~ KK PHY

Chapter -04: Dynamics (Laws of Motion)

Force
Force is a push or a pull which produces or tends to produce the motion in a body when it is at rest or stops
and tends to stop the body when it is in motion, increases or decreases the magnitude of velocity of the
moving body or changes the direction of motion of moving body.
Inertia
A body at rest wants to be at rest and a body in uniform motion wants to be in uniform motion. This
property of a body is called inertia. So, inertia is the tendency of a body to continue the state of rest or of
uniform motion until and unless an external force is applied.
Types of Inertia:
1) Inertia of Rest: The tendency of a body at rest to remain at the state of rest is called inertia of rest.
For example:
a) A passenger jerks backward when the bus suddenly moves because his legs come in the sates of motion with the
bus but his upper part of the body still continues the state of rest due to inertia of rest
b) When we hit a carpet with stick, dust particles are removed.
c) When we shake a mango tree, mangoes fall from the tree.
d) A coin placed on a cardboard just above a glass falls on the glass when the card board is suddenly pulled.
2) Inertia of Motion (uniform): The tendency of a body to continue the state of uniform motion is called
inertia of motion.
For example:
a) A passenger in a moving bus jerk forward when the bus suddenly stops because his legs come to the state of rest
as the bus stops his upper part of the body still continues the state of motion. So, he jerks forward.
b) It is dangerous to jump from a moving bus, why?
c) An athlete runs for some distance before taking a long jump, why?
d) A man jumping from a moving bus may fall down.
3) Inertia of direction
The inability of a body to change its direction of motion by itself is known as inertia of direction.
Example:
a) When a car rounds a curve suddenly, the person sitting inside is thrown outwards.
b) Rotating wheels of vehicle throw out mud, mudguard over the wheels stop this mud.
c) When a vehicle takes a sudden turn towards left, the person seated inside the vehicle is pushed towards the
right.
Linear momentum
It is defined as the product of mass of body and its velocity. It is denoted by P.
The momentum of a body of mass m moving with velocity v is given by
P=mv
It is a vector quantity and its unit is kg m s-1 or Ns
Newton’s Laws of Motion
Newton gave three laws about motion which are as:
a) First law of motion: It states that, “everybody in the universe continues in its state of rest or of
uniform motion in a straight line, if no external forces act on it”. It is also called law of inertia.
b) Second law of motion: It states that, the rate of change of linear momentum is directly proportional
to force and it takes place in the direction of force applied. If F be the force applied on a body, and dP
be the change of linear momentum, then according to Newton’s second law of motion,
F ∝ rate of change of linear momentum
dP
i.e. F ∝ dt
dP
or, F = k dt
where k is proportionality constant. Its value is k = 1

Page 1 of 34
 ~ KK PHY

dP
Then, F = dt . . . . (i)
d(mv)
or, F = dt
dv dm
or, F = m dt + v dt

dv
If v = constant, ( dt = 0), then

dm
F=v . . . . . . (ii)
dt
dm
Where represents the rate of change of mass.
dt
dm
If m = constant ( dt = 0), then

dv
F = m dt . . . . . . (iii)
or, F = ma
dv
where a = dt represents the rate of change of velocity i.e. acceleration.
Some Key Notes
▪ When n bullets each of mass m are fired per second with velocity v then force required to hold the
gun is
F = nmv
▪ Bullets hitting a wall
(i) Bullets come to rest in wall
Force on wall is Fwall = nmv
(ii) Bullets rebound elastically
Fwall = 2nmv
dm
▪ When sand is gently dropped on a horizontal conveyor belt at rate , then force required to
dt
dm
maintain its constant speed v is F = v
dt
▪ When a stone of mass m dropped from height h penetrates distance ‘S’ into sand, then average
resistance offered by the sand to the stone is
𝐡
F = mg (𝟏 + )
𝐒
c) Third law of motion:
It states that to every action, there is equal and opposite reaction i.e. the forces action and reaction
are always equal and opposite.
If FA be the force exerted by A on B (i.e. action) and FB be the force exerted by B on A (i.e. reaction). Then
according to Newton’s third law of motion,
FA = - FB
The force of action and reaction always act on different bodies and cannot be cancelled by each other.
Examples:
i) If a book is on a table, then its weight ‘W’ acts in vertically downward direction but normal reaction ‘R’
that the table gives in vertically upward direction.
∴ R = -W where –ve sign indicates that they act opposite to each other.
ii) Swimming
iii) Rocket launching
iv) Recoil of gum
v) In order to walk, we press the ground in backward direction with our foot.
vi) It is difficult to walk in sand or ice.

Page 2 of 34
 ~ KK PHY

Questions
i) Why does fan rotate sometimes after switch off?
ii) A man jumping from moving bus falls forward, why?
iii) When driver suddenly starts bus the passenger fall backward, why?
iv) Why does mango fall when we shake the mango tree?
v) Why does a gun recoil when a bullet is fired?
Second law is the real law of motion
Both the first law and the third law are contained in the second law. That’s why it is called real law of motion.
Newton’s 1st law from 2nd law.
We have from Newton’s second law of motion
F = ma
If no any external force is applied on the body i.e F = 0, then
ma = 0
Since m ≠ 0, So a = 0
Then v = constant
This means that, in the absence of force, a body at rest will remain at rest and the body moving with uniform speed
will continue its state of motion.
This verifies Newton’s first law of motion.
Newton’s 3rd law from 2nd law.
Consider two bodies A and B moving along same straight line. When they collide, their momentum will change.
Let ∆t be the time of impact.
The change in linear momentum of A is
∆PA = FB × ∆t . . . . . . (i) [∵ FB = Force exerted by A]
Change in linear momentum of B is
∆PB = FA × ∆t . . . . . . (ii) [∵ FB = Force exerted by B]
So, the total change in linear momentum of A and B is
∆P = ∆PA + ∆PB
or, ∆P = FB × ∆t + FA × ∆t
If no any external force is applied on the system, then ∆P = 0
or, FB × ∆t + FA × ∆t = 0
FB = - FA
This is Newton’s third law of motion.

Principle of conservation of linear Momentum

Statement: It states that if no external force is acting on the system of two colliding bodies, then the total
linear momentum of a system remains constant i.e. the total linear momentum before collision is equal to
the total linear momentum after collision.
If p1, p2, p3 …………pn be the linear momentum of n particles of an isolated system before collision and p1’, p2’
p3’ ……….pn’ be the linear momentum after collision then
P1 + p2 + p3 +…….. + pn = p1’+ p2’+p3’+………pn’
Proof:

Let two bodies A and B of masses m1 and m2 be moving in a straight line with initial velocities u1 and u2 such that u1
> u2. Suppose after some time they collide and after collision they move with velocities v1 and v2 in the same direction

Page 3 of 34
 ~ KK PHY

as shown in figure. Let t be the time of impact.

let FA be the force exerted by A on B then from Newton’s 2nd law of motion,
change in momentum of body B m 2 v2 - m 2 u2
FA = time = t …………..(i)
Similarly, FB be the force exerted by B on A then
change in momentum of body A m1 v1 - m1 u1
FB = time = t .................. (ii)
According to Newton's third law of motion, the forces exerted by A and B are equal and opposite,
i.e. FA = - FB .................. (iii)
m2 v2 - m2 u2 m1 v1 - m1 u1
or, = -
t t
or, m2v2 - m2 u2 = - (m1 v1 – m1 u1)
or, m2v2 - m2 u2 = - m1 v1 + m1 u1
or, m1 v1 + m2v2 = m1 u1 + m2 u2 .................. (iv)
⇒ m1 u1 + m2 u2 = m1 v1 + m2v2
i.e., Total linear momentum before collision = Total linear momentum after collision
Hence, principle of conservation of linear momentum is verified.
Numerical Problems
(1) A 40 kg shell is flying at a speed of 72 kmh-1. It explodes into two pieces. One of the two pieces of mass 15
kg stops. Calculate the speed of the other. (Ans:32m/s)
(2) A bullet of mass 7 g is fired into a block of metal weighing 7 kg. The block is free to move. After the impact,
the velocity of the bullet and the block is 70cms-1. What is the initial velocity of the bullet? (Ans:700.7 m/s)
(3) A man weighing 60 kg runs along the rails with a velocity of 18 kmh-1 and jumps into a car of mass 1 quintal
standing on the rails. Calculate the velocity with which the car will start travelling along the rails. (Ans:1.88
m/s)
(4) A machine gun of mass 10 kg fires 20 g bullets at the rate of 10 bullets per second with a speed of 500 ms-
1
. What force is required to hold the gun in position? (Ans. 100 N)
(5) A truck of mass 2×104kg travelling at 0.5 ms-1 collides with another truck of half its mass moving in the
opposite direction with a velocity of 0.4 ms-1, if the trucks couple automatically on collision, calculate the
common velocity with which they move. (Ans. 0.2 ms-1)
Impulse
If a force acts on a body for a very short interval of time, then the force is called impulsive force. The product of
impulsive force and time interval for which it acts is called impulse.
The impulsive force does not remain constant, but changes with time, first from zero to maximum and then from
maximum to zero.
Thus, impulse is measured by taking the product of average force applied and time for with the force acts.
i.e. Impulse = Fav × ∆t
From newton’s 2nd law of motion
dP
F = dt
or, F dt = dp
Integrating both sides,
𝑡 𝑃2
∫0 F dt = ∫𝑃1 𝑑𝑃 = P2 – P1
Where P1 and P2 be the initial and final momentum
𝑡
Here, ∫0 F dt represents the impulse. Then
i.e. Impulse = P2 – P1
Thus, impulse is numerically equal to the change in momentum.
Impulse is vector quantity and its direction is along the force applied. The SI unit of impulse is Ns or kg m s-1.
☞ The area under a force-time graph is equal to the impulse delivered by the force

Page 4 of 34
 ~ KK PHY

Some examples of impulse

(a) A cricketer lowers his hand while catching a cricket ball. In doing so, he increases the time of impact. As F
× t = change in momentum. When time increases, force exerted on his decreases so it huts less.
(b) A person falling on a cemented floor comes to rest in short time, so force will be greater and hence gets injured
more while a person falling on heap of sand gets injured less as there is increase in time of impact.
(c) China wares, glass wares are wrapped in paper before packing them.
(d) Cars, buses, trucks, bogies of the train etc. are provided with spring system so as to increase the time
interval of jerks. This decreases the rate of change of momentum of the passenger receiving the jerks.
Therefore, comparatively less force acts on the passengers during the jerks.

Numerical Problems
1) A body of mass 0.25 kg moving with velocity 12 m/s is stopped by applying a force of 0.6 N. Calculate the
time taken to stop the body. Also calculate the impulse of this force. (Ans:5 sec, 3 Ns)
2) A hammer weighing 1 kg moving with the speed of 10 ms-1 strikes the head of a nail driving it 10 cm into
a wall. Neglecting the mass of the nail, calculate (i) the acceleration during impact (ii) the time interval of
the impact and (iii) the impulse. (Ans:-500 ms-2, 0.02 sec, -10 Ns )
3) A ball moving with a momentum of 15 kg ms-1 strikes against the wall at an angle of 60° with the wall
and is reflected with the same momentum at the same angle. Calculate impulse. (Ans:15√3 Ns)
4) The figure shows a plot of the time dependent force Fx(t) acting on a particle in motion along the x-axis.
What is the total impulse delivered to the particle?

5) The force shown in the force vs. time diagram in figure acts on 1.5 kg object. Find (i) the impulse of
the force (ii) the final velocity of the object if it is initially at rest. (iii) the final velocity of the object if
it is initially moving along the x-axis with a velocity of 2 m/s (Ans:8 Ns, 5.33 m/s, 3.33 m/s)

6) A force acting on a body of mass 2 kg varies with time as shown in Fig. 5.11. Find (i) impulse of the
force and (ii) the final velocity of the body. (Ans:12 Ns, 6 m/s)

Numerical Problems
1. A man weighing 60 kg runs along the rail with velocity of 18 km/hr and jumps to a car mass 1 quintal
standing on the rails. Calculate the velocity with which the car starts travelling along rails. (Ans:1.88 m/s)
2. A bullet of mass 20 gm is fired horizontally into a suspended stationary wooden block of mass 380 gm with
a velocity of 200 ms-1. What is the common velocity of the bullet and block if the bullet is embedded the

Page 5 of 34
 ~ KK PHY

block? If the block and the bullet experience a constant opposing force of 2 N, Find the time taken by them
to come to rest. [3]
3. A hunter has a machine gun that can fire 25g bullets with a velocity of 200m/s. A 40kg tiger springs at
him with a velocity of 10m/s. How many bullets must the hunter fire into the tiger in order to stop him
in his tracks? [Ans: 80]
4. You’re stuck in a boat in the middle of a Phewa lake. Luckily, you
brought your physics book! You decide to use your book to propel you
back to the shore. You throw your 1 kg book overboard with a speed
of 10 m/s to propel ourself back towards the shore. Assume the
combined mass of you and the boat is 100 kg.
(i) How long would it take you to reach the shore after throwing your book? (Ignore friction
between the water and the boat.) The shore is 60 m away. [3]
(ii) Unfortunately, it starts raining as you float towards the shore. The rain falls straight down into
the boat. 10 kg of rainwater has accumulated at the bottom of your boat. What is your speed now?
5. a) State the principle of conservation of linear momentum. [2]
b) A ball of mass 0.40 kg travels horizontally and strikes a vertical wall with a
speed of 5.0 m s–1. It rebounds horizontally with a speed of 3.0 m s–1. The ball
is in contact with the wall for a time of 0.20 s. Calculate the magnitude of
average force exerted on the wall. [2]
6. The mass of the gas emitted from the rear of toy rocket is initially 0.2kg/s. If
the speed of the gas relative to the rocket is 40m/s and the mass of rocket is 4kg, what is the initial
acceleration of the rocket? (Ans:2ms-2)
7. A bullet of mass 20g is fired horizontally into a suspended stationary wooden block of mass 380g with a
velocity of 200m/s. What is the common velocity of the bullet and block if the bullet has emended in
the block? If the block and bullet experience an opposing force of 2N find the time taken by them to
come at rest. (Ans: 5ms-2, 2s)
8. A 30 kg shell is flying at 48 m s–1. When it explodes, its one part of 18 kg stops, while the remaining part
flies on. Find the velocity of the latter. (Ans: 120 m s–1)
9. A hammer weighting 1kg moving with the speed of 10ms-1 strikes the head of the head of a nail driving
it 10cm into a wall. Neglecting the mass of the nail, calculate i) the acceleration during impact ii) the
time interval during the impact and iii) the impulse. [Ans: -500ms-2, 0.02s, 10Ns]
10. A bullet of mass 20 g is fired horizontally into a suspended stationary wooden block of mass 380 g with a
velocity of 200ms–1. What is the common velocity of the bullet and block if the bullet has embedded in the
block? If the block and bullet experience an opposing force of 2N, find the time taken by them to come at
rest. [Ans.: 5 ms–2, 2s ]
11. A bullet of mass 10 g travelling horizontally with a velocity of 300ms–1 strikes a block of wood of mass
290g which rests on a rough horizontal floor. After the impact the block and bullet move together and comes
to rest when the block has travelled a distance of 15m. Calculate the coefficient of sliding friction between
the block and floor. [Ans.: 0.33]
12. A cricket ball of mass 145 gm is moving with a velocity of 14m/s and is being hit by a bat, so that the ball
is turned back with a velocity of 22m/s. The force of blow acts on the ball for 0.015 sec. Find the average
force exerted by the bat on the ball. [Ans.:348 N]
13. A vehicle having a mass of 500 kg is moving with a speed of 10 ms−1. Sand is dropped into it at the rate of
10 kg/min. What force is needed to keep the vehicle moving with a uniform peed? [Ans: F = 1.67N]

Page 6 of 34
 ~ KK PHY

Reaction on body by the surface in various cases:

1. When a block is rest on horizontal surface is pushed or pulled by a horizontal force F then ∴
R = mg
Hence reaction by the surface is equal to weight.

2. When a block is pulled by force F at an angle θ with horizontal surface.

R + F Sin θ = mg
∴ R = mg - F Sin θ

3. When a block is pushed at angle θ with horizontal surface.

∴ R = mg + F Sin θ

4. When a block rests on an inclined surface

∴ R = mg Cos θ

MCQ
1. A body of mass 10 kg is sliding on a frictionless surface with a velocity of 2 m/s. The force required to keep
it moving with a same velocity is
a. 10 N b. 5N c. 2.5 N d. zero
2. A body of mass 2 kg is moving with a velocity 8 m/s on a smooth surface. If it is to be brought to rest in 4
seconds, then the force to be applied is
a. 8 N b. 4 N c. 2 N d. 1 N
3. A 30 gm bullet initially travelling at 120 m/s penetrates 12 cm into a wooden block. The average resistance
exerted by the wooden block is
a. 2850 N b. 2200 N c. 2000 N d. 1800 N
4. A force of 100 dynes acts on mass of 5 gm for 10 sec. the velocity produced is
a. 2 cm/s b. 20 cm/s c. 200 cm/s d. 2000 cm/s
5. A force of 5 N acts on a body of weight 9.8 N. What is the acceleration produced in m/s2 is
a. 49 b. 5 c. 1.46 d. 0.51

Page 7 of 34
 ~ KK PHY

6. At a place where the acceleration due to gravity is 10 ms-2 a force of 5 kg-wt acts on a body of mass 10 kg
initially at rest. The velocity of the body after 4 seconds is
a. 5 ms-2 b. 10 ms-2 c. 20 ms-2 d. 50 ms-2
7. A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s. If the catching process be
completed in 0.1 sec, then the force of the blow exerted by the ball on the hands of the player is
a. 0.3 N b. 30 N c. 300 N d. 3000 N
8. Gravels are dropped on a conveyer belt at the rate of 0.5 kg/sec. The extra force required in newtons to keep
the belt moving at 2 m/sec is
a. 1 b. 2 c. 4 d. 0.5
9. Swimming is possible on account of
a. 1st law of motion b. 2nd law of motion c. 3rd law of motion d. Newton’s law of gravitation
10. A book is lying on an inclined plane having inclination to the horizontal θ°. What is the angle between the
weight of the book and the reaction of the plane on the book is
a. a. 0 ° b. θ ° c. 180 – θ d. 180 °
11. A cannon after firing recoils due to
a. conservation of energyb. backward thrust of gases produced
c. Newton’s 3rd law of motion d. Newton’s 1st law of motion
12. Newton’s 3rd law of motion leads to the conservation of
a. angular momentum b. energy c. mass d. momentum
13. when a horse pulls a wagon, the force that causes the horse to move forward is the force
a. he exerts on the wagon b. the wagon exerts on him
c. he exerts on the ground d. the ground exerts on him
14. The action and reaction forces referred in Newton’s 3rd law of motion
a. must act on the same body
b. must act on different bodies
c. need not equal in magnitude but must have same line of action
d. must be equal in magnitude but need not have the same line of action
15. A body, whose momentum is constant, must have constant
a. force b. velocity c. acceleration d. all of above
16. A man fires a bullet of mass 200 gm at a speed of 5 m/s. The gun is of 1 kg mass, by how much velocity
the gun rebounds backwards
a. 0.1 m/s b. 10 m/s c. 1 m/s d. 0.01 m/s
17. A gun of mass 1 kg fires a bullet of mass 1 gm with a velocity of 1 m/s. The recoil velocity of the gun is
a. 1 m/s b. 0.1 m/s c. 0.01 m/s d. 0.001 m/s
18. A bomb at rest explodes into a large number of tiny fragments. The total momentum of all the fragments
a. is zero b. depends on the total mass of all the fragments.
c. depends on the speed of various fragments d. is infinity
19. A rocket is ejecting 50 gm of gases per sec at a speed of 500 m/s. The accelerating force on the rocket will
be
a. 125 N b. 25 N c. 5 N d. zero
20. Rocket works on the principle of conservation of
a. mass b. energy c. momentum d. none of the above

Page 8 of 34