MOTION AND FORCE

LEARNING OBJECTIVES
At the end of this chapter the students will be able to:
1. Understand displacement from its definition and illustration.
2. Understand velocity, average velocity and instantaneous velocity.
3. Understand acceleration, average acceleration & instantaneous acceleration.
4. Understand the significance of area under velocity-time graph.
5. Recall and use equations, which represent uniformly accelerated motion in a straight line
including falling in a uniform gravitational field without air resistance.
6. Recall Newton’s Laws of motion.
7. Describe Newton’s second law of motion as rate of change of momentum.
8. Define impulse as a product of impulsive force and time.
9. Describe law of conservation of momentum.
10. Use the law of conservation of momentum in simple applications including elastic collisions
between two bodies in one dimension.
11. Describe the force produced due to flow of water.
12. Understand the process of rocket propulsion (simple treatment).
13. Understand projectile motion in a non-resistive medium.
14. Derive time of flight, maximum height and horizontal range of projectile motion.

INTRODUCTION:
We live in a universe of continual motion. In every piece of matter, the atoms are in a state of
never ending motion. We move around the Earth’s surface, while the Earth moves in its orbit around the
Sun. The Sun and the stars, too, are in motion. Everything in the vastness of space is in a state of
perpetual motion.
Every physical process involves motion of some sort. Because of its importance in the physical
world around us, it is logical that we should give due attention to the study of motion.
We already know that motion and rest are relative. Here, in this chapter, we shall discuss other
related topics in some more details.
DISPLACEMENT:
“The shortest distance between the two points is called displacement”.
(OR)
“Change in the position of a body from initial to final position with respect to origin through
the shortest distance is called displacement”.
97
98 Physics Intermediate Part-I

Unit:
Unit of displacement is metre (m). Typical Speeds
Speed, ms −1 Motion
Nature: 300 000 000
Light, radio waves, x-rays,
microwaves (in vacuum)
210 000 Earth-Sun travel around the galaxy
It is vector quantity. 29 600 Earth around the Sun
1 000 Moon around the Earth
Explanation: 980
333
SR-71 reconnaissance jet
Sound (in air)
267 Commercial jet airliner
→ 62 Commercial automobile (max.)
Suppose r1 is the position of a body at point A, with respect to 37 Falcon in a dive
29 Running cheetah
origin and its position changes from A to point B and then point C. 10 100-metres dash (max.)
9 Porpoise swimming
→ → 5 Flying bee
Position of B with respect to origin is r2 . Displacement of d is a 4 Human running
2 Human swimming
straight line directed from initial position A to final position B. 0.01 Walking ant

By using head to tail rule: y

→ → → A
OA + AB = OB → B C
→ d
→ → → r1 →
r1 + d = r2 r 2

→ → → x
O
d = r2 − r1
Note: For body along a straight line distance and displacement are equal.
VELOCITY:
“Time rate of change of displacement is called velocity”.
Unit and Nature:
Its unit is ms−1 and it is a vector quantity.
Mathematical Form:
→
→ ∆d
v =
∆t
Dimensions = [LT−1]
Average Velocity:
The ratio of total change in displacement to total time is called Average and instantaneous
→ velocity are equal when body
average velocity vav. moves with uniform velocity.

→
→ ∆d
vav = t

Instantaneous Velocity:
The change in the displacement of a body for such a small interval of time such that time
approaches to zero is called instantaneous velocity.
(OR)
Instantaneous velocity is defined as the limiting value of ∆d/∆t as the time internal ∆t, following
time t, approaches zero.
[Chapter-3] Motion and Force 99
Note: If instantaneous value does not change, the body is said to be moving with uniform velocity.
Velocity of body at any instant of time is given as:
 →
→ ∆ d 
v inst = Lim  
∆t → 0  ∆t 

ACCELERATION:
“Time rate of change of velocity is called acceleration”.
Mathematical Form: ∆V→ →
=V
→ 2 − →
→ → V
∆v 1
a = V1
∆t
Dimension:
→
[LT−2] V2

Nature and Unit:

• It is vector quantity and its units is ms−2.
• Acceleration produces due to following reasons:
• Change in direction of velocity.
• Change in magnitude of velocity.
• Change in both direction and magnitude of velocity.
Average Acceleration:
The ratio of total change in velocity to the total time taken is called average acceleration.
→
→ ∆v
aav =
∆t
Instantaneous Acceleration:
The change in the velocity of the body for extremely small intervals of time such that this time
approaches to zero is called instantaneous acceleration.
(OR)
Acceleration at a particular instant of time. Average and instantaneous
→ acceleration are equal when
→ ∆v body movies with uniform
a inst = Lim acceleration.
∆t → 0 ∆t

Uniform Acceleration:
If velocity changes equally in equal intervals of time then acceleration will be uniform
acceleration.
Non-uniform Acceleration:
If velocity of a body changes unequally in equal intervals of time then acceleration will be non-
uniform acceleration.
Positive Acceleration:
If velocity of body increases with time then its acceleration will be positive.
Negative Acceleration:
If the velocity of a body is decreasing with time then acceleration is said to be negative or called
deceleration.
100 Physics Intermediate Part-I

VELOCITY-TIME GRAPH:
“A continuous line that shows the variation of velocity with time is called velocity time graph”.
For velocity time graph, time is taken along x-axis and velocity along y-axis.
(i) Velocity Time Graph for Constant Velocity: y
v B
Velocity time graph for constant velocity is horizontal line
taken parallel to x-axis. It gives ∆v = 0, so a = 0. Distance covered by
body is equal to the area under the graph.
v
S = OA × OC
S = v×t t
x
(ii) For Increasing Velocity: O t C
y
→
It gives information, a is variable and positive. Slope at any Q
point on the graph gives acceleration at that instant.
→ A ∆v
→ ∆v QR
a = = PR v
∆t
(iii) Velocity Time Graph for Constant Acceleration: P ∆t R
x
→ O t
When graph between v and t is straight line inclined with Graph for instantaneous
x-axis then. acceleration
y
→
a is positive and uniform. v
S = Distance covered = Area of ∆OPQ P
1
S = 2 vt
v
Note: (i) Greater the slope larger will be acceleration.
Q
(ii) The area between velocity-time graph and the time axis is x
O t
numerically equal to the distance covered by the object.
REVIEW OF EQUATIONS OF UNIFORMLY ACCELERATED MOTION:
Suppose an object is moving with uniform acceleration a along
a straight line. If its initial velocity is vi and final velocity after a time
interval t is vf. Let the distance covered during this interval be S then • These equations are useful
only for linear motion with
we have uniform acceleration.
1st Equation of Motion: • A negative sign is assigned to
vf = vi + at qua ntities w ith directio n
nd opposite to that of initial velocity.
2 Equation of Motion:
1
S = vit + 2 at2
3rd Equation of Motion:
2 2
2aS = vf − vi
vi + vf
S =  2 ×t
 
These equations are useful only for linear motion with uniform acceleration. When the object
moves along a straight line, the direction of motion does not change in this case all the vectors can be
manipulated like scalars. In such problems, the direction of initial velocity is taken as positive.
Acceleration Due to Gravity (g):
In the absence of air resistance, all objects in free fall near the surface of the Earth, move towards
the earth with a uniform acceleration. This acceleration, known as acceleration due to gravity, is denoted
by the letter g and its average value near the Earth surface is taken as 9.8 ms−2 in the downward direction.
[Chapter-3] Motion and Force 101

Special cases of equations of motion when a = g:

Vertical displacement (m)

When the object is thrown in For downward motion (free full 20

upward direction and reaches the motion): 15

highest point: • 1st equation of motion: 10

• 1st equation of motion: vf = gt

vi = gt nd
• 2 equation of motion: 1 2 3 4
time (s)
rd
• 3 equation of motion: 1 How the displacement of a vertically
h = gt2 thrown ball varies with time?
vi = 2gh 2
20

Velocity (ms )
• 3rd equation of motion:

−1
10
vf = 2gh 2 4
0 Time (s)
NEWTON’S LAWS OF MOTION:
−10
Newton’s laws of motion are empirical laws i.e., these laws of
motion have experimental proof but no theoretical proof. These laws −20

were stated in 1687 in his famous book “Principia”. These laws hold How the velocity of a vertically
thrown ball varies with time?
good for motion of macro-objects at low speed. For very fast moving Velocity is upwards positive.
objects, such as atomic particles in an accelerator, relativistic Ans. when the body moves upward
mechanics developed by Albert Einstein is applicable. its velocity decreases continuously
and at maximum height becomes
Newton’s 1st Law of Motion: zero. When the body moves
downward its velocity increases
“A body at rest will remain at rest and a body moving with continuously and hits the ground
uniform velocity will continue to do so, unless acted upon by some with the initial velocity i.e., 20 m/s.
unbalanced external force”.
Inertia:
No body begins to move or
The property of an object due to which it opposes any change in comes to rest of itself.
its state of rest or of uniform motion is referred to as object’s inertia. ABU ALI SENA, (980-1037)
The more the inertia, the stronger is the tendency in the presence of a
force. Thus, mass of object is quantitative measure of its inertia.
Frame of Reference:
A set of coordinate system that gives time and location of an event.
(i) Inertial (non-accelerating) Frame of Reference:
The frame of reference whose acceleration is zero.
The frame of reference in which Newton’s laws of motion hold.
(ii) Non-inertial (accelerating) Frame of Reference:
The frame of reference in which Newton’s laws of motion does
not hold.
Accelerated frame of reference is called non-inertial frame of
reference. A frame of reference stationed on Earth is approximately an
inertial frame of reference.
Newton 2nd Law of Motion: At the surface of the Earth, in
situations where air friction is
→ → negligible, objects fall with the
When a force F acts upon a body, an acceleration a is same acceleration regardless of
produced, which is directly proportional to applied force and inversely their weights.
proportional to mass of the body.
102 Physics Intermediate Part-I

→ →
a ∝ F
→ 1
a ∝
m
→
→ F →
a ∝ ⇒ F = (Constant)(m)(a) All objects in free fall near the
m
surface of earth move towards
→ → earth with uniform acceleration
F = ma (Constant = 1) in absence of air resistance.

Newton’s 3rd Law of Motion:

Action and reaction are equal in magnitude but opposite in
direction.
Example:
Motion of rocket.
Note: Action and reaction never act on same body.
MOMENTUM:
“The quantity of motion in body is called momentum”.
Mathematically:
→ →
Product of mass and velocity i.e., p = m v . Momentum is
→
denoted by p . It is a vector quantity. Its direction is along the direction
of velocity.
A car accelerates along a road.
Unit of momentum is kg ms−1 or N-s. Which force actually moves the
car?

Dimension:
[ p ] = [MLT−1]
Momentum and Newton 2nd Law of Motion:
Statement:
Time rate of change of momentum is equal to applied force.
Mathematical Form:
Throwing a package onto shore
→ from a boat that was previously at
Suppose force ‘ F ’ changes velocity from initial velocity ‘vi’ to rest causes the boat to move out-
ward from shore (Newton’s third
final velocity ‘vf’ in time ‘∆t’ then from Newton’s 2nd law: law).
→
→ ∆p
F =
∆t Which will be more effective in
Derivation: knocking a bear down.
i. a rubber bullet or
Suppose the velocity of a body changes from initial velocity Ii. a lead bullet of the same
→ Momentum
value ‘vi’ to final velocity ‘vf’ in time ‘∆t’. So its acceleration a is
given by:
→ → →
→ vf − vi F m vi m vf
a =
∆t A ∆t B
[Chapter-3] Motion and Force 103

By 2nd law of motion:

→ →
F = ma
Stone
→ →
→  vf − vi  (a)
F = m 
 t  1 second
100
→ →
→ m vf − m vi
F =
∆t
Football
→ →
→ pf − pi
F = …… (a)
∆t (b)
→ → 1 second
Where, m vf = pf 10
Which hurt you in the above
→ → situations (a) or (b) and think why?
and m vi = pi In case of (a) we will be hurt
harder as the foot will experience
→ → → greater force because of short
Also pf − pi = ∆ p impulsive time.
So equation a becomes:
→
→ ∆p
F =
∆t
Hence rate of change of momentum is equal to applied force. This relation is also termed as
Newton’s second law in terms of momentum.
Impulse:
“When a very large force acts on a body for very short interval of time then the product of
large force and short ‘t’ is called impulse”.
Mathematical Form:
I = F×t Does a moving object have impulse?

Impulse I can be written mathematically as:

I = F × ∆t …… (1) Your hair acts like a crumple zone
As by 2nd law of motion: on your skull. A force of 5 N might
be enough to fracture your naked
∆p skull (cranium), but with a covering
F = of skin and hair, a force of 50 N
∆t would be needed.
So, ∆p = F × ∆t
So, eq. (1) become:
→ →
I = ∆p
So impulse is also defined as change in momentum. It has same dimensions as that of
momentum. It is a vector quantity.
Note: The force F can be regarded as the average force that acts during the time t.
104 Physics Intermediate Part-I

Law of Conservation of Momentum:

Statement:
“Total linear momentum of isolated system remains constant”.
Mathematical Form:
Law of conservation of momentum can be written mathematically as:
→ → → →
m1 v1 + m2 v2 = m1v1′ + m2v2′ F′ F

→ → → →
Where, v1 , v2 and v1′, v2′ are velocities of masses m1 m1 m2 During collision
and m2 before and after collision respectively. → →
v1 > v2
Proof:
Before collision
Consider an isolated system of two smooth hard balls
of masses “m1” and “m2” moving with velocities ‘v1’ and ‘v2’
where v1 > v2 as shown in figure. →
→ m m2 v2′
The impulse of mass m1 is: v1′ 1
After collision
→ →
I1 = F′ × t
→ →
= m1v1′ − m1 v1 …… (1)
Similarly for mass m2:
→ → What is the effect on the speed of
I2 = F ×t a fighter plane chasing another
when it opens fire? What happens
→ → to the speed of pursued plane
= m2v2′ − m2 v2 …… (2) when it returns the fire?

Adding eq. (1) and (2):

→ → → → → →
F′ × t + F × t = m1v1′ − m1 v1 + m2v2′ − m2 v2 …… (3)
From Newton’s 3rd law:
→ →
F′ = −F (Action and reaction)
→ →
F′ + F = 0 …… (4)
When a moving car stops quickly,
So (3) can be written as: the passengers move forward
towards the windshield. Seat
→ → → → → → belts change the forces of motion
( F′ + F ) × t = m1v1′ − m1 v1 + m2v2′ − m2 v2 and prevent the passengers from
moving. Thus the chance of injury
From eq. (4): is greatly reduced.

→ → → →
0 × t = m1v1′ − m1 v1 + m1v2′ − m2 v2
→ → → →
0 = m1v1′ − m1 v1 + m2v2′ − m2 v2
[Chapter-3] Motion and Force 105

→ → → →
m1v1′ + m2v2′ = m1 v1 + m2 v2
Total linear momentum Total linear momentum
 after collision  =  before collision 
Isolated System:
A system on which no external agency exerts any force.
For Example:
The molecules of a gas enclosed in a glass vessel at constant
temperature constitute an isolated system. The molecules can collide A motorcycle's safety helmet is
padded so as to extend the time of
with one another because of their random motion but, being enclosed any collision to prevent serious
injury.
by glass vessel, no external agency can exert a force on them.
ELASTIC AND INELASTIC COLLISION:
“When two bodies are approaching each other a force comes into play between them for a
finite time and brings about a measurable changes in their velocities, momentum and energy,
according to the respective law of conservation, a collision is said to have take place”.
Types of Collision:
There are two types of collisions:
(i) Elastic collision (ii) Inelastic collision
(i) Elastic Collision:
The collision in which kinetic energy and momentum are conserved before and after collision.
Example:
Collision of gas molecules enclosed in a vessel is elastic collision.
(ii) Inelastic Collision:
The collision in which momentum is conserved but the kinetic energy is not conserved is called
in-elastic collision.
Examples:
Collision between two cars is an example of inelastic collision.
Elastic Collision in One Dimension:
Consider two balls of masses m1 and m2 moving with v1 and v2 respectively. Such that v1 > v2
having same direction. After collision their velocities becomes v1′ and v2′.

v1 v2 v1 v1′ v2 v2′

m1 m2 m1 m2 m1 m2
Before collision During collision After collision
106 Physics Intermediate Part-I

According to law of conservation of K.E.

Total K.E before Total K.E after
 collision  =  collision 
1 2 1 2 1 2 1 2
2 m 1v1 + m2v2 =
2 2 m 1v′1 +
2 m 2v′2

2 2 2 2
∴ m1v1 + m2v2 = m1v′1 + m2v′2
2 2 2 2
m1v1 − m1v′1 = m2v′2 − m2v2
2 2 2 2
m1(v1 − v′1) = m2(v′2 − v2)
m1(v1 − v1′)(v1 + v1′) = m2(v2′ − v2)(v2′ + v2) …… (1)
Now by law of conservation of momentum:
Total momentum Total momentum
 before collision  =  after collision 
m1v1 + m2v2 = m1v1′ + m2v2′
m1v1 − m1v1′ = m2v2′ − m2v2
m1(v1 − v1′) = m2(v2′ − v2) …… (2)
Dividing eq. (1) by (2):
m1(v1 − v1′)(v1 + v1′) m2(v2′ − v2)(v2′ + v2)
=
m1(v1 − v1′) m2(v2′ − v2)
(v1 + v1′) = (v2′ + v2) …… (a)
v1 − v2 = v2′ − v1′
v1 − v2 = −(v1′ − v2′)
Where (v1 − v2) is relative velocity of ball of mass m1 with respect to ball m2 before collision and
(v1′ − v2′) is relative velocity of ball of mass m1 with respect to ball of mass m2 after collision.
or (Velocity of approach) = −(Velocity of separation)
m1 − m2 2m2v2
v1 ′ =  m + m  v1 + m + m …… (x)
 1 2 1 2

2m1v1 m2 − m1

v2 ′ = m + m +  m + m  v2 …… (y)
1 2  1 2

FOR YOUR INFORMATION:

Derivation of Velocity after Collision:
From eq. (a):
v2 ′ = v1 + v1′ − v2
So, eq. (2) becomes:
m1(v1 − v1′) = m2(v1 + v1′ − v2 − v2)
m1(v1 − v1′) = m2(v1 + v1′ − 2v2)
[Chapter-3] Motion and Force 107

m1v1 − m1v1′ = m2v1 + m2v1′ − 2m2v2

m1v1 − m2v1 + 2m2v2 = m2v1′ + m1v1′

or m1v1′ + m2v1′ = m1v1 − m2v1 + 2m2v2

v1′(m1 + m2) = v1(m1 − m2) + 2m2v2

v1(m1 − m2) 2m2v2

v1 ′ = m1 + m2 + m1 + m2

m1 − m2 2m2v2

v1 ′ =  v + If another car crashes into back of
m1 + m2 1 m1 + m2 yours, the head-rest of the car
seat can save you from serious
neck injury. It helps to accelerate
m2 − m1 2m1v1 your head forward with the same
Similarly, v2 ′ = m + m  v2 + m + m rate as the rest of your body.
 1 2  1 2

Special Cases:
Case-I: Two particles have the same mass:

In this case m1 = m2 = m.

So, eq. (x) becomes:

(m − m) v1 2(m) v2 v1 v2
v1 ′ = +
m+m m+m
m1 m2
0 + 2mv
v1 ′ = 2m

v1 ′ = 0 + v2

v1 ′ = v2

Eq. (y) becomes:

(m − m) v2 2mv1
v2 ′ =
(m + m) + m + m

2mv1
v2 ′ = 0 + 2m

v2 ′ = 0 + v1

v2 ′ = v1

Result:
In this case colliding bodies will interchange their velocities.
108 Physics Intermediate Part-I

Case-II: Two bodies have same mass and one of the body is at rest:

m1 = m2 = m , v2 = 0

So, eq. (x) becomes:

v1 v2 = 0
m − m 2m m1 m2
v1 ′ = m + m v1 + (0)
  m+m
Before collision
v1 ′ = 0
v1′ = 0 v2′ = v1
So, eq. (y) becomes: m1 m2

m − m 2mv1 After collision

v2 ′ = m + m v2 +
  m+m

2mv1
v2 ′ = 2m

v2 ′ = v1

Result:

Velocity of balls will interchange. The ball which was initially at rest will start moving. And the
ball which was moving initially will come to rest after collision.

Case-III: Mass at rest is very heavy:

When light body collides with massive body at rest i.e., m2 >> m1 and m1 ≈ 0, v2 = 0.

So eq. (x) is given by:

0 − m2 2m2(0) v1 v2 = 0
v1 ′ =   v1 +
0 + m2 m1 + m2
m1 m2
 m2
v1 ′ =  − m  v1 + 0
 2
Before collision

v1 ′ = −v1
v1′ = −v1 v2′ = 0

Also eq. (y) becomes as: m1 m2

m2 − 0 2(0)(v1)
v2 ′ = m + 0 (0) + 0 + m After collision
 2  2

v2 ′ = 0

Result:

Thus lighter body bounces back while heavy body does not change its state of rest.
[Chapter-3] Motion and Force 109
Case-IV: Mass at rest is very light:
In this case m1 >> m2 and v2 = 0, m2 ≈ 0.
So eq. (x) becomes:
m1 − 0 2(0)v2 v1 v2 = 0
v1 ′ = m  v1 +
 1 + 0 m1 + 0
m1 m1 m2
v1 ′ =  m  v1 + 0
 1
Before collision
~ v1
v1 ′ −
v1′ = v1 v2′ ~
− 2v1
Also eq. (y) becomes:
(0 − m1)(0) 2m1v1 m1 m2
v2 ′ = (m1 + 0) + m1
After collision
v2 ′ = 0 + 2v1

v2 ′ = 2v1

Result:
Thus velocity of incident ball does not change. While the target ball moves or bounces off in the
forward direction with velocity double of velocity of incident ball.
FORCE DUE TO FLOW OF WATER:
Force exerted by water during its flow is equal to product of
mass per unit time and change in velocity.
In thrill mac hine rid es at
Formula: amusement parks, there can be
an acceleration of 3g or more. But
without head rests, acceleration
m
F =  t ×v like this would not be safe. Think
  why not?

Proof:
Suppose ‘m’ mass of water flows in a horizontal pipe with initial velocity vi = v. It strikes the
wall and comes to rest i.e., vf = 0.
Initial momentum of water = pi = mv
Final momentum of water = pf = 0
Change in momentum = ∆p = pf − pi
∆p = 0 − mv → → Wall
vi = v
∆p = −mv F

Time rate of change of momentum: Force

∆p mv exerted
= − …… (1) by wall
∆t ∆t on water
∆p
As = F′
∆t
110 Physics Intermediate Part-I

Because force exerted opposite to flow of water:

So, eq. (1) becomes:
mv
F′ = − …… (2)
∆t
That force decreases the momentum with respect to time that when vf = 0.
If tf = t and
ti = 0 then
∆t = tf − ti
∆t = t
So, eq. (2) becomes:
m
F′ = − t v

It is the force exerted by wall on the water. According to 3rd law of motion, water exerts a force
on wall of same magnitude in opposite direction. So force due to water flow is:
F = −F′

mv
F =
t

MOMENTUM AND EXPLOSIVE FORCES:

Momentum of system may be changed by explosive forces as given in the following examples:
Exploding Shell:
Consider a shell of mass M is dropped from a bomber jet shell explodes in air into two fragments
of mass m1 and m2 having momentum p1 and p2. Vector addition of momenta after explosion is given by if:
Mass of shell = M
Momentum of shell before explosion = pb
Momentum of fragment 1 after explosion = p1
Momentum of fragment 2 after explosion = p2
Momentum after explosion = p1 + p2 p1 p2
From figure: pb
pb = p1 + p2 p2

 Momentum   Momentum 
before explosion = after explosion
This property of conservation of momentum will be valid even the shell explodes into many
particles. In this case the sum of momenta of all the fragments will be equal to momentum of shell
before explosion.
[Chapter-3] Motion and Force 111
Firing of Gun:
Consider a system of a gun loaded with a bullet.
Total momentum before firing = PT = 0
Momentum of bullet after firing = Pb = mbvb
Momentum of gun after firing = PG = mGvG
Total momentum after firing = Sum of momentum of bullet and gun
By law of conservation of momentum:
Momentum before firing = [Momentum after firing]
0 = mbvb + mGvG
mbvb = −mGvG
mGvG
vb = − m
b

Negative sign indicates velocity of recoil of gun and

velocity of bullet are opposite in direction.
ROCKET PROPULSION:
Rockets move by expelling burning gas throw engines at
their rear. The ignited fuel turns to high pressure gas, which is
expelled with extremely high velocity from rocket engines.
The momentum of rocket is equal to momentum of
expelled gases. Hence rocket gets more momentum during its
upward flight. fuel
(liquid
→ → hydrogen)
P rocket = − P gas
liquid
The rocket engines continue to expel gases after the rocket oxygen
has begun moving and hence rocket continues to gain more and
more momentum. So instead of travelling at steady speed the Combustion
Chamber
rocket gets faster and faster so long the engines are operating.
A rocket carries its own fuel in the form of a liquid or
solid hydrogen and oxygen. It can, therefore work at great heights
where very little or no air is present.
Facts about Rocket:
• A typical rocket consumes 10000 kg s−1 of fuel.
Fig. Fuel and oxygen mix in the
• Ejects the burnt gases at speed of 4000 ms−1. combustion chamber. Hot gases
• Almost 80% of launch mass of rocket is fuel. exhaust the chamber at a very
high ve lo city. T he g ain in
• When one part of rocket has done its job, it is discarded momentum of the gases equals.
which give greater speed. The gain in momentum of the
rocket. The gas and rocket push
Acceleration of Rocket: against each other and move in
Let M = Mass of rocket opposite directions.
m = Mass of gases ejected per second at speed v relative to rocket
So, ∆v = v
∆t = 1s
nd
By 2 law of motion in terms of momentum:
∆p ∆(mv) m∆v
F = = =
∆t ∆t ∆t
mv
F = 1 = mv …… (1)
112 Physics Intermediate Part-I

By 2nd law of motion:

F = Ma …… (2)
Comparing eq. (1) and (2):
Ma = mv

→
→ mv
a =
M

1
a ∝
M

The mass of rocket decreases and hence, acceleration of rocket increases as it moves upward.
PROJECTILE MOTION:
“Two dimensional motion under constant acceleration due to gravity is called projectile
motion”.
(OR)
“Projectile motion is defined as a motion in which body covers horizontal as well as vertical
distance only under gravitational acceleration”.
Examples:
(i) The motion of ballistic missile. If a projectile is thrown at angle of
O
45 . At highest point its K.E
(ii) Motion of bullet fired from gun. becomes half of initial value.

(iii) Motion of foot ball kicked by players into air.

Projectile:
An object that perform projectile motion is called projectile.
Examples:
Foot ball, bullet, ballistic missile are examples of projectiles.
Trajectory:
Path followed by projectile in air is called trajectory.
The trajectory of projectile is parabolic.
Horizontal Distance Covered by Projectile:
Consider a body of mass m which is thrown with initial velocity = vix with x-axis as shown in
figure. According to Newton’s first law of motion, there will be no acceleration in horizontal direction,
unless a horizontally directed force acts on the ball. Ignoring the air friction, only force acting on the ball
during flight is the force of gravity. There is no horizontal force acting on it. So its horizontal velocity
will remain unchanged and will be vx, until the ball hits something.
vix = vfx
Thus, horizontal acceleration ax = 0.
[Chapter-3] Motion and Force 113

Distance travelled = S = x y

Time = t vx
A
vy
Now from 2nd equation of motion: vx
1
S = vit + at2 Y vy
2 vx
1 vy
X = vixt + 2 axt2 vx
x
B C v
y
X = vixt = vi cos θt X

Vertical Distance Covered by Projectile:

Initial vertical velocity = viy = 0
Vertical acceleration = ay = g
Because gravitational acceleration acts in the downward direction:
Distance = S = y
nd
From 2 equation of motion: Relation between the maximum
1 height and time of flight is given
S = vit + ayt2 by H = (gt2/8).
2
1
y = viyt + 2 ayt2
1
y = (0) t + 2 gt2

1
y = 2 gt2
y
Magnitude of Resultant Velocity:
Suppose that a projectile is fixed in a direction at angle θ
with horizontal by initial velocity vi. vy →
v
→
Magnitude of resultant velocity in terms of final velocity. v i
2 2 2 vx
v = vfx + vfy (Pythagoras theorem) vy
vx
→
(1) Considering horizontal motion: θ v
O x
vfx = vix = vi cos θ
(2) Considering vertical motion, the initial vertical component of velocity is vi sin θ in upward direction.
So, vfy = vi sin θ − gt
y
Taking square root on both sides: P
v = vfx2 + vfy2
From ∆OPA: vf
vfy
vfy
tan φ = v
fx φ
O vfx x
Direction of ‘vf’: A

vfy
φ = tan−1 v 
 fx
114 Physics Intermediate Part-I

Height of Projectile:
y
The maximum vertical distance covered by a projectile
during its flight is called height of projectile. It is represented by ‘h’. vi

Mathematical Form:
h
Mathematically it is given by: θ
x
2 2
vi sin θ
h = 2g

Derivation:

Consider a projectile is thrown at an angle θ with

horizontal. Let viy = vi sin θ be its initial velocity.

vfy = 0

ay = −g

From 3rd equation of motion:

2 2
2aS = vf − vi (Q S = Height = h)

In the vertical direction:

2 2
2ayS = vfy − viy
A photograph of two balls released
2(−g)h = 0 − (vi sin θ)2 simultaneously from a mechanism
that allows one ball to drop freely
2 while the other is projected
−2gh = −vi sin2 θ horizontally. At any time the two
balls are at the same level, i.e.,
2 their vertical displacements are
vi sin2 θ equal.
h =
2g

Time of Flight:
The time for which projectile remains in air is called time of flight of projectile.

(OR)
y
The time taken by the projectile to cover a distance from the
point of projection to the place where it hits the ground at the same vi
level is called time of flight.

Mathematical Form: θ
x
Time of flight can be written as:
2vi sin θ
t =
g
[Chapter-3] Motion and Force 115

Derivation:
Suppose a projectile is thrown in air at an angle of ‘θ’ with
horizontal with initial velocity.
viy = vi sin θ
Since projectile returns to earth so it covers no vertical distance
i.e.,
Water is projected from two rubber
y = S = 0 (S = h = 0) pipes at the same speed-from one
at an angle of 30O and from the other
O
Acceleration due the gravity = ay = −g at 60 . Why are the ranges equal?

Time of flight = t
From 2nd equation of motion:
1 Time to reach maximum height
S = vit + 2 at2 t
t′ =
2
1
y = viyt + 2 ayt2 t′ =
vi sin θ
g

1
0 = vi sin θt + 2 (−g)t2

1
vi sin θt = 2 gt2

1
vi sin θ = 2 gt

2vi sin θ
t = g

Range of Projectile:
Maximum horizontal distance travelled by a projectile is called
range of projectile.
• Height and range are related by
Mathematical Form: R tan θ = 4H
• Range and time of flight are
Range of projectile is given by: related by
2 R tan θ = ½ gt2
vi sin 2θ
R =
g
Derivation:
Since range of projectile is horizontal distance covered by projectile. So,
y
X = vixt …… (1)
X = Range of projectile = R
vix = vi cos θ vi

2vi sin θ
and t = θ x
g R=x
116 Physics Intermediate Part-I

So eq. (1) becomes:

2
vi cos θ × 2vi sin θ vi 2 sin θ cos θ
R = =
g g
2
vi sin 2θ
R = (∴ 2 sin cos θ = sin 2θ)
g

Maximum Range:
Range of projectile will be maximum when
sin 2θ = 1 or For two angles whose sum is 90O
will have same range. e.g., 60O and
2θ = sin−1 (1) = 90° O
30 have same range.

90°
θ = 2 = 45°
2
vi
So, Rmax. = g

Height
O
60 O
So range is maximum for θ = 45°. 45
O
30
Range will be maximum 2θ = 90°.
Range
Ballistic Missile: For an angle less than 45 , the
O

height reached by the projectile

An un-powered and un-guided missile is called ballistic missile. and the range both will be less.
When the angle of projectile is
These missiles have short range and less precision. O
larger than 45 , the height attained
will be more but the range is again
Remote Control Guided Missile: less.

A powered and guided missile is called remote control guided

missile. These missiles have long range and greater precision.
Ballistic Trajectory: Ideal Path
The path followed by ballistic missiles is called ballistic
trajectory.
Actual Path
Ballistic Flight:
In the presence of air friction the
A ballistic flight is that in which a projectile is given an initial trajectory of a high speed projectile
push and then allowed to move freely due to inertia and under the fall short of a parabolic path.
action of gravity.
[Chapter-3] Motion and Force 117

SOLVED EXAMPLES
EXAMPLE 3.1
The velocity time graph of a car moving on a straight road is shown in figure. Describe the
motion of the car and find the distance covered.

SOLUTION
Motion of car from A to B:
The graph shows that car starts from rest i.e., Vi = 0 and its
velocity becomes 20 m/s in time 5 sec. Then
Average acceleration is given by
∆V 20
a = = 5 = 4 m/s2
∆t
Motion of car from B to C:
The graph shows that car moves with uniform velocity of 20
2
m/s . Here acceleration is zero.
Motion of Car From C to D:
The graph shows that the acceleration decreases during last four seconds
∆V – 20
and a = = 4 = – 5 m/s2
∆t
Negative sign shows that velocity is decreasing.
Distance Covered by Car:
Distance covered = Area of ∆ABF + Area of rectangle BFEC + Area of ∆ECD
1 1
= 2 (5)(20) + (10)(20) + 2 (5)(20)

= 50 + 200 + 50
= 300 m

EXAMPLE 3.2
A 1500 kg car has its velocity reduced from 20 ms–1 to 15 ms–1 in 3.0 s. How large was the
average retarding force.
Data:
Mass = m = 1500 kg
Initial velocity = Vi = 20 ms–1
118 Physics Intermediate Part-I

Final velocity = Vf = 15 ms–1

Time taken = t = 3.0 s
To Find:
Force = F = ?

SOLUTION
We know that
m Vf – m Vi
F =
t
m (Vf – Vi)
=
t
Putting values, we get
1500 (15 – 20)
F = 3
1500 x (– 5)
= 3
= – 2500 N
F = – 2500 N

Result:
Force = F = −2500 N
= −2.5 kN
The negative sign indicates that the force is retarding.

EXAMPLE 3.3
Two spherical balls of 2.0 kg and 3.0 kg masses are moving towards each other with
velocities of 6.0 ms–1 and 4 ms–1 respectively. What must be the velocity of the smaller ball after
–1
collision, if the velocity of the bigger ball is 3.0 ms ?
Data:
Mass of 1st ball = m1 = 2 kg
Mass of 2nd ball = m2 = 3 kg
Velocity of 1st ball = V1 = 6 m s–1
Velocity of 2nd ball = V2 = – 4 m s–1

Velocity after collision = V2′ = – 3 m s–1

To Find:
Velocity of 1st ball after collision = V1′ = ?
[Chapter-3] Motion and Force 119

SOLUTION
Using law of conservation of momentum.
Momentum of the system before collision = Momentum of the system after collision

m1 V1 + m2 V2 = m1 V1′ + m2 V2′
Putting values

2 (6) + 3 (– 4) = 2 V1′ + 3 (– 3)

12 – 12 = 2 V1′ – 9

0 = 2 V1′ – 9

2 V1′ = 9

V1′ = 4.5 m / s
Result:
Velocity of 1st ball after collision = V′1 = 4.5 m/s

EXAMPLE 3.4
A 70 g ball collides with another ball of mass 140 g. The initial velocity of the first ball is
–1
9 ms to the right while the second ball is at rest. If the collisions were perfectly elastic what
would be the velocity of the two balls after the collision?
Data:
Mass of 1st ball = m1 = 70 g = 0.07 kg
Mass of 2nd ball = m2 = 140 g = 0.14 kg
Velocity of 1st ball = V1 = 9 m s–1
Velocity of 2nd ball = V2 = 0
To Find:
Velocity of 1st ball after collision = V1′ = ?

Velocity of 2nd ball after collision = V2′ = ?

SOLUTION
m1 – m2 2 m 2 V2
Using V1′ = m + m V1 + m + m
1 2 1 2

0.07 − 0.14
V1′ = 0.07 + 0.14 × 9 + 0

V1′ = – 3 m/s
120 Physics Intermediate Part-I

2m1 m2 – m1
Now, V2′ = m + m V1 + m + m V2
1 2 1 2

2 x 0.07
V2′ = 0.07 + 0.14 x 9 + 0

V2′ = 6 ms–1
Result:
Velocity of 1st ball after collision = V′1 = −3 m/s

Velocity of 2nd ball after collision = V′2 = 6 m/s

EXAMPLE 3.5
A 100 g golf ball is moving to the right with a velocity of 20 ms–1. It makes a head on
collision with a 8 kg steel ball, initially at rest. Compute velocities of the ball after collision.
Data:
Mass of 1st ball = m1 = 100 g
100
= 1000 kg

= 0.1 kg
nd
Mass of 2 ball = m2 = 8 kg
Velocity of 1st ball = V1 = 20 ms–1
Velocity of 2nd ball = V2 = 0
To Find:
Velocity of 1st ball after collision = V1′ = ?
Velocity of 2nd ball after collision = V2′ = ?

SOLUTION
m1 – m2 2m2 V2
Using V1′ = m + m V1 + m + m
1 2 1 2

0.1 – 8
V1′ = 0.1 + 8 x 20 + 0

V1′ = –19.5 ms–1

2m1 m2 – m1
Now, V2′ = m + m V1 + m + m V2
1 2 1 2

2 x 0.1
V2′ = 0.1 + 8 x 20 + 0

V2′ = 0.49 m s–1

V2′ = 0.5 m s–1

[Chapter-3] Motion and Force 121
Result:
Velocity of 1st ball after collision = V1′ = −19.5 m/s

Velocity of 2nd ball after collision = V2′ = 0.5 m/s

EXAMPLE 3.6
A hose pipe ejects water at a speed of 0.3 ms–1 through a hole of area 50 cm2. If the water
strikes a wall normally, calculate the force on the wall, assuming the velocity of the water normal
to the wall is zero after striking.
Data:
v = 0.3 m s–1
A = 50 cm2
= 50 x 10–4 m2
To Find:
Force on the ball = F = ?

SOLUTION
mv
Using F = …… (1)
t
m
As ρ =
v
m = ρv
V
∴ (volume of water per second) = AV
t
= 50 x 10–4 x 0.3
= 0.0015 m3
m V
Since = ρ t
t
m
t = 1000 x 0.0015

= 1.5 kg/s
Putting eq. (1)
∴ F = 1.5(0.3)
= 0.45 N

Result:
Force on the ball = F = 0.45 N
122 Physics Intermediate Part-I

EXAMPLE 3.7
–1 o
A ball is thrown with a speed of 30 ms in a direction 30 above the horizontal. Determine
the height to which it rises, the time of flight and horizontal range.

Data:
Angle with horizontal = θ = 30o
Initial speed = Vi = 30 m s–1

To Find:
Vertical height = h = ?
Time of flight = t = ?
Horizontal range = R = ?

SOLUTION
For Vertical Height:
2
V i sin2 θ
h = 2g
(30 sin 30)2 (30)2 × (sin 30)2
h = 2 x 9.8 or h =
2 × 9.8
= 11.5 m
For Range of Projectile:
2
V i sin 2θ
R =
g
(30)2 sin (2 x 30)
R = 9.8
(30)2 sin 60
R =
9.8
(30)2 (0.866)
R =
9.8
R = 79.5 m
For Time of Flight:
2 Vi sin θ
t =
g
2 x 30 x sin 30
t = 9.8
t = 3.1 sec
[Chapter-3] Motion and Force 123
Result:
Vertical height = h = 11.5 m
Horizontal range = R = 79.5 m
Time of flight = t = 3.1 sec.

EXAMPLE 3.8
In example 3.7 calculate the maximum range and the height reached by the ball if the
angles of projection are (i) 45o (ii) 60o.
Data:
Initial speed = Vi = 30 m/s
Angle with horizontal = θ = 30o
To Find:
Maximum range = R = ?
Vertical height = h = ?
When: (i) θ = 45°
(ii) θ = 60°

SOLUTION
As we know that
2
V i sin 2θ
R = g
(i) When θ = 45°°
(30)2 sin (2 x 45o)
R = 9.8
900 x sin 90
= 9.8
900 x 1
=
9.8
R = 91.8 m
and vertical height
Using formula
2
V i sin2 θ
h =
2g
(30)2 (sin 45)2
=
2 (9.8)
900 (0.707)2
=
19.6
= 23 m
h = 23 m
124 Physics Intermediate Part-I

(ii) When θ = 60°°

We know that
2
V i sin 2θ
R =
g
(30)2 sin (2 x 60o)
= 9.8
900 x sin 120
=
9.8
900 x 0.866
=
9.8

R = 80 m
and vertical height is
Using formula
2
V i sin2 θ
h = 2g
(30)2 (sin 60)2
= 2 (9.8)
900 (0.866)2
=
19.6
= 34.4 m
h = 34.4 m

Result:
(i) When θ = 45°°
Maximum range = R = 91.8 m
Vertical height = h = 23 m
(ii) When θ = 60°°
Maximum range = R = 80 m
Vertical height = h = 34.4 m
[Chapter-3] Motion and Force 125

SHORT QUESTION ANSWERS

3.1 What is the difference between uniform and variable velocity. From the explanation of
variable velocity, define acceleration. Give SI units of velocity and acceleration.
Ans. Uniform Velocity: A body is said to be moving with uniform velocity if it covers equal
displacements in equal intervals of time.
Variable Velocity: A body is said to be moving with variable velocity if it covers unequal
displacements in equal intervals of time. Velocity of a body may change due to change in
magnitude or direction or both.
Acceleration: It is defined as “time rate of change of velocity”.
→
→ ∆v
a =
∆t
In SI-units, velocity is measured in m/s and acceleration is measured in m/s2.
3.2 An object is thrown vertically upward. Discuss the sign of acceleration due to gravity,
relative to velocity, while the object is in air.
Ans. (i) When an object is thrown vertically upward, then the sign of v
acceleration due to gravity relative to velocity or direction
of motion of the object will be negative because the object
will be moving vertically upward and acceleration due to −g +g
v
gravity will be vertically downward. (i) (ii)
(ii) When the object will be falling downward, the sign of acceleration due to gravity relative
to velocity will be positive because both velocity and acceleration due to gravity are in the
same direction.
3.3 Can the velocity of an object reverse direction when acceleration is constant? If so, give an
example.
Ans. Yes, a body can reverse its direction of motion, when it is moving with constant negative
acceleration for example when an object is thrown vertically upward, it moves with a constant
acceleration due to gravity, its velocity decreases with time. The object stops at the highest point
and then reverses its direction of motion and starts falling downward with the same acceleration.
3.4 Specify the correct statements:
(a) An object can have a constant velocity even its speed is changing.
(b) An object can have a constant speed even its velocity is changing.
(c) An object can have a zero velocity even its acceleration is not zero.
(d) An object subjected to a constant acceleration can reverse its velocity.
Ans. Statement (a) is incorrect. Whereas Statements (b), (c) and (d) are correct.
(i) Reason: “a” is incorrect. Because for constant velocity, a body must move with same
speed and direction. If speed is changing velocity cannot remain constant.
(ii) Reason: “b” is correct. Velocity of object may change even if its speed is constant. e.g.,
when an object moves in circular path its speed is constant but direction of motion changes
at each point.
126 Physics Intermediate Part-I

(iii) Reason: “c” is correct. When a car starts its motion from rest its initial velocity is zero but
acceleration is not zero.
(iv) Reason: “d” is correct. When an object is thrown in upward direction its acceleration
remains constant throughout the motion but velocity reverses its direction on reaching the
highest point.
3.5 A man standing on the top of a tower throws a ball straight up with initial velocity vi and at
the same time throws a second ball straight downward with the same speed. Which ball
will have larger speed when it strikes the ground? Ignore air friction.
Ans. Both the balls will have the ‘same speed’ on striking the ground. This is because the ball which
is thrown vertically upward with velocity vi will come back to the level of projection with the
same velocity vi in the downward direction. Hence, both the balls start the downward motion
with the same velocity, with the same acceleration but with different times. So both balls will
have same speed when they strike ground but strike at different points.
→ →
3.6 Explain the circumstances in which the velocity v and acceleration a of a car are:
(i) Parallel (ii) Anti-parallel
→ →
(iii) Perpendicular to one another (iv) v is zero but a is not
→ →
(v) a is zero but v is not zero
→ →
Ans. (i) v and a are parallel: When velocity of a car is increasing and it is moving on a straight
road.
→ →
(ii) v and a are anti-parallel: When velocity of a car moving on a straight road decreasing.
i.e., when brakes are applied.
→ →
(iii) v and a are perpendicular: When a car is moving in a circular path. v is along direction
of tangent and a is towards the centre along the radius.
→ →
(iv) v is zero but a is not zero: When a car is accelerated from rest, its initial velocity is zero
→
but a is not zero. Or when brakes are applied the final velocity of car becomes zero but
acceleration is not zero.
→ →
(v) a is zero but v is not zero: When a car moves on a straight road with constant speed, its
→
acceleration will be zero, but v will have a constant value.
3.7 Motion with constant velocity is a special case of motion with constant acceleration. Is this
statement true? Discuss.
Ans. Yes, the statement is true, because when a body moves with constant velocity, its acceleration is
zero and remains constantly zero during the motion i.e., 0 ms−2.
[Chapter-3] Motion and Force 127
3.8 Find the change in momentum for an object subjected to a given force for a given time and
state law of motion in terms of momentum.
→
Ans. Suppose an object of mass ‘m’ moving with velocity ‘vi’. An external force F acts on it for time
‘t’ and its velocity becomes ‘vf’ then acceleration produced is given by:
→ →
→ vf − vi
a =
t

By Newton’s 2nd law:

→
→ F
a =
m

Comparing the above equations:

→ → →
F vf − vi
m =
t

→ →
→ m vf − m vi
F =
t

→
→ ∆p
F =
t

Statement: Force acting on a body is equal to time rate of change of momentum of that body.
This is Newton’s 2nd law in terms of momentum.

3.9 Define impulse and show that how it is related to linear momentum?

Ans. “When a very large force acts on a body for very short interval of time then the product of large
force and short ‘t’ is called impulse.”
→
Impulse = F × t

Relation between impulse and momentum:

→
→ ∆p
As, F = t

→ →
⇒ F ×t = ∆p
→ →
F ×t = Impulse = ∆ p

Impulse is equal to change in momentum of an object.

128 Physics Intermediate Part-I

3.10 State the law of conservation of linear momentum, pointing out the importance of isolated
system. Explain, why under certain conditions, the law is useful even though the system is
not completely isolated?
Ans. This law states, that total linear momentum of an isolated system remains constant. An isolated
system is a system upon which no external force acts. This law is useful even when the system is
not completely isolated when internal forces of bodies on each other are much larger than
external forces. Hence, external forces are ignored and the system is considered to be an isolated
system. e.g., In rocket propulsion, firing of a bullet, explosion of forces.
3.11 Explain the difference between elastic and inelastic collisions. Explain how would a
bouncing ball behave in each case? Give plausible reasons for the fact that K.E is not
conserved in most cases?
Ans. In an ‘elastic collision’, both K.E and momentum remains constant before and after collisions.
Whereas, in an inelastic collisions, K.E of the colliding bodies does not remain constant but
momentum is constant. When a bouncing ball makes an elastic collision it bounces back to the
same height, from where it was dropped on marble. But in case of inelastic collision, it does not
bounce back to the same height. K.E is not conserved and will be converted into heat and sound
energy.
3.12 Explain what is meant by projectile motion. Derive expressions for
(a) The time of flight (b) The range of projectile.
Show that the range of projectile is maximum when projectile is thrown at an angle of 45˚
with the horizontal.
Ans. (a) Time of Flight: The time for which projectile remains in air y
is called time of flight of projectile. vi
(OR)
The time taken by the projectile to cover a distance from the
point of projection to the place where it strikes the ground at θ
x
the same level is called time of flight.
2vi sin θ
t = g
Derivation: Suppose a projectile is thrown in air at an angle of ‘θ’ with horizontal with
initial velocity.
viy = vi sin θ
Since projectile returns to earth so it covers no vertical distance i.e.,
y = S = h = 0
Acceleration due the gravity = ay = −g
Time of flight = t
nd
From 2 equation of motion:
1
S = vit + 2 at2

1
y = viyt + 2 ayt2
[Chapter-3] Motion and Force 129
1
0 = vi sin θt + 2 (−g)t2

1
vi sin θt = 2 gt2

1
vi sin θ = gt
2
2vi sin θ
t =
g
(b) Range of Projectile: Maximum horizontal distance travelled by a projectile is called range
of projectile.
Derivation: Since range of projectile is horizontal distance y
covered by projectile. So,
X = vixt …… (1)
vi
X = Range of projectile = R
vix = vi cos θ θ x
R=x
2vi sin θ
and t =
g
So eq. (1) becomes:
2
vi cos θ × 2vi sin θ vi 2 sin θ cos θ
R = g = g
2
vi sin 2θ
R = (∴ 2 sin cos θ = sin 2θ )
g
3.13 At what point or points in its path does a projectile have its minimum speed, its maximum
speed?
Ans. The projectile has the maximum speed at the point of projection y
and just before reaching the ground. Its minimum at the highest
point. Because, speed of a projectile change due to its vertical
vmin
component vfy.
∴ vfy is minimum at the highest point, and vfy has maximum θ
magnitude at the point of projection and a point just before vmax vmax
x
reaching the ground.
3.14 Each of the following questions is followed by four answers, one of which is correct answer.
Identified that answer.
(i) What is meant by a ballistic trajectory?
(a) The path followed by an un-powered and unguided projectile is called ballistic
trajectory.
(b) The path followed by the powered and unguided projectile is called ballistic
trajectory.
(c) The path followed by un-powered but guided projectile.
(d) The path followed by powered and guided projectile.
130 Physics Intermediate Part-I

(ii) What happens when two-body system undergoes elastic collision?

(a) The momentum of the system changes.
(b) The momentum of the system does not change.
(c) The bodies come to rest after collision.
(d) The energy conservation law is violated.
Ans. (i) (a) is correct.
(ii) (b) is correct.

SIDE INFORMATION
3.15 A car accelerates along a road which force actually moves the car?
Ans. When a car moves on road, several forces acting on it are:
(a) Force ‘F’ exerted by engine in the forward direction.
(b) Friction between road and tyre ‘f1’.
(c) Air friction ‘f2’.
Therefore, the resultant of these forces will accelerate the car.
Net force = F − (f1 + f2)
3.16 Does a moving object have impulse?
Ans. A moving object may or may not have impulse because
I = ∆p
If momentum does not change impulse remain zero.
3.17 Which will be more effective in knocking a bear down?
(a) Rubber bullet (b) Lead bullet of same momentum
Ans. The rubber bullet will be more effective in knocking a bear down as impulse is equal to change
in momentum.
i.e., I = ∆p
Hence rubber bullet will be more effective in knocking a bear down, because change in
momentum of rubber bullet is greater than that of lead bullet.
3.18 (a) What is effect on speed of a fighter plane chasing another when it opens fire?
(b) What happens to the speed of pursued plane when it returns the fire?
Ans. (a) When the chasing plane opens fire, than its speed decreases. According 3rd law of motion
firing is an action its reaction will be in opposite direction which reduces the speed of plane.
(b) When pursued plane opens back fire it will get reaction in forward direction. Hence speed
of pursued plane increases.
3.20 Do objects of different masses fall with same acceleration on earth? (Neglect air resistance)
Ans. Yes, objects fall with the same acceleration regardless of their masses at the surface of earth, by
neglecting air resistance.
[Chapter-3] Motion and Force 131
3.21 Why do we wear seat belts?
Ans. When a moving car stops, the passengers move forward to maintain their inertia. Seat belts
provide an unbalanced force to prevent the passengers from moving and thus avoiding injury.
3.22 Why do we wear helmet?
Ans. A motor cyclist wear helmet so as to extend the time of any collision (impact) to prevent serious
injury.
→ →
I = F ×t
3.23 Identify the answer:
(i) What is meant by ballistic trajectory?
(ii) What happens when two body system undergoes elastic collision?
Ans. (i) The path followed by an un-powered and un-guided projectile is called ballistic trajectory.
(ii) The momentum of the system does not change. K.E also remains same.

EXTRA SHORT QUESTIONS

Q. Can action and reaction forces cancel each other?
Ans. No, it is true that action and reaction forces are equal and opposite but they cannot cancel each
other. It is because they act on different bodies. For this reason, when you are considering the
forces on a body, only one of them (action or reaction) will be involved.
Q. Where do we apply the concept of impulse?
Ans. Often we wish to apply the concept of impulse (i.e., change in linear momentum) to situations
where the applied force in not constant for example. Consider the case that a bat strikes the ball.
In this case, it is fairly easy to measure the change in linear momentum. Therefore, impulse can
be evaluated even though the average force and the time of contact may be extremely difficult to
determine.
Q. What are position dependent forces?
Ans. Forces between two bodies which depend on the position of the bodies are called position
dependent forces e.g. gravitational forces, electric force etc.
Q. The linear momentum of a system of particles is always conserved is it true or false?
Ans. No, it is false. It is because linear momentum of a system is conserved only if the net external
force on the system is zero.
Q. Body is dropped freely from the window of a train. Will the time of the free fall be equal, if
(i) the train is stationary (ii) the train moves with a constant velocity (iii) the train moves
with an acceleration?
Ans. In all the three cases, the time of free fall of the body will be the same. It is because the motion
of train affects only the horizontal motion and not motion along the vertical. For vertically
downward motion the parameters (g and h) are the same in the three cases.
132 Physics Intermediate Part-I

Q. Person sitting in a moving train throws a ball vertically upwards. How does the ball appear
to move to an observer (i) inside the train (ii) outside the train?
Ans. (i) To the observer sitting inside the train, the ball will appear to move straight vertically
upwards and then down wards. It is because in this case the ball has only one velocity
acting vertically.
(ii) To an observer outside the train the ball will appear to move along the parabolic path.
Because in this case the ball will be acted upon by the vertical as well as horizontal
velocities.
Q. If the displacement of a body is zero, is the distance covered by it necessarily zero? Comment.
Ans. No, the distance covered by the body may or may not be zero. For example, when a body
moving in a circle of radius r completes one revolution. The displacement of the body is zero but
distance travelled by it is 2πr.
Q. Explain that rest and motion are always relative.
Ans. Rest and motion are relative terms. For Example, a book on the table is at rest w.r.t table and
other objects in the room. If an observer is located on the moon. He will observe that the book
and other objects in the room are moving. Thus the book is at rest if view from the room but is
moving if viewed from the moon.
[Chapter-3] Motion and Force 133

PROBLEMS WITH SOLUTIONS

PROBLEM 3.1
A helicopter is ascending vertically at the rate of 19.6 ms−1. When it is at a height of 156.8 m
above the ground, a stone is dropped. How long does the stone take to reach the ground?
Data:
Initial velocity = vi = 19.6 ms−1
Weight = h = 156.8 m
Required:
Time = t = ?
Calculations:
t = t1 + t2 …… (1) B
t1
For t1: h1
A
Using 1st equation of motion:
vf = vi + at …… (2)
h2
Where, vf = 0 ms−1 h t2

vi = 19.6 ms−1
a = −g = −9.8 m/s2 C
t = t1 From A → B
So eq. (2) becomes: Stone moves upward initially
just because of inertia.
(0) = (19.6) − (9.8)(t1)
9.8 t1 = 19.6
19.6
t1 = 9.8

t1 = 2 seconds

For t2:
By 3rd equation of motion:
2 2
2aS = vf − vi
Here, a = −g = 9.8 m/s2
S = h1
vf = 0 ms−1
vi = 19.6 ms−1
134 Physics Intermediate Part-I

Putting the values:

2(−9.8)(h1) = (0)2 − (19.6)2
−(19.6)h1 = −384.16
19.6 h1 = 384.16
384.16
h1 =
19.6

h1 = 19.6 m
As, h2 = h + h1
h2 = 19.6 + 156.8
h2 = 176.4 m
1
So, S = vit + 2 at2

S = h2 = 176.4 m
vi = 0 m/s
a = g = 9.8 m/s2
t = t2
1
h2 = 0 + 2 (9.8)(t2)2

176.4 = 4.9 (t2)2

(t2)2 = 36
t2 = 36
t2 = 6 seconds
So, (1) becomes:
t = t1 + t2
t = (2 + 6) seconds
t = 8 seconds

Result:
The stone will take 8 seconds to reach to the ground.
(ALTERNATE SOLUTION)
Data:
Initial velocity = vi = 19.6 ms−1
Weight = h = 156.8 m
Required:
Time = t = ?
[Chapter-3] Motion and Force 135
Calculations:
1
As s = vit + at2 …… (1)
2
Here, s = −h = −156.8 m
a = −g = −9.8 ms−2
vi = 19.6 ms−1
So, equation (1) becomes:
1
−156.8 = (−19.6)t − (9.8)t2
2
−32 = 4t − t2
t2 − 4t − 32 = 0
t2 − 8t + 4t − 32 = 0
t(t − 8) + 4(t − 8) = 0
(t + 4)(t − 8) = 0
⇒ t+4 = 0 or t = −4s
⇒ t−8 = 0 or t = 8s
As time cannot be negative.
So, t = 8s

PROBLEM 3.2
Using the following data, draw a velocity-time graph for a short journey on a straight road
of a motorbike.
Velocity (ms−1) 0 10 20 20 20 20 0
Time (s) 0 30 60 90 120 150 180
Use the graph to calculate
(a) The initial acceleration
(b) The final acceleration
(c) The total distance travelled by the motorcyclist.
Data:
C
B a=0 D
20 vi = 20 ms−1
Velocity (ms−1)

15
ai af
10
vf = 0 ms−1
5
C E F
A
0 30 60 90 120 150 180
ti t1
Time (s)
136 Physics Intermediate Part-I

Required:
(a) Initial acceleration ai = ?
(b) Final acceleration af = ?
(c) Distance travelled S = ?
Calculations:
For ai:
∆v
ai =
∆t
vf − vi
ai =
tf − ti
Putting the values we get:
20 − 0
ai =
60 − 0
20
ai = 60

ai = 0.33 ms−2
For af:
∆v
af =
∆t
vf − vi
af =
tf − ti
0 − 20
af =
180 − 150
−20
af = 30

af = −0.67 ms−2
(Negative sign shows that acceleration is decreasing / retarding).
For Distance:
S = Area of ∆ABC + Area of rectangle BDEC + Area of ∆DEF
1 1
S = 2 (B)(H) + (B)(H) + 2 (B)(H)

Putting the values:

1 1
S = 2 (60)(20) + (90)(20) + 2 (30)(20)

S = (30)(20) + 1800 + (30)(10)

S = 600 + 1800 + 300
S = 2700 m or 2.7 km
[Chapter-3] Motion and Force 137
Result:
Initial acceleration = 0.33 ms−2
Final acceleration = −0.67 ms−2
Distance = 2700 m = 2.7 km

PROBLEM 3.3
A proton moving with speed of 1.0 × 107 ms−1 passes through a 0.02 cm thick sheet of paper
and emerges with a speed of 2.0 × 106 ms−1. Assuming uniform deceleration, find retardation and
time taken to pass through the paper.
Data:
Initial velocity = vi = 1 × 107 ms−1
Distance = S = 0.020 cm
= 0.020 × 10−2 m
Final velocity = vf = 2 × 106 ms−1
Required:
(a) Acceleration = a = ?
(b) Time = t = ?
Calculations:
For a:
By 3rd equation of motion:
2 2
2aS = vf − vi
2 2
vf − vi
a =
2S
(2 × 106)2 − (1 × 107)2
a =
2(0.020 × 10−2)
a = −2.4 × 1017 ms−2
a = −24 × 1016 ms−2
For t:
vf = vi + at
vf − vi
t =
a
(2 × 106) − (1 × 107)
t =
2.4 × 1017
t = 3.33 × 10−11 seconds

Result:
(a) Acceleration = a = −2.4 × 1017 ms−2
(b) Time = t = 3.33 × 10−11 seconds
138 Physics Intermediate Part-I

PROBLEM 3.4
Two masses m1 and m2 are initially at rest with a spring compressed between them. What is
the magnitude of ratio of their velocities after the spring has been released?
Data:
v1 v1 v2
= ?
v2
Let m1 and m2 are the masses attached with two ends of m1 m2
the spring.
Calculations:
By law of conservation of momentum:
(Initial momentum) = (Final momentum)
As system was initially at rest, so
m1(0) + m2(0) = m1v1 + m2v2
0 = m1v1 + m2v2
−m1v1 = m2v2
v1 m2
= −m
v2 1

For magnitude, we ignore negative sign. Therefore,

v1 m2
v2 = m
1

1
Velocity ∝
Mass

PROBLEM 3.5
An amoeba of mass 1.0 × 10−12 kg propels itself through water by blowing a jet of water
through a tiny orifice. The amoeba ejects water with a speed of 1.0 × 10−4 ms−1 and at a rate of
1.0 × 10−13 kgs−1. Assume that the water is being continuously replenished so that the mass of the
amoeba remains the same.
(a) If there were no force on amoeba other than the reaction force caused by the
emerging jet, what would be the acceleration of the amoeba?
(b) If amoeba moves with constant velocity through water, what is force of surrounding
water (exclusively of jet) on the amoeba?
Data:
Mass of amoeba = m = 1 × 10−12 kg
m′
Mass flow rate = = 1 × 10−13 kgs−1
t
Velocity = v = 1 × 10−4 ms−1
[Chapter-3] Motion and Force 139
Required:
(a) Force = ?
(b) Acceleration = ?

Calculations:
For F:
m′
F = t ×v

F = (1 × 10−13)(1 × 10−4)
F = 1 × 10−17 N
For a:
By 2nd law of motion:
F = ma
F
a = m

1 × 10−17
a =
1 × 10−12

a = 1 × 10−5 ms−2

Result:
(a) Force = F = 1 × 10−17 N
(b) Acceleration = a = 1 × 10−5 m/s2

PROBLEM 3.6
A boy places a fire cracker of negligible mass in an empty can of 40 g mass. He plugs the
end with a wooden block of mass 200 g. After igniting the firecracker, he throws the can straight
up. It explodes at the top of its path. If the block shoots out with a speed of 3 ms−1, how fast will
the can be going?

Data:
Mass of can = m1 = 40 g = 40 × 10−3 kg
Mass of block = m2 = 200 g = 200 × 10−3 kg
Speed of block = v2 = 3 ms−1

Required:
Speed of can = v1 = ?
140 Physics Intermediate Part-I

Calculations:
By law of conservation of momentum:
(Initial momentum) = (Final momentum)
The initial momentum of system is zero. So,
0 = m1v1 + m2v2
−m1v1 = m2v2
m2v2
v1 =
−m1
−(200 × 10−3)(3)
v1 =
40 × 10−3

v1 = −15 ms−1

Negative sign shows that can will move with a velocity opposite to that of the block.

Result:
Velocity of the can = v1 = 15 ms−1

PROBLEM 3.7
An electron (m = 9.1 × 10−31 kg) traveling at 2.0 × 107 ms−1 undergoes a head on collision
with a hydrogen atom (m = 1.67 × 10−27 kg) which is initially at rest. Assuming the collision to be
perfectly elastic and a motion to be along a straight line, find the velocity of hydrogen atom?

Data:
Mass of electron = m1 = 9.1 × 10−31 kg
Velocity = v1 = 2 × 107 ms−1
Mass of hydrogen = m2 = 1.67 × 10−27 kg
Velocity = v2 = 0 ms−1

Required:
Velocity of hydrogen atom after collision = v2′ = ?

Calculations:
Using formula:
2m1v1 v2(m2 − m1)
v2 ′ = m +m + m +m
1 2 1 2

2m1(2 × 107) (0)(m2 − m1)

v2 ′ =
m1 + m2 + m1 + m2
[Chapter-3] Motion and Force 141
2m1(2 × 107)
v2 ′ = m1 + m2 + 0
2(9.1 × 10−31)(2 × 107)
v2 ′ =
(9.1 × 10−31) + (1.67 × 10−27)
36.4 × 10−24
v2 ′ =
1.67091 × 10−27

v2 ′ = 2.2 × 104 ms−1

Result:
The velocity of H-atom after collision is 2.2 × 104 ms−1.

PROBLEM 3.8
A truck weighing 2500 kg and moving with a velocity of 21 ms−1 collides with a stationary
car weighing 1000 kg. The truck and the car move together after the impact. Calculate their
common velocity.
Data:
Mass of truck = m1 = 2500 kg
Mass of car = m2 = 1000 kg
Velocity of truck = v1 = 21 ms−1
Velocity of car = v2 = 0 ms−1
Required:
Combined velocity after collision = v = ?
Calculations:
By law of conservation of momentum:
(Initial momentum) = (Final momentum)
m1v1 + m2v2 = m1v1′ + m2v2′
As, v1 ′ = v2 ′ = v
∴ m1v1 + m2v2 = m1v + m2v
m1v1 + m2v2 = v(m1 + m2)
m1v1 + m2v2
v =
m1 + m2
(2500 × 21) + (1000 × 0)
v =
2500 + 1000

v = 15 ms−1

Result:
The combined velocity after collision is 15 ms−1.
142 Physics Intermediate Part-I

PROBLEM 3.9
Two blocks of masses 2.0 kg and 0.5 kg are attached at the two ends of a compressed
spring. The elastic potential energy stored in the spring in 10 J. Find the velocities of the blocks if
the spring delivers its energy to the blocks when released.
Data:
P.E = 10 J v1 v2

m1 = 2 kg
m1 m2
m2 = 0.5 kg
Required:
v1 = ?
v2 = ?
Calculations:
By law of conservation of momentum:
(Initial momentum) = (Final momentum)
As the system was initially at rest:
∴ m1(0) + m2(0) = m1v1 + m2v2
m1v1 = −m2v2
−m1v1
v2 = m2
−(2)v1
v2 =
0.5
v2 = −4v1 …… (1)
By law of conservation of energy:
Kinetic energy = Potential energy
∆K.E = ∆P.E
1 2 1 2
2 m1v1 + 2 m2v2
= 10

1 2 2
2 (m1v1 + m2v2)
= 10
2 2
m1v1 + m2v2 = 10 × 2
2 2
m1v1 + m2v2 = 20
2(v1)2 + (0.5)(−4v1)2 = 20
2 2
2v1 + 8v1 = 20
2 20
10v1 = 10
[Chapter-3] Motion and Force 143
2
v1 = 2 ms−1

v1 = 2 ms−1 or v1 = 1.4 m/s

Putting in eq. (1)

v2 = −4( 2)

v2 = −5.7 ms−1

Negative sign shows that the two masses with move in opposite direction.
Result:
v1 = 2 ms−1 = 1.4 ms−1
v2 = −5.7 ms−1

PROBLEM 3.10
A football is thrown upward with an angle of 30o with respect to the horizontal. To throw a
40 m pass what must be the initial speed of the ball?
Data:
θ = 30°
R = 40 m
Required:
vi = ?
Calculations:
2
vi sin 2θ
As, R = g
2
gR = vi sin 2θ
2 gR
vi =
sin 2θ
gR
vi =
sin 2θ
(9.8)(40)
vi =
sin 2(30)
392
vi =
0.866

vi = 21 ms−1

Result:
The initial velocity is 21 ms−1.
144 Physics Intermediate Part-I

PROBLEM 3.11
A ball is thrown horizontally from a height of 10 m with velocity of 21 ms−1. How far off it
hit the ground and with what velocity?
y
Data:
h = 10 m
vix = 21 ms−1
Required: h

(a) x = ?
(b) vf = ? x
x
Calculations:
For x:
x = vixt …… (1)
For t:
Using 2nd equation of motion:
1
S = vit + 2 at2 …… (2)

S = h = 10 m
vi = viy = 0 m/s
a = ay = g = 9.8 m/s2
So, (2) becomes:
1
(10) = (0)(t) + 2 (9.8)(t2)

10 = 4.9t2
10
t2 = 4.9

t2 = 2.04
t = 2.04
t = 1.430 s
Now (1) becomes:
x = (21)(1.43)
x = 30.03 m
x = 30 m
For vf:
2 2
vf = vfx + vfy …… (3)
As, vfx = vix = 21 ms−1
[Chapter-3] Motion and Force 145
For vfy:
vfy = viy gt
vfy = 0 ₊ (9.8)(1.43)
vfy = 14.01 m/s
So, eq. (3) becomes:

vf = (21)2 + (−14.01)2

vf = 25 ms−1

Result:
x = 30 m
vf = 25 ms−1

PROBLEM 3.12
A bomber dropped a bomb at a height of 490 m when its velocity along the horizontal was
300 kmh−1.
(a) At what distance from the point vertically below the bomber at the instant the bomb
was dropped, did it strike the ground?
(b) How long was it in air?
Data:
h = 490 m
vix = 300 kmh−1
300 × 103
=
3600
vix = 83.33 ms−1
Required:
(a) x = ?
(b) t = ?
Calculations:
For t:
1
S = vit + 2 at2

S = h = 490 m
vi = viy = 0 m/s
a = ay = g = 9.8 ms−2
146 Physics Intermediate Part-I

So, 490 = 4.9t2

490
t2 = 4.9

t = 10 seconds
For x:
x = vixt
x = (83.33)(10)
x = 833 m

Result:
(a) x = 833 m
(b) t = 10 seconds

PROBLEM 3.13
Find the angle of projection of a projectile for which its maximum height and horizontal
range are equal.
Data:
θ = ?
H = R
Calculations:
2
vi sin2 θ
As, H = 2g …… (1)
2
vi sin 2θ
and R = …… (2)
g
By given condition:
H = R
2 2
vi sin2 θ vi sin 2θ
2g = g
2 2
vi sin2 θ vi (2 sin θ cos θ)
=
2g g
sin θ 2 cos θ
2g = g
sin θ
2 = 2 cos θ

sin θ = 4 cos θ
sin θ
= 4
cos θ
[Chapter-3] Motion and Force 147
sin θ
Q = tan θ
cos θ
So, tan θ = 4
θ = tan−1 (4)
θ = 76°
Result:
Angle θ = 76°
PROBLEM 3.14
Prove that for angles of projection, which exceed or fall short of 45o by equal amounts, the
ranges are equal.
Show that:
R = R′
Let φ = 15°
For R:
2
vi sin 2θ
As, R = g …… (1)

Now, θ = 45° + 15° = 60°

So, eq. (1) becomes:
2
vi sin 2(60°)
R =
g
2
vi sin (120°)
R =
g
2
vi
R = (0.866) …… (A)
g
For R′′:
2
vi sin 2θ′
R′ = g …… (2)

θ′ = 45° − 15° = 30°

So, eq. (2) becomes:
2
vi sin 2(30°)
R′ = g
2
vi sin (60°)
R′ = g
2
vi
R′ = g (0.866) …… (B)

Comparing eq. (A) and (B):

R = R′
148 Physics Intermediate Part-I

PROBLEM 3.15
A SLBM (submarine launched ballistic missile) is fired from a distance of 3000 km. If the
Earth were flat and the angle of launch is 45o with horizontal, find the time taken by SLBM to hit
the target and the velocity with which the missile is fired.
Data:
R = 3000 km
R = 3 × 106 m
θ = 45°
Required:
vi = ?
t = ?
Calculations:
For vi:
Using formula:
2
vi sin 2θ
R = g
2
vi sin 2(45)
R =
g
2
vi sin (90)
R =
g
2
vi (1)
R = g
2
vi
R =
g
2
vi = Rg
vi = R×g
vi = 9.8 × 3 × 106
vi = 5.42 × 103 ms−1
vi = 5.42 kms−1
For t:
2vi sin θ
t = g
2(5.42 × 103) sin (45)
t =
9.8
2
t = 7.8 × 10 s
7.8 × 102
t = 60
t = 13 min.
Result:
(a) vi = 5.42 kms−1
(b) t = 13 min.
[Chapter-3] Motion and Force 149

MULTIPLE CHOICE QUESTIONS

Note: You have four choices for each objective type question as a, b, c and d. Tick the
correct choice.
1. The rate of change of distance is called:
(a) Velocity (b) Displacement (c) Acceleration (d) Speed
 →
∆ v 
2. The expression Limit   represent:
∆t → 0  ∆t 

(a) Average velocity (b) Variable velocity

(c) Instantaneous velocity (d) Instantaneous acceleration
3. The area under the curve of velocity-time graph give:
(a) Acceleration (b) Displacement (c) Distance (d) Direction
4. Inertia of a body is measured in terms of its:
(a) Weight (b) Force (c) Mass (d) Acceleration
5. Distance covered by a freely falling object in 2 seconds is:
(a) 20 m (b) 25 m (c) 24.5 m (d) 34.5 m
6. When total energy and momentum remain conserved then the collision will be:
(a) Elastics collision (b) In-elastic collision (c) Both (a) and (b) (d) None of these
7. An ball is dropped from a height of 10 m. its velocity at the moment its touches the ground is:
(a) 10.0 m/sec. (b) 14.0 m/sec. (c) 19.6 m/sec. (d) 19.8 m/sec.
8. A body is thrown vertically with initial velocity of 10 m sec−1. It will reach the height of:
(a) 3.0 m (b) 5.0 m (c) 10.0 m (d) 100 m
9. If the acceleration of a body is negative, then slope of the velocity-time graph will be:
(a) Zero (b) Positive (c) Negative (d) Infinity
10. The instantaneous and average velocity are equal:
(a) If a body moves with uniform velocity (b) If a body moves with variable velocity
(c) If a body moves with increasing velocity (d) If a body moves with increasing acceleration
11. When a bullet is fired from a gun, the gun moves backward with a:
(a) Velocity equal to that of bullet (b) Velocity less than that of bullet
(c) Velocity greater than of bullet (d) None of these
12. At the highest point, we can claim that:
(a) Resultant velocity is zero (b) Only horizontal component of velocity is zero
(c) Only vy is zero (d) Nothing of above
150 Physics Intermediate Part-I

13. The force which might be enough to fracture the naked skull is:
(a) 50 N (b) 10 N (c) 15 N (d) 5N
14. If the displacement of a particle is zero, then the distance covered:
(a) must be zero (b) cannot be zero
(c) is negative (d) may or may not be zero
15. A body moves 4 m towards east and then 3 m north. The displacement and distance
covered by the body are:
(a) 7 m, 6m (b) 6 m, 5 m (c) 5 m, 7 m (d) 4 m, 3 m
16. A moving body is covering the distance directly proportional to the square of the time. the
acceleration of a body is:
(a) Increasing (b) Decreasing (c) Zero (d) Constant
17. Which of the following changes when a particle is moving with uniform velocity?
(a) Speed (b) Velocity (c) Acceleration (d) Position vector
18. A moving particle finally comes to rest. What will be angle between acceleration and
displacement during motion?
π π
(a) (b) (c) π (d) 0
2 4
19. When the velocity of the body is constant then its velocity time graph is:
(a) Horizontal straight line (b) Curve
(c) Sinusoidal (d) None of these
20. Action and reaction forces act on the .
(a) Different bodies (b) Same bodies (c) Both (a) and (b) (d) None of these
21. A projectile is thrown an angle of 45° with horizontal. If K is the K.E with which the
projectile was thrown, then the K.E at the top of the trajectory is:
K K
(a) 2K (b) 4K (c) 2 (d)
4
22. The water flows out from pipe at 1 kg s−1 and its velocity changes from 10 ms−1 to zero on
striking the wall, then force exerted by water on wall is:
(a) 5N (b) 10 N (c) 20 N (d) Zero
23. Two projectiles P and Q are projected at an angle of 20° and 70° respectively with
horizontal . Which one will reach earlier to ground?
(a) P (b) Q
(c) Both will reach at same time (d) None of these
24. During projectile motion, the quantities that remain unchanged are:
(a) Force and vertical component of velocity
(b) Acceleration and horizontal component of velocity
(c) Kinetic energy and acceleration
(d) Acceleration and momentum
[Chapter-3] Motion and Force 151
25. A shell in flight explodes into four unequal parts. Which of the following is conserved?
(a) Momentum and K.E. (b) Momentum and total energy
(c) K.E (d) Neither momentum nor K.E
26. A projectile will cover same horizontal range when the initial angles of projection are:
(a) 30°, 40° (b) 30 , 45° (c) 30°, 60° (d) 45°, 60°
27. If distance of a particle is zero, the displacement:
(a) must be zero (b) may or may not be zero
(c) cannot be zero (d) depends upon the practical
28. The numerical value of the ratio of distance to displacement is:
(a) always less than 1 (b) equal or more than one
(c) always more than 1 (d) equal to or less than one
29. The numerical value of the ratio of velocity to speed is:
(a) always less than 1 (b) equal or more than one
(c) always more than 1 (d) equal to or less than one
30. A force of 10 N acts on a body of mass 1 kg for 2 sec. to a distance of 10 m, the rate of
change of momentum is:
(a) 10 N (b) 20 N (c) 50 N (d) 100 N
31. A fighter plane is chasing another plane, when its open fire its speed:
(a) Increases (b) Decreases (c) Remains same (d) None of these
32. A car covers its journey in two halves the first half distance at a speed of 40 km h−1 and the
 2v1v2 
other half at 60 km h−1 then its average speed is: Hint: Vav = 
 v1 + v2
(a) 45 km h−1 (b) 48 km h−1 (c) 40 km h−1 (d) None of these
33. During the projectile motion, the horizontal component of velocity:
(a) decreases with time (b) becomes zero
(c) remains constant (d) increases with time
34. Newton first law of motion gives definition of:
(a) Mass (b) Force (c) Acceleration (d) Speed
35. Motion of rocket is based upon:
(a) Newton’s thirds law of motion (b) Law of conservation of momentum
(c) Newton’s law of gravitation (d) Both (a) and (b)
36. A force “F” gives an acceleration “a” to an object of mass “m”. For the same force if mass
is doubled than acceleration produced will be:
(a) Double (b) Half (c) Quartered (d) Remain same
152 Physics Intermediate Part-I

37. Change in the momentum is equal to:

(a) Force (b) Pressure (c) Impulse (d) Tension
38. The angle of projection for which the maximum height and the horizontal range of a
projectile are equal is:
(a) 45° (b) 60° (c) 76° (d) 83°
39. A body is dropped from a tower which reaches ground in 2 second, the height of the tower
is about:
(a) 10 m (b) 20 m (c) 40 m (d) 60 m
40. The angle of projection to cover maximum horizontal distance is:
(a) 90° (b) 70° (c) 60° (d) 45°
41. The ballistic missile are useful only for:
(a) Long range (b) Short range (c) Both (a) and (b) (d) None of these
42. For long range, the missile used are called:
(a) Unguided missile (b) Ballistic missiles (c) Guided missiles (d) Un-powered missiles
43. Velocity of projectile at maximum height is .
(a) Zero (b) Minimum (c) Maximum (d) None of these
44. Ballistic trajectory is the path followed by:
(a) Powered and unguided missile (b) Un-powered an guided
(c) Un-powered an unguided missile (d) Powered and guided
45. When angle of projection increases beyond 45°°:
(a) Range and height both decreases (b) Range decreases but height increases
(c) Range increases but height decreases (d) None of these
46. A ballistic flight is under:
(a) Inertia (b) Gravity (c) Both (a) and (b) (d) None of these
47. The laws of motion are not valid in a system which is:
(a) Moving with uniform velocity (b) At rest
(c) Isolated (d) Non-inertial
48. The famous book “Principia” is written by:
(a) Newton (b) Galileo (c) Einstein (d) Joule
49. Direction of impulse is same that of:
(a) Change in momentum (b) Velocity
(c) Acceleration (d) No direction
50) Which bullet of same momentum is more effective in knocking a bear down?
(a) Rubber bullet (b) Lead bullet
(c) Both are equally effective (d) None of these
[Chapter-3] Motion and Force 153
51. The motor cycle’s safety helmet prevent the serious injury due to padding because it:
(a) Extends the time of impact (b) Increases impulsive force
(c) Decreases the impulsive force (d) Both (a) and (c)
52. A mass of fuel consumed by a typical rocket to overcome earth’s gravity is:
(a) 10000 kg s−1 (b) 1000 kg s−1 (c) 100 kg s−1 (d) 10 kg s−1
53. A typical rocket ejects the burnt gases at speeds of over:
(a) 400 ms−1 (b) 4000 ms−1 (c) 4 km s−1 (d) Both (b) and (c)
54. The acceleration of a rocket  as it gains height.
(a) Increases (b) Decreases
(c) Remains constant (d) Becomes equal to “g”
55. For the projectile motion angle of projection must satisfy the condition:
(a) 0° ≤ θ ≤ 90° (b) 0° ≥ θ ≥ 90° (c) 0° < θ < 90° (d) 0° ≤ θ ≤ 180°
56. An object is projected with velocity v at an angle θ. The angle φ of vf at any instant P is:
(a) greater than θ (b) less than θ
(c) equal to θ (d) may be lesser or greater to θ
57. Range of projectile is half of its maximum range at:
(a) 45° (b) 30° (c) 22.5° (d) 15°
58. Two masses m1 and m2 are initially at rest with a spring compressed between them. What is
the ratio of magnitude of their velocities after the spring has been released?
(a) m1 : m2 (b) m2 : m1 (c) m1 : m2 (d) m2 : m1
59. Impulse of body is  when it moves from A to B with a speed 100 m/s provided
that A and B are 10 m apart.
(a) 0 Ns (b) 100 Ns (c) 10 Ns (d) 1000 Ns
60. If the time of flight of projectile is doubled, than the maximum height attained by the
 gT2
projectile becomes: Hint: H =
 2 
(a) 2 times (b) 4 times (c) 6 times (d) 8 times
61. At the highest point during the projectile motion the angle between velocity and
acceleration is:
(a) 90° (b) 60° (c) 45° (d) 0°
154 Physics Intermediate Part-I

ANSWERS
Sr. Ans. Sr. Ans. Sr. Ans. Sr. Ans. Sr. Ans.
1. (d) 2. (d) 3. (c) 4. (c) 5. (a)
6. (c) 7. (b) 8. (b) 9. (c) 10. (a)
11. (b) 12. (c) 13. (d) 14. (d) 15. (c)
16. (d) 17. (d) 18. (c) 19. (a) 20. (a)
21. (c) 22. (b) 23. (a) 24. (b) 25. (b)
26. (c) 27. (a) 28. (b) 29. (d) 30. (a)
31. (b) 32. (b) 33. (c) 34. (b) 35. (d)
36. (b) 37. (c) 38. (c) 39. (b) 40. (d)
41. (b) 42. (c) 43. (b) 44. (c) 45. (b)
46. (c) 47. (d) 48. (a) 49. (a) 50. (a)
51. (d) 52. (a) 53. (d) 54. (a) 55. (c)
56. (b) 57. (d) 58. (b) 59. (a) 60. (b)
61. (a)