Class IX Physics
Force & Laws of motion
Force
It is the force that enables us to do any work.
To do anything, either we pull or push the object. Therefore, pull or push is called force.
Example – to open a door, either we push or pull it. A drawer is pulled to open and pushed to
close.
Effect of Force:
Force can make a stationary body in motion. For example a football can be set to move by
kicking it, i.e. by applying a force.
Force can stop a moving body – For example by applying brakes, a running cycle or a running
vehicle can be stopped.
Force can change the direction of a moving object. For example; By applying force, i.e. by
moving handle the direction of a running bicycle can be changed. Similarly by moving steering
the direction of a running vehicle is changed.
Force can change the speed of a moving body – By accelerating, the speed of a running vehicle
can be increased or by applying brakes the speed of a running vehicle can be decreased.
Force can change the shape and size of an object. For example -– By hammering, a block of
metal can be turned into a thin sheet. By hammering a stone can be broken into pieces.
Forces are mainly of two types:
1. Balanced Forces
2. Unbalanced Forces
Balanced Forces
If the resultant of applied forces is equal to zero, it is called balanced forces.
Example : - In the tug of war if both the teams apply similar magnitude of forces in opposite
directions, rope does not move in either side. This happens because of balanced forces in
which resultant of applied forces become zero.
Balanced forces do not cause any change of state of an object. Balanced forces are equal in
magnitude and opposite in direction.
Balanced forces can change the shape and size of an object. For example - When forces are
applied from both sides over a balloon, the size and shape of balloon is changed.
Unbalanced Forces
If the resultant of applied forces are greater than zero the forces are called unbalanced forces.
An object in rest can be moved because of applying balanced forces.
Unbalanced forces can do the following:
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Move a stationary object.
Increase the speed of a moving object.
Decrease the speed of a moving object.
Stop a moving object.
Change the shape and size of an object.
Laws of Motion:
Galileo Galileo: Galileo first of all said that object move with a constant speed when no forces
act on them. This means if an object is moving on a frictionless path and no other force is acting
upon it, the object would be moving forever. That is there is no unbalanced force working on the
object.
But practically it is not possible for any object. Because to attain the condition of zero
unbalanced force is impossible. Force of friction, force air and many other forces always acting
upon an object.
Newton’s Laws of Motion:
Newton studied the ideas of Galileo and gave the three laws of motion. These laws are known
as Newton’s Laws of Motion.

Newton's First Law of Motion - Any object remains in the state of rest or in uniform
motion along a straight line, until it is compelled to change the state by applying external
force.
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Newton's Second Law of Motion - The rate of change of momentum is directly
proportional to the force applied in the direction of force.

Newton's Third Law of Motion - There is an equal and opposite reaction for evrey action
Newton’s First Law of Motion:
Any object remains in the state of rest or in uniform motion along a straight line, until it is
compelled to change the state by applying external force.
Explanation: If any object is in the state of rest, then it will remain in rest untill a exernal force is
applied to change its state. Similarly an object will remain in motion untill any exeternal force is
applied over it to change its state. This means all objects resist to in changing their state. The
state of any object can be changed by applying external forces only.
Newton’s First Law of Motion in Everyday Life:
a. A person standing in a bus falls backward when bus is start moving suddenly. This
happens because the person and bus both are in rest while bus is not moving, but as the
bus starts moving the legs of the person start moving along with bus but rest portion of
his body has tendency to remain in rest. Because of this person falls backward; if he is
not alert.
b. A person standing in a moving bus falls forward if driver applies brakes suddenly. This
happens because when bus is moving, the person standing in it is also in motion along
with bus. But when driver applies brakes the speed of bus decreases suddenly or bus
comes in the state of rest suddenly, in this condition the legs of the person which are in
the contact with bus come in rest while the rest parts of his body have tendency to
remain in motion. Because of this person falls forward if he is not alert.
c. Before hanging the wet clothes over laundry line, usually many jerks are given to the
cloths to get them dried quickly. Because of jerks droplets of water from the pores of the
cloth falls on the ground and reduced amount of water in clothes dried them quickly. This
happens because, when suddenly cloth are made in motion by giving jerks, the water
droplets in it have tendency to remain in rest and they are separated from cloths and fall
on the ground.
d. When the pile of coin on the carom-board hit by a striker; coin only at the bottom moves
away leaving rest of the pile of coin at same place. This happens because when the pile
is struck with a striker, the coin at the bottom comes in motion while rest of the coin in
the pile has tendency to remain in the rest and they vertically falls the carom board and
remain at same place.
e. Seat belts are used in car and other vehicles, to prevent the passengers being thrown in
the condition of sudden braking or other emergency. In the condition of sudden braking
of the vehicles or any other emergency such as accident, the speed of vehicle would
decrease or vehicle may stop suddenly, in that condition passengers may be thrown in
the direction of the motion of vehicle because of the tendency to remain in the state of
motion.
f.
The head of hammer is tightened on a wooden handle by banging the handle against a
hard surface. When handle of the hammer is struck against a surface, handle comes in
rest while hammer over its head has tendency to remain in motion and thus after some
jerks it tightens over the handle.
Mass and Inertia:
The property of an object because of which it resists to get disturbed its state is called Inertia.
Inertia of an object is measured by its mass. Intertia is directly proportional to the mass. This
means inertia increases with increase in mass and decreases with decrease in mass. A heavy
object will have more inertia than lighter one.
In other words, the natural tendecny of an object that resists the change in state of motion or
rest of the boject is called intertia.
Since a heavy object has more intertia, thus it is difficult to push or pul a heavy box over the
ground than lighter one.
Momentum
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Before discussing about second law of motion we shall first learn about momentum of a
moving object.
From our daily life experiences like during the game of table tennis if the ball hits a
player it does not hurt him. On the other hand, when a fast moving cricket ball hits a
spectator, it may hurt him.
This suggests that impact produced by moving objects depends on both their mass and
velocity.
So, there appears to exist some quantity of importance that combines the object's mass
and its velocity called momentum and was introduced by Newton.
Momentum can be defined as "mass in motion". All objects have mass; so if an object is
moving, then it has momentum - it has its mass in motion.
The momentum, p of an object is defined as the product of its mass, m and velocity, v.
That is, Momentum p=mv
(1)
Momentum has both direction and magnitude so it is a vector quantity. Its direction is the
same as that of velocity, v.
The SI unit of momentum is kilogram-meter per second (kg m s-1).
Since the application of an unbalanced force brings a change in the velocity of the
object, it is therefore clear that a force also produces a change of momentum.
We define the momentum at the start of the time interval is the initial momentum and at
the end of the time interval is the final momentum.
When the object moves then it gains momentum as the velocity increases. Hence
greater the velocity greater is the momentum.
Q1: Define inertia.
Answer: Inertia is the resistance of a body to change in its state of rest or motion. It is an inherent
property of the body. The property of inertia is due to the mass of the body. Greater the mass, higher will
be the inertia.
There are three kinds of inertia:
1.
Intertie of rest
2.
Inertia of motion
3.
Inertia of direction.
(Some physicists consider inertia of direction is part of inertia of motion).
Q2: Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
(b) A bicycle and a train?
(c) A five-rupee coin and a one-rupee coin?
Answer: The property of inertia is due to the mass of the body. Greater the mass, higher will be the
inertia.
(a) The stone will have more inertia because it has more mass than the ball.
(b) Mass of a train is more than the mass of a bicycle, thus the train will have more inertia.
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, the five rupee coin has greater
inertia than the one-rupee coin.
Q3: State Newton's First Law of motion.
Answer: Newton's First Law of motion states if a body is in state of rest, it will remain in the state of rest
and if it is in the state of motion it will remain in the state of the motion with same velocity and same
direction unless an external force is applied on it.
Q4: State why Newton's first law of motion is called law of inertia.
Answer: Inertia is a tendency of the object to resist change in its state. Newton's first law of motion also
states similar i.e. the object will remain its present state unless an external force is applied. That's why
Newton's first law is called Law of inertia.
Q5: Justify Newton's first law gives notion (or definition) about force.
Answer: According to Newton's first law of motion, an object tends to continue it is present state unless
and external force is applied to change its state. For example, when a ball is rolled over a glass surface, it
stops after covering a distance. It is because the frictional forces between the ball and the glass are being
applied here. If there is frictionless surface in a vacuum chamber, the ball will continue to run and cover a
longer distance. It implies the external force (frictional force in this case) is almost absent and the ball
continues to move its present state.
Thus Newton's first law provides a notion about force.
Q6: In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the
goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of
his own team”.
Also identify the agent supplying the force in each case.
Answer: We need to apply first law of motion and identify what force is applied which changes the
velocity (state i.e. rest, direction, motion) of the ball. A force when applied can produce acceleration in
body.
Event
A football player kicks a
football
to another player of his
team who
kicks the football towards
the
goal.
The goalkeeper of the
opposite team
collects the football
and kicks it towards a
player of
his own team
Agent
football
player
force and Action
push force, ball moves
from rest.
another
player
force, changes direction
of ball.
force to stop ball
goal keeper (ball velocity is zero
now)
force moves ball from
rest to
goalkeeper
motion in opposite
direction.
Q7: What a force can do?
Answer: A force is a push or pull which produces acceleration in the body on which it acts. A force acting
on a body can cause:
1.
it can alter the speed (or velocity) of the moving object.
2.
it can change the direction of motion of a body.
3.
it can change the shape of an object
Q8: Explain why some of the leaves may get detached from a tree if we vigorously shake its
branch.
Answer: Some leaves of a tree get detached when we shake its branches vigorously isdue to inertia of
rest. When the tree moves to and fro, leaves tend to remain in the rest state and resist the change. Due
to this, some leaves gets detached and fall down.
Q9: Study the following scenarios and identify what type of inertia tends to resist the change.
(a) When a bus starts suddenly, the bus passengers standing in the bus tend to fall backwards.
(b) When a passenger jumps out of moving bus, he falls down.
(c) A cyclist on a levelled road does not come to rest immediately even he stops pedalling.
(d) When the playing card is flicked with the finger the coin placed over it falls in the tumbler.
(e) An athlete often jumps before taking a long jump.
(f) When a car enters a curved path, the car passengers tend to tilt outwards.
Answer:
(a) Inertia of rest
(b) Inertia of motion.
(c) Inertia of motion.
(d) Inertia of rest.
(e) Inertia of motion.
(f) Inertia of direction.
Q10: Why do you fall in the forward direction when a moving bus brakes to a stop and fall
backwards when it accelerates from rest?
Answer: Inertia tend to resist the change in state. When a moving bus brakes to stop, our body is in
inertia of motion state. It tends to oppose the change of state of rest and we fall forward.
Similarly when the bus is at rest and starts, our body is in inertia of rest state. It opposes the forward
motion of the bus and hence we fall backwards.
Q11: Is force required to keep a moving object in motion (yes/no)?
Answer: No.
Q12: Define momentum.
Answer: Momentum is defined as the product of its mass and velocity.
i.e. Momentum (P) = mass(m) × velocity(v).
Momentum is a measure of the quantity of motion of a body. It is a vector quantity. It has both direction
and magnitude. and its direction is same as the direction of velocity. The SI unit of momentum is kg-m/s.
Q13: A body of mass 25 kg has momentum of 125 kg m/s. What is the velocity of the body?
Answer: Since P = m × v
⇒ v = P / m = 125 / 25 = 5m/s
Q14: Why is it easier to stop a tennis ball in comparison to a cricket ball moving with the same
speed ?
Answer: Tennis ball is lighter than (less mass) than a cricket ball. Tennis ball moving with same speed
has less momentum ( mass × velocity) than a cricket ball. ∴ It is easier to stop tennis ball having less
momentum.
Q15: Which is having higher value of momentum ? A bullet of mass 10 g moving with a velocity of
400 m/s or a cricket ball of mass 400 g thrown with the speed of 90 km/hr.
Answer: Momentum (P) = mass(m) × velocity(v).
mass of bullet = 10g = 10 × 10-3kg = 10-2kg
velocity of bullet = 400 m/s
momentum of bullet = 10-2kg × 400 m/s = 4 kg m/s
mass of cricket ball = 400g = 400 × 10-3kg = 0.400 kg
velocity of ball = 90 km/hr = 90 × 1000m / 3600s = 25m/s
momentum of ball = 0.400 × 25 = 10kg m/s
∴ The cricket ball has higher momentum.
Following questions asked in CBSE examination (2010) are similar to Q14 and Q15.
Q: Two similar vehicles are moving with the same velocity on the roads such that one of them is loaded
and the other one is empty. Which of the two vehicles will require larger force to stop it. Give reasons?
Q: Which one has greater inertia : a stone of mass 1 kg or a stone of mass 5 kg ?
Q16: The term 'mass' is analogous to physical quantity (a) Weight
(b) Intertia
(c) Force
(d) Acceleration
Answer: (b) Intertia
Q17: What are balanced and unbalanced forces?
Answer: If a set of forces acting on a body does not change the state of rest or of motion of an object
such forces are called balanced forces. In this case object is said to be inequilibrium state. The vector
sum of all forces is zero i.e. ∑F x = 0.
If set of forces acting on body results in change in state i.e. either changes the speed or direction, such
forces are called unbalanced forces. In this case the object is in non-equilibrium state and the resultant
force is non-zero. i.e. ∑Fx ≠ 0.
Q18: The object shown below moves with a constant velocity. Two forces are acting on the object.
Considering negligible friction, the resultant force will be:
(a)
(b)
(c)
(d)
17N leftwards
10N leftwards
3N leftwards
7N rightwards
Answer: (c) 3N leftwards.
Q19: If the set of forces acting on an object are balanced, then object
(a) must be at rest.
(b) must be moving
(c) must not be accelerating.
(d) none of these.
Answer: (c) must not be accelerating. If the forces are balanced, the object will remain in its present state
i.e. if it is moving, it will keep on moving with uniform speed. If it is at rest, it will remain at rest.
Q20(NCERT): An object experiences a net zero external unbalanced force. Is it possible for the
object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on
the magnitude and direction of the velocity. If no, provide a reason.
Answer: According to Newto's first law, in order to change the state of motion of a body, an external
unbalanced force must be applied.
If an object is moving with a uniform speed and the object experiences a net zeroexternal unbalanced
force. It implies the object is in equilibrium state and it will remain in its present state i.e. it will keeps on
moving with the same speed and direction.
Q21: When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer: Due to inertia of rest. When a carpet is beaten with a stick, dust particles tend to remain in inertia
of rest and resist the change. Hence, the dust particles come out of the carpet.
Q22: Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: While the bus is moving, luggage tends to remain in inertia of motion state. When the bus stops,
the luggage tends to resist the change and due to inertia of motion it moves forward and may fall off.
That's why it is advised to tie any luggage kept on the roof of a bus with a rope.
Q23: State Newtons second law of motion.
Answer: The second law of motion states that the rate of change of momentum of an object is
proportional to the applied unbalanced force in the direction of force.
Mathematically, the law tells that force is the product of mass and acceleration.
i.e. F = ma
where m = mass of the object, a = acceleration and
F is the net force (i.e. F = ∑Fx = F1 + F2 + ... + Fn)
Q24: Which of the following graph represents unbalanced force.
Answer: An unbalanced force results in acceleration. Figure (b) represents this condition.
In fig. (a), v-t graph represents uniform motion ⇒ acceleration (a) = 0.
Fig. b, v-t graph shows the velocity is not uniform, it is decreasing ⇒ acceleration is non-zero.
Fig. c is a Distance-time graph which shows uniform motion ( a straight line).
Q25: How Newton's Second law of motion is different from First law?
Answer: Newton’s first law of motion deals with the behaviour of objects on which all existing forces are
balanced. It gives us notion of external force required to change the inertia state of the object.
While the second law of motion deals with the behaviour of objects on which all existing forces are not
balanced. It gives us a measure of the force.
Q26: Name the categories of forces based on interaction.
Answer: Force exists only as a result of an interaction of two objects. Based on it, forces are categorised
as:
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contact forces and
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non-contact-forces
Q27: What are contact forces? Give examples
Answer: Contact forces are forces in which the two interacting objects are physically in contact with each
other. Examples of contact forces are:
1.
Tension force
2.
Normal force
3.
Air resistance
4.
Buoyant force
5.
Friction force
Q28: What are non-contact forces? Give examples.
Answer: Non-contact forces are forces in which the two interacting objects are not in physical contact
which each other, but are able to exert a push or pull. These are also called Action-at-a-distance forces.
Examples are:
1.
Gravitational force
2.
Magnetic Forces
3.
Electric Force
4.
Nuclear Force
Q29: Using second law of motion, derive the relation between force and acceleration.
Answer: Let an object of mass, m is moving along a straight line with an initial velocity, u.
Let the final velocity = v m/s
Acceleration = a ms-2
Time taken = t s
and Force applied = F.
Initial momentum (before the force applied) = P1 = mu
Final momentum (after the force applied) = P2 = mv
Change in momentum ∝ P2 - P1
and Rate of Change in momentum ∝ (P2 - P1)/t
⇒
∝ (mv - mu)/t
∝ m(v - u)/t
According to Newton's second law, Rate of Change in momentum = F
⇒ F ∝ m(v - u)/t
⇒ F ∝ ma
(∵ v = u + at)
⇒ F = kma
(where k is a constant of proportionality).
The SI units for m is 1 kg, a is ms-2 , thus k = 1
⇒ F = ma
SI unit of force is Newton (N) = kg-ms-2
Q30: Define 1N (Newton).
Answer: One Newton is the force acts on a body of mass 1 kg and it produces an acceleration of
1 ms-2.
⇒ 1N = 1kg × 1 ms-2
Q31: What is the unit of force in c.g.s system?
Answer: dyne (g cm s-2)
Q32: What's the relation between 1N and 1 dyne?
Answer: 1 N = 1kg × 1 ms-2
and 1 dyne = 1g × 1 cms-2
1 N = 1kg × 1 ms-2 = 1000g × 100 cms-2
⇒ 1N = 105 dyne
Q33: Is Newton's second law of motion is consistent with the first law? How?
Answer: Yes Newton's second law is consistent with first law. The first law of motion can be
mathematically stated from the mathematical expression for the second law of motion.
Since F = ma = m(v - u)/t
or Ft = mv - mu.
⇒ F = 0 when v = u for any given time t.
⇒ The object will continue moving with uniform velocity, u throughout the time, t.
Similarly when object is at rest, u = v = 0. ⇒ F = 0. No external force is being applied, the object will
remain at rest.
Q34: A batsman hits a cricket ball which then rolls on a level ground. After covering a short
distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer: (c) there is a force on the ball opposing the motion.
Frictional force exerted by the ground opposes the motion due to which the ball comes to rest.
(Note: Frictional force always acts in the direction opposite to the direction of motion.)
Q35: A truck starts from rest and rolls down a hill with a constant acceleration. It travels a
distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric
tonnes. (Hint: 1 metric tonne = 1000 kg.)
Answer: Given, Truck starts from rest, initial velocity (u) = 0 m/s
Distance covered (S) = 400m
Time taken (t) = 20s
mass = 7 metric tonne = 7 × 1000 = 7000 kg
final velocity(v) = ?
acceleration (a) = ?
Force (F) = ?
Using equation of motion, S = ut + ½at2, acceleration (a) is
i.e. 400 m = 0 + ½(a)(20)2
⇒ 400 × 2 = a × 400
⇒ a = 2 m/s2
...(answer)
Using equation v = u + at
final velocity (v) = 0 + 2 × 2 = 4 m/s
...(answer)
Since Force (F) = mass(m) × acceleration (a)
⇒ F = 7000 kg × 2 = 14000 N
...(answer)
Q36: A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a
very short time of 0.01 s. What is the force of the nail on the hammer ?
Answer: Given, mass of hammer(m) = 500g = 0.5 kg
initial velocity of hammer(u) = 50 m/s
final velocity of hammer (v) = 0 m/s
Duration (t) = 0.01s
Using second law of motion, F = ma = m(v - u)/t
⇒ F = 0.5(0 - 50)/0.01 = (0.5 × -50)/0.01 = -2500N ... (answer)
The -ve sign shows the force of 2500N acts in the opposite direction of motion.
Q37: What is the momentum of an object of mass m, moving with a velocity v?(a) (mv)2
(b) mv2
(c) ½ mv2
(d) mv
Answer: (d) mv
Q38: A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and
comes to rest after travelling a distance of 50 m. What is the force of friction between the stone
and the ice? (Calculate time taken and acceleration also).
Answer: mass of stone (m) = 1kg
initial velocity of stone (u) = 20 m/s
final velocity of stone (v) = 0 m/s
Distance covered (S) = 50m
time taken (t) = ?
acceleration (a) = ?
Friction force between stone and ice (F) = ?
Using equation of motion, v2 - u2 = 2aS , we have
(0)2 - (20)2 = 2 × a × 50m
-400 = 100a
⇒ a = -4 m/s2
Negative sign indicates that it is retardation acting in the opposite side of motion.
Since Force (F) = mass(m) × acceleration (a)
F = 1kg × (-4m/s2)
⇒ F = -4 kg m/s2 = -4 N
The -ve sign shows the force acts in the opposite direction of motion.
using equation v = u + at
i.e. 0 = 20 - 4t
⇒ -4t = -20
⇒ t = 5s
Q39: An object of mass 5 kg is moving with a velocity 4 m/s. A constant force of 20 N acts on the
object. What will be the velocity after 3 s.
Answer: Give, mass (m) = 5 kg
Initial velocity (u) = 4 m/s
Force (F) = 20 N
Time taken (t) = 3 s
acceleration (a) = ?
final velocity (v) = ?
Since F = ma ⇒ a = F/m
∴ a = 20/5 = 4 m/s2
Using equation v = u + at
v = 4 + 4×3 = 4 + 12 = 16 m/s
Q40: While catching a fast moving ball, fielder gradually pulls his hand backwards. Give reasons.
Answer: By doing so, the fielder increases the time during which the high velocity of the moving ball
decreases to zero. The acceleration of the ball is decreased and therefore the impact of catching the fast
moving ball is also reduced.
We can prove it mathematically,
Let u be the initial velocity of ball before it reaches fielder hands. Let m be the mass of the ball.
Initial momentum of the ball = mu.
When the ball reaches field hands, it stops. ∴ final velocity (v) = 0.
And the final momentum = mv = 0
The ball will exert force (F) on fielder hands = Change in momentum / time(t 1)
In this case F1 = (0 - mu)/ t1 = -mu/t1
If fielder increases the time (say t2) i.e. t2 > t1 , the force will reduce
i.e. F2 = - mu/ t2 and F2 < F1
Q41: A man pushes a box of mass 50 kg with a force of 80 N. What will be the acceleration of the
box due to this force? What would be the acceleration if the mass were doubled ?
Answer: Given,
mass of the box (m) = 50kg
Force (F) acting on box = 80N
acceleration (a) = ?
Since Force (F) = mass(m) × acceleration (a)
⇒ a = F/m = 80/50 = 1.6 m/s2
If mass becomes m = 2 × 50 = 100 kg
acceleration (a) = 80 / 100 = 0.8 m/s2
Thus, the acceleration will be halved if the mass were doubled.
Q42: An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle
and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms -2.
Answer: Given, mass of vehicle (m) = 1500 kg
negative acceleration (a) = -1.7 m/s2
According to Newton's 2nd law of motion,
Force (F) = mass(m) × acceleration (a)
F = 1500 × -1.7 m/s2 = -2550N
∴ The negative sign indicates a force of 2550N is to be applied in the opposite direction of motion.
Q43: State Newton's third law of motion.
Answer: It states that to every action there is always an equal and opposite reaction.
When two objects A and B act on each other, the force exerted by A on B (F AB) is equal to the force
exerted by object B on A (FBA) in magnitude but are in opposite directions.
i.e. FAB = -FBA
or FAB + FBA = 0
One of the force (say FAB) can be said action force then other one is reaction force (i.e. F BA). Actionreaction forces always act on different bodies and their line of action is the same.
Q44: When two objects act on each other, does action-reaction force pair cancel out each other?
Answer: No. The action-reaction forces act on different bodies.
Q45: While driving on a highway, an insect strikes on the car windshield and splatters. Which
experiences greater impact (force): an insect or the windshield?
Answer: It is an example of third law of motion. Both insect and the car windshield experiences same
amount of force but in opposite direction. But the force is too great for the insect that its body splatters
while the windshield is able to withstand the impact.
Q46: If action is always equal to the reaction, explain how a horse can pull a cart.
Answer: According to Newton's third law of motion, action force is equal to reaction but acts on two
different bodies and in opposite directions. When a horse pushes the ground, the ground reacts and
exerts a force on the horse in the forward direction. The force is able to overcome friction force of the cart
and it moves.
(Note: A detailed explanation on house cart problem can be read here)
Q47: Identify the action and reaction forces in the following cases:
(a) A television (TV) lying on a table
(b) Pushing a wall with your hand
(c) Firing a bullet from a gun.
(d) Walking of a person on a ground.
Answer:
(a) A TV lying on a table: TV is interacting with table surface. Applying Newton's third law of motion,
action force acts on table due to TV (FTV-Table) and a reaction force acts on TV by the table i.e. (FTable-TV).
Both these forces are equal in magnitude but in opposite directions.
⇒ FTable-TV = - FTV-Table
(b) Pushing a wall with your hand: When you pushes a wall, you exert a force (action) on the wall.
Similarly, the wall exerts a (reaction) force on your hands in contact with the wall. According to Newton;s
third law of motion, both these forces form action-reaction pair and act on different bodies. Both are equal
in magnitude but opposite in direction.
(c) Firing a bullet from a gun: When a gun fires a bullet, a force is exerted (action) on a bullet and the
gun experiences an equal recoil (reaction) force in opposite direction.
(d) Walking of a person on a ground: When a person pushes the ground (action) backwards by his foot,
the ground also exerts an equal force (reaction) on the his foot in forward direction.
Q48: A TV set is lying on a table. The TV set experiences a gravitational force (pull) downwards
(FW) by the Earth. It also experiences a force on it due to table in contact i.e. (FTable-TV) in upward
direction. Do these forces form an action-reaction pair? Do these forces follow Newton's third Law
of motion?
Answer: No because these two forces are acting on the same body, these are not due to Newton's third
law of motion. These two forces do not form action-reaction pair.
In fact, Newton's first law of motion does apply. Since the TV set is at rest, because these forces are
balanced. i.e. FW = - FTable-TV
A similar question (as Q48) was asked in CBSE 2010 examination, hope you may answer it!
Q 49: According to the third law of motion when we push on an object, the object pushes back on
us with an equal and opposite force. If the object is a massive truck parked along the road side, it
will probably not move. A student justifies this by answering that the two opposite and equal
forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer: Two cases arise here:
Case I: No external force is applied. In this case, the truck weight (gravitational force) acting downwards,
is cancelled by the contact force applied on the truck by the ground.
i.e.
FWeight = - Fground
Case II: If a person or group of persons pushes truck, the static friction acting on the truck horizontally,
cancels this push-force. Static friction is a self-adjusting force. More push is applied, more static friction
will oppose it (upto a limit) and forces are balanced. In order move the truck, the push-force must
overcome static friction.
i.e. FWeight = - Fground
and FPush = - FFriction
Q50: Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at
a high velocity.
Answer: Large amount of water coming out from the nozzle of the hose at high velocity will have large
momentum. Due to a large momentum, water comes out exerting a large force. According to Newton's
third law of motion, the hose will also experience an equal reaction force but in opposite direction.
Because of this reaction force, it is difficult for a fireman to hold a hose.
Q51 (CBSE 2010): It is difficult to balance our body when we accidentally slip on a peel of banana.
Explain why?
Answer: A frictional force always acts parallel to the surface and is directed to oppose sliding. Banana
skin reduces friction (or frictional force) and thus brings body in unbalanced state and we tend to fall.
Q52(NCERT): Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor
at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer: A friction force (here it is kinetic or sliding friction) always acts parallel to surface but in direction
opposing motion. Since the wooden cabinet is moving with uniform speed and a horizontal force of 200N
acts on it. According to Newton's first law of motion, forces are balanced. Hence, a frictional force of 200
N is exerted on the cabinet.
Q53: State Law of conservation of momentum.
Answer: According to law of conservation of momentum, in the absence of external unbalanced force the
total momentum of a system of objects remains unchanged or conserved by collision.
i.e. Total
Initial Momentum Before Reaction = Total Initial Momentum After
Reaction
This law holds good for any number of objects.
Q54: Law of conservation of momentum is applicable to an isolated system. What do you mean by
an isolated system?
Answer: Isolated system refers to absence of external unbalanced force or net force is zero.
Q55: Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed
decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and
attains the speed v again before striking the ground. It implies that the magnitude of initial and
final momentums of the ball are same. Yet, it is not an example of conservation of momentum.
Explain why ?
Answer: The reason is because there exists external unbalanced force i.e. gravitation pull of the earth
due to which ball falls down and is accelerating. Therefore it is not an example of conservation of
momentum.
Q56: What is collision? What are the types of collision?
Answer: The interaction between two or more bodies causing the exchange of momentum is called
collision. It is categorised as of two types:

elastic collision (both momentum and kinetic energy are conserved)

inelastic collision (only momentum is conserved)
Q57: From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s.
Calculate the initial recoil velocity of the rifle.
Answer: Initially both rifle and bullet are at rest. Given,
mass of rifle (m 1) = 4 kg
initial velocity of rifle (u1) = 0 m/s
final velocity of rifle (v1) = ?
mass of bullet (m2) = 50g = 50 × 10-3 kg = 0.05 kg
initial velocity of bullet (u2) = 0 m/s
final velocity of bullet (v2) = 35 m/s
Applying law of conservation of momentum,
total initial momentum = total final momentum
i.e. (m1u1) + (m2u2) = (m1v1) + (m2v2)
⇒ 0 + 0 = (4v1) + (0.05 × 35)
⇒ - 4v1 = 0.05 × 35
⇒ v1 = - 0.4375 m/s
... (answer)
The negative sign indicates that rifle recoils in opposite direction to bullet i.e. backwards
Q58: Two objects of masses 100 g and 200 g are moving along the same line and direction with
velocities of 2 m s-1 and 1 m s-1, respectively. They collide and after the collision, the first object
moves at a velocity of 1.67 m s-1. Determine the velocity of the second object.
Answer:
mass of the first object (m 1) = 100g = 0.1 kg
velocity of the first object (u1) before collision = 2 m/s
velocity of the first object (v1) after collision = 1.67 m/s
mass of the second object (m 2) = 200g = 0.2 kg
velocity of the second object (u2) before collision = 1 m/s
velocity of the second object (v2) after collision = ? m/s
Since no unbalanced external force is acting on the system, total momentum is conserved.
i.e. total initial momentum = total final momentum
⇒ (m1u1) + (m2u2) = (m1v1) + (m2v2)
⇒ (0.1 × 2) + (0.2 × 1) = (0.1 ×1.67) + (0.2 × v2)
⇒ 0.2 + 0.2 = 0.167 + 0.2v2
⇒ 0.2v2 = 0.4 - 0.167 = 0.233
⇒ v2 = 0.233 / 0.2 = 1.165 m/s
Q59: Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite
directions. The velocity of each object is 2.5 m/s before the collision during which they stick
together. What will be the velocity of the combined object after collision?
Answer: Given
mass of the first object (m 1) = 1.5 kg
velocity of the first object (u1) before collision = 2.5 m/s
mass of the second object (m 2) = 1.5kg
velocity of the second object (u2) before collision = -2.5 m/s
The negative sign indicates the second object moves in direction opposite to the first one.
mass of the combined objects after collision (m 3) = m1+ m2= 3.0 kg
velocity of the combined objects after collision (v3) =?
Being an isolated system, total momentum is conserved i.e.
(m1u1) + (m2u2) = (m3v3)
⇒ (1.5 × 2.5) + (1.5 × -2.5) = 3.0 × v3
⇒ 0 = 3.0 × v3
⇒ v3 = 0 m/s
Q60: What is impulse?
Answer: The effect of force applied for a short duration is called impulse. It is the product of force (F) and
the time duration (t) for which the force is applied.
i.e. Impulse = F × t
Impulse in a vector quantity and its SI unit is N-s.
Since F = ma = m(v-u)/t
⇒ F × t = mv - mu
⇒ Impulse is equal to change in momentum.
Q61: A hockey ball of mass 200 g travelling at 10 m s-1 is struck by a hockey stick so as to return it
along its original path with a velocity at 5 m s-1. Calculate the change of momentum occurred in
the motion of the hockey ball by the force applied by the hockey stick.
Answer: Given, mass of ball = 200g = 0.2 kg
initial velocity of ball (u) = 10 m/s
final velocity of ball (v) = -5 m/s
(When ball struck by Hockey stick, it returns to its original position, means the direction of ball reverses.
Hence -ve sign of final velocity).
∴ initial momentum Pi = mu = 0.2 × 10 = 2 kg m/s
and final momentum (Pf) = mv = 0.2 × - 5 = -1 kg m/s
Change in momentum (ΔP) = Pf - Pi = -1 - 2 = -3 kg m/s
Momentum is a vector quantity. The -ve sign indicates the direction of momentum i.e. same as final
velocity.
Q62: Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an
expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran
started pondering over the situation.
(a) Kiran suggested that the insect suffered a greater change in momentum as compared to the
change in momentum of the motorcar (because the change in the velocity of the insect was much
more than that of the motorcar).
(b) Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force
on the insect. And as a result the insect died.
(c) Rahul while putting an entirely new explanation said that both the motorcar and the insect
experienced the same force and a change in their momentum.
Comment on these suggestions.
Answer: Let us consider the following scenario,
mass of the car (mc) = 1500 kg
speed of car before collision (uc) = 90 km/hr = 25 m/s
mass of the insect (mi) = 10g = 0.01 kg
speed of insect before collision (ui) = -5 m/s
(-ve sign indicates its direction is opposite to car's motion)
final speed of (car + insect) after collision (v) = ? m/s
Since both car and insect are moving with uniform speed, and there is not external force applied. Hence,
total momentum is conserved, i.e.
(mcuc) + (miui) = (mc+mi)v
⇒ (1500 × 25) + (0.01 × -5) = (1500 + 0.01) × v
⇒ 37500 - 0.05 = 1500.01 × v
⇒ 37499.95 = 1500.01 × v
⇒ v = 37499.95 / 1500.01= 24.9998 m/s
So velocity of insect changes from -5 m/s to 24.9998 m/s (big change)
Velocity of car changes from 25 m/s to 24.9998 m/s (almost a negligible change)
Change in momentum of the car = (m cv) - (mcuc) = mc(v - uc)
= 1500 × (24.9998 - 25) = - 0.3 N-s
Change in momentum of the insect = (miv) - (miui) = mi(v - ui)
= 0.01 × ( 24.9998 + 5) = 0.3 N-s
⇒ The momentum gained by the insect is equal to the momentum lost by the car.
(a) Kiran is incorrect in terms of change in momentum. However s/he is correct, the change in velocity in
insect is higher than that of the car.
(b) Akhtar observation is correct that speed of car is higher than that of insect. However he is incorrect
that the insect experiences larger force. According to Newton's third law of motion both experienced same
amount of force.
(c) Rahul is correct that both experienced same force. If he says the amount of change in momentum in
both object is same, then he is correct.
Q63: A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary
wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into
the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer: Given,
mass of the bullet (m) = 10g = 0.01 kg
initial velocity of the bullet (u) = 150 m/s
final velocity of the bullet (v) = 0 m/s
time taken (t) = 0.03s
Distance (S) = ? m
Force (F) = ? N
acceleration (a) = ? m/s2
Using equation v = u + at
0 = 150 + a × 0.03
⇒ a = -150/0.03 = -5000 m/s2
The -ve sign of acceleration indicates that it is retardation.
Using equation of motion v2 - u2 = 2aS, we have
(0)2 - (150)2 = 2 × (-5000) × (S)
- 22500 = -10000 × (S)
⇒ S = 22500/10000 = 2.25 m
Using Newton's second law of motion i.e. F = ma
F = 0.01 × (-5000) = -50N
The -ve sign of force indicates that the force opposes the motion.
Q64:A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the
engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer:
(a)Force generated by engine = 40000N
Friction force by track = -5000N
∴ Net force (Fnet) acting on wagons = 40000 - 50000 = 35000N
(b) acceleration of wagons = F /(total mass of wagons)
Note here we are assuming net force acting on wagons by engine and the mass of the engine is not
considered in the system.
Total mass of 5 wagons = 5 × 2000kg = 10000kg
acceleration (a) = 35000N / 10000kg = 3.5 m/s2
⇒ all wagons are moving with the same acceleration i.e. 3.5 m/s 2
(c) Consider the free body diagram of the first wagon
Two forces are acting on the first wagon and it is moving with acceleration 'a'
Fnet is the force applied by the engine.
T1 is the action-reaction force (Newton's third law) on wagon-1 due to wagon-2 . Same force T1 is also
acting on wagon-2 by the first wagon. Since there is acceleration,
⇒ Fnet - T1 = m1a
(where m1 is the mass of wagon-1)
⇒ T1 = Fnet - m1a = 35000 - (2000 × 3.5 ) = 35000 - 7000 = 28000 N
Note: If you consider engine part of the system, total mass = mass of engine + wagons = 18000N.
Acceleration (a) = 35000/18000 = 1.944 m/s2. Then engine and wagons will move with same acceleration.
In some books two accelerations are computed. In some books, two accelerations are computed. One
including engine plus wagons. And another for wagons separately, which I think is incorrect.
Q65: An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, and
sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the
same straight line. Calculate the total momentum just before the impact and just after the impact.
Also, calculate the velocity of the combined object.
Time in seconds
Distance in meters
0
1
2
3
4
5
6
7
0
1
8
27
64
125
216
343
Answer: Given, mass of object (m 1)
= 1 kg
initial velocity of object before
collision (u1) = 10 m/s
mass of wooden block (m 2) = 5kg
velocity of wooden block before
collision (u2) = 0 m/s
final velocity of object+wooden
black (v) after collision = ? m/s
Since momentum is conserved,
total initial momentum = total final
momentum
i.e. (m1u1) + (m2u2) = (m1+ m2)v
⇒ 1 × 10 + 0 = (1 + 5)v
⇒ v = 10/6 m/s = 5/3 m/s
Initial momentum = (m1u1) + (m2u2) = 1 × 10 + 0 = 10 kg m/s
Final momentum = (m1+ m2)v = (1 + 5) × (5/3) = 6 × (5/3) = 10 kg m/s
Q66: The following is the distance-time table of an object in motion:
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
Answer: (a) Since the distance travelled in in equal intervals of time is not equal rather it is increasing. It
means the object is moving with a non-uniform velocity. It is accelerating.
By looking at the table we find distance (S) ∝ t3
...(I)
Let us check if acceleration is constant. We know that for fixed 'a', equation S = ut + ½at2 is valid.
In this case, u = 0, (since at t = 0, S = 0), ⇒ S = ½at2 or (S) ∝ t2
... (II)
But from the table, we get S ∝ t3, we conclude, that 'a' is increasing.
(b) Since the object is in accelerated condition, According to Newton's Second Law, F ∝ a.
We can say, unbalanced force is acting on the object.
Q67: Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level
road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2m s -2.
With what force does each person push the motorcar?
(Assume that all persons push the motorcar with the same muscular effort.)
Answer: Let each person applies force = F N
Initially car is moving with uniform speed. When two persons apply force, there is no change in state, car
moves with same uniform speed. It means the force of the two persons is cancelled by the kinetic friction
of the car.
⇒ 2F + Fkinetic-friction = 0
... (I)
When the third person applies there occurs change in state, due to which acceleration (a) occurs.
⇒ 3F + Fkinetic-friction = ma
where m is the mass of the car = 1200 kg and a is acceleration = 0.2 m s -2.
Using eq. I, we get
F = ma = 1200 × 0.2 = 240N
∴ Each person applies 240N force.
Q68: A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90
km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the
acceleration and change in momentum. Also calculate the magnitude of the force required.
Answer: Given,
mass of the car (m) = 1200 kg
initial velocity of car (u) = 90 km/h = 90 × 1000m/3600s = 25 m/s
final velocity of car (v) = 18 km/h = 5 m/s
time taken (t) = 4s
acceleration (a) = ? m s-2.
Force applied (F) = ? N
Using equation of motion v = u + at
⇒ 5 = 25 + 4a
⇒ 4a = -25+ 5 = -20
⇒ a = -5 m s-2
The -ve sign indicates it is retardation.
Change in momentum = mv - mu = m(v - u) = 1200 × (5 - 25) = 1200 × -20 = -24000 kg m s-1.
According to 2nd law of motion, Force(F) = mass (m) × acceleration (a)
⇒ F = 1200 × -5 = -6000 N
-ve sign indicates a force of 6000N acts on car opposing its motion.
Q69: A spring scale reads 20 N as it pulls a 4.0 kg object across a table. What is the magnitude of
the force exerted by object on the spring scale?
(a) 40 N
(b) 20 N
(c) 4 N
(d) 5 N
Answer: (b) 20 N (Hint: Newton's 3rd Law Action-reaction force).
Q70: A boy throws a ball upwards with a speed of 10m/s. The mass of the ball is 400g. Calculate
(i) Weight of the ball.
(ii) Initial momentum of the ball.
(iii) How high the ball will go?
(iv) Momentum of the ball at its highest point.
(Take acceleration towards gravity (g) = 10 m/s2)
Answer: Given,
mass of the ball (m) = 400g = 0.4 kg
initial speed of the ball (u) = 10 m/s
When ball will reach at its highest point, its velocity will become zero and it will start to fall.
Velocity at the highest point (v) = 0 m/s
If we take upward direction positive(+ve), the downward direction will be taken as (-ve).
∴ acceleration due to gravity (g) = -10 m/s2
(a) Weight of the ball = mass × acceleration due to gravity(g) = 0.4 × -10 = -4N
-ve sign indicates weight (force) acts downwards.
(b) Initial momentum = mass(m) × initial velocity(u) = 0.4 × 10 = 4 kg m/s2
(c) Consider equation of motion, v2 - u2 = 2aS,
The height (h) will be,
(0)2 - (10)2 = 2 (-10)h
⇒ -100 = -20h
⇒ h = 100/20 = 5m
(d) Momentum at the highest point = mass (m) × final velocity(v) = 10 × 0 = 0 kg m/s2
Q71: Are there any limitations to Newton's Laws of motion?
Answer: Newton's laws of motion hold true for common practical applications. However it has certain
limitations:
1.
Second Law (F = ma) assumes mass is constant. When an object travels at the speed of
the light, its mass is also affected. Newton's law is not applicable in this case.
2.
Newton's Laws fail to explain motion of electrons around nucleus.
3.
These law also fail to explain about black holes and bending of gravity due to star light.
4.
It also fails to explain why a leaf falls in river stream, its path cannot be determined
(chaos theory)
Later these challenges were explained by Einstein's theory of relativity and quantum physics which
became the basis of modern physics.
Q72: Two objects having their masses in ratio 3:5 are acted upon by two forces each on one
object. The forces are in the ration of 5:3. Find the ratio in their accelerations.
Answer: Since masses are in ratio 3:5.
Let the mass of the objects be 3x and 5x.
Let F1 and F2are the two forces with a1 and a2 accelerations.
∴ F1 = m1a1 = 3xa1
and F2 = m2a2 = 5xa2
Since F1:F2 = 5:3, we have
3xa1:5xa2 :: 5:3
⇒ a1:a2 = 5×5:3×3
⇒ a1:a2 = 25:9
Q73: The speed-time graph of a car is shown below. The car weighs 1000kg.
(i) Find the distance covered by the in fits two seconds.
(ii) Find how much force is applied by the car brakes in the fifth second so that the car comes to
an halt by sixth second.
Answer: Given mass of the car(m) = 1000kg
(i) As shown in graph, Distance covered in first 2 secs = Area of Δ inscribed in first 2s.
= ½ × base × height = ½ × 2 × 15 = 15m
(ii) Force (F) = mass(m) × acceleration (a)
Time taken by force to stop the car = 6 - 5 = 1 second
Initial Velocity (i.e. velocity at B) = 15 m/s
Final velocity (at point C) = 0 m/s
Using equation v = u + at
a = (v - u)/t = (0 - 15)/1 = -15 m/s2 (-ve sign indicates it is retardation)
Force applied by brakes = 1000 × -15 = -15000N = -15KN
LAW OF CONSERVATION OF MOMENTUM
Consider two objects ( two balls A & B ) of mass m1 and m2 are travelling in the same
direction along a straight line at different velocities u1 & u2 .Let u1> u2 and the two balls
collide with each other . During collision which last for a time ‘t’ the ball exert a force F AB on ball
B and the ball B exerts a force F BA on ball A .Suppose v1 and v2 are the final velocities. Of the ball after
collision.
Momentum of body A
= m1u1
Momentum of body B
= m2 u2
Initial momentum of the system = m1u1+ m2 u2
Final momentum of the system = m1 v1+ m2v2
F AB =Rate of change of momentum of body B
= m1 ( v1-u1 )
-------(1)
t
F BA = Rate of change of momentum of body A
= m1 ( v1- u1)
t
since
(2)
ACTION = - REACTION ( Newton’s third law of motion )
F AB =
− − F BA
From 1 & 2
m1(v1-u1)
t
m1v1-m1u1
= -- m2 ( v2- u2)
t
= --m2 v2 + m2u2
m1v1 + m2v2
= m1u1 + m2 u2
total momentum before collision = total momentum after collision
This is the law of conservation of momentum.
According to law of conservation of momentum states that “the sum of momenta of the two objects before
collision is equal to the sum of momenta after after the collision provided there is no external unbalanced
force acting on them “.