book Example 3.4 and numerical 3.5. Not asked in 2026, so it is due.
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Elastic collision in one dimension: describe it and prove that the relative speed of approach equals the relative speed of separation; the special cases of the final velocities. (long)
2026 Set D Q.4 (6 marks, split 1+5; the paper tags it Unit 5 but it is section 3.8). 2025 Set D Q.2 (iii) OR (m_1\ll m_2) and MCQ 3. Comprehensive question 3.3 and numerical 3.7.
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Use the equations of motion for uniform acceleration: distance, final velocity, acceleration and time (use 2aS=v_f^2-v_i^2 when time is not given). (numerical)
2026 Set B MCQ 4 (distance in 5 s), Set B Q.2 (vii) (shell in a barrel, 3 marks) and Set D MCQ 5 (velocity after 5 s). 2025 Set B Q.2 (ii). Book numericals 3.1, 3.3 and Assignment 3.1.
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Effect of air resistance on projectile motion (range, maximum height, velocity components, path shape), with a graph. (short)
2026 Set D Q.2 (vii), 3 marks (2 for the effect and 1 for the graph). Short question 3.6 and an SLO of the unit.
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Concept questions on projectile motion: the time of flight depends on the vertical component; the horizontal velocity stays constant; the vertical component is zero at the top. (mcq)
2026 Set B MCQ 3 (time of flight) and Set D MCQ 4 (horizontal velocity). Book exercise MCQs 1, 4, 5, 8 and 9 and short questions 3.2 to 3.5.
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Derive the three equations of motion for uniform acceleration, from the definitions and from the velocity-time graph. (long)
Comprehensive question 3.1 and an SLO of the unit. Not asked as a long question in 2025 or 2026, but the equations are used in the projectile and numerical questions of both years.
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State the law of conservation of momentum and explain it with the handball bouncing off a wall; why kinetic energy can change in a closed system. (long)
Comprehensive questions 3.2 and 3.6. Momentum conservation was not examined in its own question in 2025 or 2026 (only through elastic collisions).
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