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# NCERT solutions of class 9 science chapter 9-Force and Laws of motion

## NCERT solutions of class 9 science chapter 9-Force and Laws of motion

NCERT Solutions for Class 9 Science Chapter 9- Force and Laws of Motion are set up here with the determination of boosting the student’s preparation for clearing their questions of textbook and concept of the topic altogether. NCERT solutions of class 9 science chapter 9-Force and Laws of motion are the solutions of each questions mentioned in the entire chapter 9. Class 9 NCEERT Solutions of Science is a valuable reference study material that assists CBSE students of 9 class with clearing questions right away in a compelling manner. NCERT Solutions for Class 9 Science approaches students in a tuned way and is stacked with questions, exercises, and activities that are the school board and competitive entrance exam-oriented.

NCERT Solutions for Class 9 Science are prepared in a planned way and created by the specialists of the science and checked and edited by a team having experience in instructing students. It is created keeping in view the concept-based methodology, alongside the exact strategy of answering the questions in assessments.
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### Download PDF -NCERT solutions of class 9 science chapter 9-Force and Laws of motion

PDF-NCERT solutions of class 9 science chapter 9-Force and laws of motion

## NCERT Solutions of class 9 science

### NCERT solutions of class 9 science chapter 9-Force and Laws of motion

Page 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?

Ans.(a) Inertia of an object is directly proportional to its mass, so the inertia of a stone will be more than a rubber ball because the mass of a stone is more than a rubber ball of the same size.
(b) A train will have more inertia because its mass is more than the mass of the bicycle.
(c) A five rupee coin is heavier than the one rupee coin so five rupee coin will have more inertia.

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.

Ans. The initial velocity of football is zero which is changed to v1 when kicked by the player of a team towards another player of the same team

Another player kicked the ball towards the goal, the velocity of the football changed from v1 to v2 .

Once again velocity of football changes from vto zero when the Goalkeeper stops the ball by applying an opposite force

Then the goalkeeper changed the velocity of football from 0 to v3 by kicking it to the player of his own team.

The velocity of football changed 4 times as follows.

 Change in velocity 0 →V1 V1→V2 V2→0 0→V3

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

Ans. When we shake the branch of a tree it moves due to an external force acts on the branch, Due to the more inertia of branch as compared with leaves the branch of the tree gets into rest but leaves continue to be in the effect of that an external force and detached from the tree.

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

Ans. When moving bus brakes to a stop, the velocity of the bus and the passenger seats become zero moreover since the lower part of our body  attached to the seat so it also stops along the bus but due to the inertia upper part of the body still carried on the same motion so we fall in the forward direction.

When the bus accelerates from the rest our lower part attached with the seat becomes in motion but the upper part of the body still in zero velocity due to the inertia so we fall backward.

See the video of question 3 and 4

### NCERT solutions of class 9 science chapter 9-Force and Laws of motion

Page. 126

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

Ans. In this situation the force applied by the horse is equal to the reactive force produced on the cart due to the frictional force of the earth, horse pushes the earth backward and the earth exerted the same force on the horse in forward direction which overcomes the frictional force on the cart, it is that’s why horse becomes able to pull a cart.

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

Ans. The water ejected from the hose due to a certain amount of force. An equal and opposite reactive force exerted on the hose due to the third law of motion, due to which the fireman faces difficulty in holding it.

Q3.From a rifle of mass 4 kg . a bullet of mass 50 gm is fired with an initial velocity of 35 m/s.Calculate the initial recoil velocity of the rifle.

Ans.Let the initial recoil velocity of the rifle is = v

initial velocity of  bullet = 35 m/s

mass of rifle = 4kg , mass of bullet =50 gm = 0.05 kg

The momentum  of rifle and bullet before the fire = 0( both of them have their velocities 0) ……(i)

The momentum  of rifle and bullet  after the fire

= 35 × 0.05 + 4 × v

=1.75 + 4v…..(ii)

According to the rule of conservation of the momentum

Inial momentum  =  final momentum

The momentum of rifle and bullet before the fire =The momentum  of rifle and bullet  after the fire

0 =1.75 + 4v

4v = -1.75

v = -1.75/4 = – 0.4375≡ -0.44

Therefore the recoil velocity of the rifle is -0.44 m/s, negative sign shows that the motion of the rifle is in the direction opposite to the bullet.

Q4. Two objects of masses 100 gm and 200 gm are moving along the same line and direction with velocities of 2 m/s and 1 m/s respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

Ans. According to the conservation of momentum we have

m1u1 + m2u = m1v1 + m2v2,

Where mass of the  first object,m1 =100gm = 0.1 kg, mass of the second object is, m2 = 200 gm = 0.2 kg, initial velocity of the first object,u = 2 m/s, initial velocity of the second object,u = 1 m/s, final velocity of the first object,v = 1.67 m/s, we have to determine the final velocity of the second object,v2 = ?

Substituting these values in the above equation

0.1×2 + 0.2 × 1 = 0.1 × 1.67 + 0.2 × v2

0.2 + 0.2 = 0.167 + 0.2 v2

0.4 0-167 = 0.2 v2

0.2 v2= 0.233

v2= 1.165

Therefore the velocity of the second object will be 1.165 m/s.

### NCERT solutions of class 9 science chapter 9-Force and Laws of motion

EXERCISES

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.

Ans. Net external unbalanced force  = 0, acceleration = a. Mass of the object = m

According to Newton’s second law of motion,the relationship among mass,acceleration and force is

F = ma, where F is applied force on the object,m is the mass of the object and a is the acceleration produced on the object by the force

a = F/m = 0/m = 0

Applying the first equation of motiion if u = 0 then v also be =0 and if u ≠0 then v≠o = u

Therefore if an object experiences a net zero external unbalanced force .  it is  possible for the object to be travelling wth a constant  non zero velocity

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

Ans. When a carpet is beaten with a stick, the surface of the carpet moves outward and due to inertia it stops but particles of dust due to their low mass continue to move outward.

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

Ans. When a brake is applied by the driver to a stop, the bus stops but due to inertia, the luggage on the roof continues to move with the velocity of the bus so to prevent it from falling down it is advised to tie it 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.

Ans. (c) ) there is a force exerted by the surface of the ground known as the frictional force on the ball opposing the motion, so after covering a short distance, the ball comes to rest.

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 tonnes ( 1 tonne = 1000 kg.)

Ans. Initial velocity of truck u = 0

Distance traveled by the truck, s = 400 m

Time taken to cover the distance,t = 20 s

Acceleration,a = ?

s = ut + ½ at²

400 = 0 ×20 + ½ (a × 20²)

a = 2

Therefore the truck  is traveling with the constant acceleration of  2 m/s²

Aplying Newton’s second law of motion

F = ma, where F is the force acting on the truck, m is the mass of the truck and a is the acceleration prodced on the truck

mass of the truck,m = 7 tonne = 7 × 1000 =7000 kg

F = ma

F = 7000 ×2 = 14000

The force  acting on the truck = 14000 N

Q6.A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50m. What is the force of friction between the stone and the ice ?

Ans. Mass of stone,m = 1kg, its initial velocity,u = 20 m/s

s= 50 m, force of friction between the stone and the ice,F = ?

Final velocity of the stone = 0 since stone rests after a certain distance

Applying the third equation of motion for evaluating acceleration  of the stone

v²  = u² + 2as

0 = 20² + 2 × a × 50

100a = -400

a = -4

Acceleration produuced by the frictional force on the stone = -4 m/s²

Aplying Newton’s second law of motion

F = ma, where F is the force acting on the stone, m is the mass of the stone and a is the acceleration prodced on the stone

F = 1 × -4 m/s²

F = -4

Therefore 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;

(b) the acceleration of the train : and

(c) the force of wagon 1 on wagon 2.

Ans.(a)  the engine exert a force = 40000N

Frictional force by the track on the train= -5000 N

So, the net accelerating force

= 40000 – 5000 N

=35000 N

(b) mass of the train,m = 5 × 2000 + 8000 = 18000 kg(of 5 wagons +engine)

Net force, F = 35000 N

Aplying Newton’s second law of motion

F = ma

a = F/m = 35000/18000 = 1.944

So, acceleration of the train = 1.944 m/s²

(c) The force applied by the engine to the wagon 1,F = the force of wagon 1 on wagon 2,F

Total mass of the train including the engine,m = 8000 + 5 ×2000 = 18000 kg, acceleration of train,a = 1.944 m/s²

Aplying Newton’s second law of motion

F = ma

= 18000 × 1.944 =15556N

So the force of wagon 1 on wagon 2 = 15556 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/s² ?

Ans. mass of automobile vehicle =1500 kg

Acceleration =  -1.7 m/s²

Accordiing to Newton’s secoond law,the force,F between the vehicle and road = ma= 1500 × -1.7 m/s²

= -2550 N

A negative sign shows that the direction of the force will be opposite to the direction of the vehicle.

Q9. 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

Ans. the momentum  of an object of mass m, moving with a velocity v = mass of object x velocity of an object

= mv

Q10. Using a horizontal force of 200 N, we intend to move wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Ans. F = 200 N applied on the wooden cabinet,

The friction force that will be exerted on the cabinet in oppposite direction of applied horizontal force

= -200 N

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

Ans. Let after the collision their combined velocity is = v

Momentum = mv

Their momentum before collision = 1.5 × 2.5 + 1.5 × 2.5= 7.5 kg.m/s,

Their momentum after collision = 1.5 × v= 1.5v

1.5v = 7.5

v = 3

So the velocity of the combined object after collision = 3m/s

Q12.According to the third law of motion when we push on an object the object pushes back on us with 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 .

Ans., when we push on an object the object pushes back on us with equal and opposite force this, is true, we also have the law of inertia to illustrate the situation, the truck has more inertia due to its heavy mass, according to the law of inertia a body always opposes the change in itself. The applied force is not sufficient to overcome the inertia of the truck so it does not move.

Q13. A hockey ball of mass 200g traveling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.

Ans.mass of hocky ball( m) = 200 g =0.2 kg, initial velocity of hocky ball = 10 m/s

Momentum of  the ball  before it was struck by the hocky = 0.2 × 10 = 2 kg.m/s

Momentum of  the ball  after it was struck by the hocky = -10 × 0.2 = -2 kg.m/s

Change in the momentum = -2 -2 = -4 kg.m/s

Q14.A bullet of mass 10 g traveling 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.

Ans. mass of bullet = 10 g = 10/1000 = 0.01 kg

Initial velocity of the bullet at the time when it strike the wooden block (u) = 150 m/s

Final velocity when the bullet come to rest = v = 0

Applying the first equation of motion , v =u + at

0 = 150 + ax0.03

a =-150/0.03 = -5000

Acceleration of bullet produced during the penetration of bullet  =-5000s²(opposite to the direction of bullet)

The distance of penetration of the bullet into the block = s

Applying the third equation of the motion

v² = u² + 2as
0 = 150² + 2 × -5000 × s

10000s = 22500

s = 2.25

Hence the distance of penetration of the bullet into the block = 2.25 cm

The force exerted by the wooden block on the bullet = mass x acceleration = 0.01 × -5000 = -50 N

Q15. An object of mass 1 kg traveling 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.

Ans. Momentum = mass × velocity

Total momentum of the object and the wooden block before the impact = 1 × 10 + 5 × 0( block is in stationary position, so its velocity =0)

= 10 kg.m/s

Total momentum of the object and the wooden block after the impact

= 5 × v ( v= their combined velocity)

According to the rule of conservation of the momentum

5v = 10

v = 2

Their combined velocity = 2 m/s

Their total momentum before the impact = 10kg.m/s

Their total momentum  after the impact = 5v = 5× 2 =10kg.m/s

Q16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Ans. initial momentum of the object = mv = 100 × 5= 500 kg.m/s

Final momentum of the object = 100 × 8 = 800 kg.m/s

For evaluating the acceleration applying the first equation of the motion ,

v = u + at

8 = 6 + 6a, a = 1/3 m/s²

So the force = ma = 100 x 1/3 = 100/3 = 33.33 N

Q18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm ? Take its downward acceleration to be 10 m/s².

Ans. Initial velocity of dumbbell ,u = 0

Height,h = 80 cm = o.80 m

Let the  velocity at which it strike the floor = v

For calculating v applying the third equation of the motion

v² =  u² + 2gh

v² = 0 + 2 × 10 × 0.8

v = 4

The velocity (final velocity) at which it strike the floor = 4 m/s

The moment at which it strike the floor = mv = 10 × 4 = 40 kg.m/s.

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