NCERT Solutions for Class 9 Physics : Force and Laws of Motion

Class 9 is the first stepping stone for a student in the competitive world. With the introduction of the CBSE Board Exam for class 10 a few years back, this has become an important gateway for a student. Based on the results of class 9th a student selects his future stream of Science, Commerce or Arts suiting his interest.

Takshila Learning is providing NCERT Solutions for Class 9 Physics as per the latest syllabus by CBSE. Class 9 is the building block for the CBSE Class 10 Board Exams, not only for your exams but also for your higher studies and career. Physics is the most essential subject and the knowledge in this field opens up wider career opportunities for the students.

Below you can find the NCERT solution for Class 9th Physics. You can get a Solution for the all-important question of “Motion”

Important Formulae

(1) v=u+at

(2) s= ut +1/2at²

(3) v²=v²+at

(4)momentum=mass × velocity

(5)force=mass ×acceleration

(6) acceleration = change in velocity /time take

(7)speed = distance /time

(8)velocity = displacement /time

 

Pg 118

 

Q1. 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-rupees coin and a one-rupee coin?

 

Solution:

Inertia is the measure of the mass of the body. The greater is that the mass of the body; the greater is its inertia and vice-versa.

 

(a) Mass of a stone is more than the mass of a rubber ball for the identical size. Hence, inertia of the stone is more than that of a rubber ball.

 

(b) Mass of a train is more the mass of a bicycle. Hence, inertia of the train is larger than that of the bicycle.

 

(c) Mass of a five rupee coin is more that of a one-rupee coin. Hence, inertia of the five rupee coin is more than that of the one-rupee coin.

 

Q2. 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.

 

Solution:

The velocity of the ball changes four times.

 

As a participant kicks the football, its speed changes from zero to a particular value. As a result, the speed of the ball gets changed. During this case, the player applied a force to alter the speed of the ball. Another player kicks the ball towards the goal post. As a result, the direction of the ball gets changed. As a result, its velocity also changes. During this case, the player applied a force to alter the speed of the ball. The goalkeeper collects the ball. In other words, the ball reaches to rest. Thus, its speed reduces to zero from a particular value. The speed of the ball has changed. During this case, the goalkeeper applied an opposite force to stop/change the speed of the ball. The goalkeeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a particular value. Hence, its velocity changes once more. During this case, the goalkeeper applied a force to alter the speed of the ball.

 

 

Q3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

 

Solution:

Some leaves of a tree get detached after we shake its branches vigorously. This is often because when the branches of a tree are shaken, it moves to and fro, but its leaves tend to stay at rest. This is often because the inertia of the leaves tend to resist the to and fro motion. Due to this reason, the leaves dip from the tree when shaken vigorously.

 

Q4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

 

Solution:

Due to the inertia of the passenger

 

Every body tries to keep up its state of motion or state of rest. If a body is at rest, then it tries to stay at rest. If a body is moving, then it tries to stay in motion. In a moving bus, a passenger moves with the bus. As the driver applies brakes, the bus involves rest. But, the passenger tries to keep up his state of motion. As a result, a forward force is exerted on him. Similarly, the passenger tends to fall backwards when the bus accelerates from rest. This is often because when the bus accelerates, the inertia of the passenger tends to oppose the progression of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.

 

Pg 126

 

Q1. If action is always equal to the reaction, explain how a horse can pull a cart.

 

Solution:

A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the cart on the horse within the forward direction. As a result, the cart moves forward.

 

Q2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

 

Solution:

Due to the backward reaction of the water being ejected

 

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, a reaction force is exerted on him by the ejecting water within the backward direction. This is according to Newton’s third law of motion. As a results of the backward force, the stabilty of the fireman decreases. Hence, it’s difficult for him to stay stable while holding the hose.

 

Q3.  From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−1. Calculate the initial recoil velocity of the rifle.

 

Solution:

M = 4 kg              m  = 50 gms

v  = 35 m/sec

Initial momentum of rifle and bullet together = 0.

So, as per the conservation of linear momentum,

0 = M * (-V) +  m  v

=>     0   = 4 * (-V) + 50/1000  * 35

 

v =  0.4375  m/sec is the recoil velocity.

 

Q4. 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.

 

Solution:

Mass of first objects, m₁= 100 g = 0.1 kg

Mass of second object, m₂ = 200 g = 0.2 kg

Velocity of m₁ before collision, v₁= 2 m/s

Velocity of m₂ before collision, v₂ = 1 m/s

Velocity of m₁ after collision, v₃ = 1.67 m/s

Velocity of m₂ after collision = v₄

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m₁v₁ + m₂v₂ = m₁v₃+ m₂v₄

⇒ 0.1 * 2 + 0.2 * 1 = 0.1 * 1.67 + 0.2 * v₄

⇒ 0.4 = 0.67 + 0.2 * v₄

⇒ v₄  = 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

 

Pg 128

 

Q1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling 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.

 

Solution:

Yes. Even when an object experiences a net zero external unbalanced force, it’s possible that the article is traveling with a non-zero velocity. This will be possible only the article has been moving with a relentless velocity in an exceedingly particular direction. Then, there is not any net unbalanced force applied on the body. The article will keep moving with a non-zero velocity. To vary the state of motion, a net non-zero external unbalanced force must be applied on the article.

 

Q2. When a carpet is beaten with a stick, dust comes out of it. Explain.

 

Solution:

Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet involves motion. But, the dust particles attempt to resist their state of rest. In accordance with Newton’s first law of motion, the dust particles stay in an exceedingly state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

 

Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

 

Solution:

When the bus accelerates and moves forward, it acquires a state of motion. However, the bags kept on the roof, due to its inertia, tends to stay in its state of rest. Hence, with the forward movement of the bus, the bags tends to stay at its original position and ultimately falls from the roof of the bus. To avoid this, it’s advised to tie any luggage kept on the roof of a bus with a rope.

 

Q4. 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.

 

Solution:

(c) A batsman hits a cricket equipment, which then rolls on the ground. After covering a brief distance, the ball reaches to rest because there’s frictional force on the ball opposing its motion.

 

Frictional force always acts within the direction opposite to the direction of motion. Hence, this force is to blame for stopping the cricket equipment.

 

Q5. 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).

 

Solution:

The truck starts from rest, so initial velocity = 0

 

Distance travelled = 400 m

 

Time taken = 20 s

 

We know the equation of motion, S = ut + ½ at2

 

400 = 0 + ½ a * (20)2

 

a = 2 m/s2

 

Now, mass = 7 metric tonnes

 

= 7 * 1000

 

= 7000 kg

 

So force = mass * acceleration

 

F = 7000 * 2

 

= 14000 N

 

Q6. 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 traveling a distance of 50 m. What is the force of friction between the stone and the ice?

 

Solution:

Initial velocity of the stone, u = 20 m/s

 

Final velocity of the stone, v = 0 (finally the stone comes to rest)

 

Distance covered by the stone, s = 50 m

 

According to the third equation of motion:

 

v2 = u2 + 2as

 

Where,

 

Acceleration, a

 

(0)2 = (20)2 + 2 × a × 50

 

a = −4 m/s2

 

The negative sign indicates that acceleration is acting against the motion of the stone.

 

Mass of the stone, m = 1 kg

 

From Newton’s second law of motion:

 

Force, F = Mass × Acceleration

 

F = ma

 

F = 1 × (− 4) = −4 N

 

Hence, the force of friction between the stone and the ice is −4 N.

 

 

Q7. 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 and

 

(b) the acceleration of the train.

 

Solution:

(a) Force exerted by the engine, F = 40000 N

Frictional force offered by the track, F_f = 5000 N

Net accelerating force, Fa = F – F_f = 40000 – 5000 = 35000 N

Hence, the net accelerating force is 35000 N.

 

(b) Acceleration of the train = a

The engine exerts a force of 40000 N on all the five wagons.

Net accelerating force on the wagons, F_a = 35000 N

Mass of the wagons, m = Mass of a wagon x Number of wagons

Mass of a wagon = 2000 kg

Number of wagons = 5

m = 2000 — 5 = 10000 kg

Total mass, M = m = 10000 kg

From Newton’s second law of motion:

Fa= Ma

a=Fam = 35000 10000 = 3.5 ms^(-2)

Hence, the acceleration of the wagons and the train is 3.5 m/s2.

 

(c) Mass of all the wagons except wagon 1 is 4 — 2000 = 8000 kg

Acceleration of the wagons = 3.5 m/s²

Thus, force exerted on all the wagons except wagon 1

= 8000 — 3.5 = 28000 N

Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 N.

Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.

 

Q8. 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 m s⁻²?

 

Solution:

Mass of the automobile vehicle, m = 1500 kg

 

Final velocity, v = 0 (finally the automobile stops)

 

Acceleration of the automobile, a = −1.7 ms⁻²

 

From Newton’s second law of motion:

 

Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N

 

Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.

 

Q9. What is the momentum of an object of mass m, moving with a velocity v?

 

• (mv)² (b)mv²     (c) ½ mv²  (d) mv

 

Solution:

(d) mv

 

Mass of the object = m

 

Velocity = v

 

Momentum = Mass × Velocity

 

Momentum = mv

 

Q10. 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?

 

Solution:

A force of 200 N is applied in the forward direction. According to Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet.

 

Q11. 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−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

 

Solution:

Mass of one of the objects, m₁ = 1.5 kg

 

Mass of the other object, m₂ = 1.5 kg

 

Velocity of m₁ before collision, v₁ = 2.5 m/s

 

Velocity of m₂, moving in opposite direction before collision, v₂ = −2.5 m/s

 

(Negative sign arises because mass m₂ is moving in an opposite direction)

 

After collision, the two objects stick together.

 

Total mass of the combined object = m₁ + m₂

 

Velocity of the combined object = v

 

According to the law of conservation of momentum:

 

Total momentum before collision = Total momentum after collision

 

m₁v₁ + m₂ v₂ = (m₁ + m₂) v

 

1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v

 

3.75 − 3.75 = 3 v

 

v = 0

 

Hence, the velocity of the combined object after collision is 0 m/s.

 

Q12. 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 roadside, 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.

 

Solution:

The truck has a large mass.So, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. So, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction.

 

Therefore, the student is right in justifying that the two opposite and equal cancel each other.

 

Q13. A hockey ball of mass 200 g traveling 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.

 

Solution:

Mass of the hockey ball, m = 200 g = 0.2 kg

 

Hockey ball travels with velocity, v1 = 10 m/s

 

Initial momentum = mv1

 

Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s

 

Final momentum = mv2

 

Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s⁻¹

 

Hence, the change in momentum of the hockey ball is 3 kg m s⁻¹.

 

Q14. A bullet of mass 10 g traveling horizontally with a velocity of 150 m s⁻¹ 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.

 

Solution:

Given that :

 

u = 150 m/s

 

v = 0

 

t = 0.03 sec

 

m = 10 g = 0,01 Kg

 

We know that , S = ( v² – u²) / 2a  ————( 1 )

 

To calculate acceleration we use ,  v = u + at

 

a =  ( v – u ) / t

 

a = ( 0 – 150 ) / 0.03

 

a = -5000 m/s² —————( 2 )

 

Substituting equation ( 2 ) in ( 1 ) gives ,

 

S = { 0 – (150)² } / – ( 2 x 5000 )

 

S= 22500 / 10000

 

S = 2.25 m

 

Hence, the distance penetrated by the bullet is 2.25 m

 

To calculate force ,we use

 

F = m x a

 

⇒ F = 0.01 x 5000

 

⇒ F = 50 N

 

The force exerted by the wooden block on the bullet  is 50 N.

 

Q15. An object of mass 1 kg traveling in a straight line with a velocity of 10 m s⁻¹ 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. 

 

Solution:

Mass of the object, m₁ = 1 kg

 

Velocity of the object before collision, v₁ = 10 m/s

 

Mass of the stationary wooden block, m₂ = 5 kg

 

Velocity of the wooden block before collision, v₂ = 0 m/s

 

∴ Total momentum before collision = m₁ v₁ + m₂ v₂

 

= 1 (10) + 5 (0) = 10 kg m s⁻¹

 

It is given that after collision, the object and the wooden block stick together.

 

Total mass of the combined system = m₁ + m₂

 

Velocity of the combined object = v

 

According to the law of conservation of momentum:

 

Total momentum before collision = Total momentum after collision

 

m₁ v₁ + m₂ v₂ = (m₁ + m₂) v

 

1 (10) + 5 (0) = (1 + 5) v

 

The total momentum after collision is also 10 kg m/s.

 

Total momentum just before the impact = 10 kg m s⁻¹

 

Total momentum just after the impact = (m₁ + m₂) v = 6 X (5/3) = 10 kg m s⁻¹

 

Hence, velocity of the combined object after collision =

5/3  m s⁻¹

 

Q16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s⁻¹ to 8 m

s⁻¹ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

 

Solution:

Initial momentum=mass*initial velocity=(100*5)=500kg m/s

Final momentum=mass*final velocity=(100*8)=800 kg m/s

Force=Rate of change of momentum

Force=m(v-u)/t

=100(8-5)/6

=100*3/6=50 N