# CBSE Class 9 Physics Force And Laws Of Motion Worksheet Set C

Read and download free pdf of CBSE Class 9 Physics Force And Laws Of Motion Worksheet Set C. Students and teachers of Class 9 Physics can get free printable Worksheets for Class 9 Physics in PDF format prepared as per the latest syllabus and examination pattern in your schools. Standard 9 students should practice questions and answers given here for Physics in Grade 9 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 9 Physics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems provided here with solutions on daily basis to get more score in school exams and tests

## Force And Laws Of Motion Class 9 Physics Worksheet Pdf

Class 9 Physics students should refer to the following printable worksheet in Pdf for Force And Laws Of Motion in standard 9. This test paper with questions and answers for Grade 9 Physics will be very useful for exams and help you to score good marks

### Class 9 Physics Worksheet for Force And Laws Of Motion

CBSE Class 9 Physics Worksheet - Force and Laws of Motion (2)

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Exam Questions NCERT Class 9 Science Chapter 9 Force and Laws of Motion

Question. A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of 100 msl. Calculate the velocity of recoil of the gun.
Mass of bullet, m = 100 g – 1000 kg
Velocity of bullet, u = 100 ms–1
Mass of gun, M = 20 kg
Let recoil velocity of gun = V
Step 1. Before firing, the system (gun + bullet) is at rest, therefore, initial momentum of the system = 0
Final momentum of the system
= momentum of bullet + momentum of gun
= mu + MV = 10
1 × 100 + 20 V
V = 10 + 20 V
Step 2. Apply law of conservation of momentum Final momentum = Initial momentum
i.e. 10 + 20 V = 0
20 V = 10
or V = – 0.5 ms–1
Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.

Question. Define force. What are different types forces?
Answer. Force : It is a push or pull on an object that produces acceleration in the body on which it acts. The S.I. unit of force is Newton.
Types of forces :
Balanced force : When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
Unbalanced force : When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced forces.
Frictional force : Force of friction is the force that always opposes the motion of object.

Question. Why are road accidents at high speeds very much worse than accidents at low speeds?
Answer. The time of impact of vehicles is very small at high speed. So, they exert very large forces on each other.
Hence, road accidents at high speeds are highly fatal.

Question. Name two factors which determine the momentum of a body.
Answer. Two factors on which momentum of a body depend is mass and velocity. Momentum is directly proportional to the mass and velocity of the body.

Question. What decides the rate of change of momentum of an object?
Answer. The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

Question. Why do athletes run some distance before jumping?
Answer. Athlete has the inertia of motion and thus continues to move past the line.

Question. Is force a scalar quantity or a vector quantity?
Answer. Force is a vector quantity. It has both magnitude and direction.

Question. Why action and reaction do not cancel each other?
Answer. Action and reaction act simultaneous but on different objects. Hence, they do not cancel each other.

Question. What is inertia? Explain different types of inertia.
Answer. Inertia : The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of itsinertia. Its S.I. unit is kg.
Types of inertia :
Inertia of rest : The object remain in rest unless acted upon by an external unbalanced force.
Inertia of motion : The object in the state of uniform motion will continue to remain in motion with same speed and direction unless external force is not appliedon it.

Question. What is the acceleration produced by a force of 5 N exerted on an object of mass 10 kg?
Here F = 5 N; m = 10 kg; a = ?
Now F = ma or a = F/m
a = 0.5 ms–2

Question. Which would require greater force : accelerating a 10 g mass at 5 ms–2 or 20 g mass at 2 ms–2?
In first case m1 = 10 g = kg = 0.010 kg;
Now al = 5 ms–2 ; F1 = ?
F1 = m1a1 = 0.010 × 5
F1 = 0.050 Newton
In second case, m2 = 20 g = 0.020 kg
a2 = 2 ms–2; F2 = ?
Now F2 = m2a2 = 0.020 × 2
or F2 = 0.04 Newton
We find that F1 > F2, hence more force is required to accelerate 10 g at 5 ms–2 than accelerating 20 g at 2 ms–2.

Question. State Newton’s second law of motion.
Answer. The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Question. 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. Yes, it is possible. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.

Question. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer. When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet. Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them in motion in the opposite direction. This is why the dust comes out of the carpet when beaten.

Question. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer. When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force (in the backward direction) on the luggage. In a similar manner, when a bus which is initially in a state of motion suddenly comes to rest due to the application of brakes, a force (in the forward direction) is exerted on the luggage.
Depending on the mass of the luggage and the magnitude of the force, the luggage may fall off the bus due to inertia. Tying up the luggage will secure its position and prevent it from falling off the bus.

Question. 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. When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c).
If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.

Question. 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 it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Given, distance covered by the truck (s) = 400 meters
Time taken to cover the distance (t) = 20 seconds
Initial velocity of the truck (u) = 0 (since it starts from a state of rest)

Question. A stone of 1 kg is thrown with a velocity of 20 ms-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?
Given, Mass of the stone (m) = 1kg
Initial velocity (u) = 20m/s
Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)
Distance travelled by the stone (s) = 50 m

Question. An 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
(a) Given, force exerted by the train (F) = 40,000 N
Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg.
As per the second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)
= 35,000/18,000 = 1.94 ms-2
The acceleration of the train is 1.94 m.s-2.

Question. 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?
Given, mass of the vehicle (m) = 1500 kg
Acceleration (a) = -1.7 ms-2
As per the second law of motion, F = ma
F = 1500kg × (-1.7 ms-2) = -2550 N
Therefore, a force of 2550 N must act on the vehicle in a direction opposite to that of its motion.

Question. 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
Since momentum is defined as the product of mass and velocity, the correct answer is (d), mv.

Question. 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?
Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.

Question. 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 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Given, mass of the objects (m1 and m2) = 1.5kg
Initial velocity of the first object (u1) = 2.5 m/s
Initial velocity of the second object which is moving in the opposite direction (u2) = -2.5 m/s
When the two masses stick together, the resulting object has a mass of 3 kg (m1 + m2)
Velocity of the resulting object (v) =?
As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Total momentum before the collision = m1u1 + m2u2
= (1.5kg) (2.5 m/s) + (1.5 kg) (-2.5 m/s) = 0
Therefore, total momentum after collision = (m1+m2) v = (3kg) v = 0
Therefore v = 0
This implies that the object formed after the collision has a velocity of 0 meters per second.

Question. 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.
Since the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction (since the person pushing the truck is not displaced when the truck doesn’t move). Therefore, the student’s logic is correct.

Question. A hockey ball of mass 200 g travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Given, mass of the ball (m) = 200g
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = 5m/s
Initial momentum of the ball = mu = 200g × 10 ms-1 = 2000 g.m.s-1
Final momentum of the ball = mv = 200g × 5 ms-1 = 1000 g.m.s-1
Therefore, the change in momentum (mv – mu) = 1000 g.m.s-1 – 2000 g.m.s-1 = -1000 g.m.s-1
This implies that the momentum of the ball reduces by 1000 g.m.s-1 after being struck by the hockey stick.

Question. A bullet of mass 10 g travelling 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.
Given, mass of the bullet (m) = 10g (or 0.01 kg)
Initial velocity of the bullet (u) = 150 m/s
Terminal velocity of the bullet (v) = 0 m/s
Time period (t) = 0.03 s
To find the distance of penetration, the acceleration of the bullet must be calculated.

Therefore, force exerted by the wooden block on the bullet (F) = 0.01kg × (-5000 ms-2)
= -50 N

This implies that the wooden block exerts a force of magnitude 50 N on the bullet in the direction that is opposite to the trajectory of the bullet.

Question. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–1collides 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.
Given, mass of the object (m1) = 1kg
Mass of the block (m2) = 5kg
Initial velocity of the object (u1) = 10 m/s
Initial velocity of the block (u2) = 0
Mass of the resulting object = m1 + m2 = 6kg
Velocity of the resulting object (v) =?
Total momentum before the collision = m1u1 + m2u2 = (1kg) × (10m/s) + 0 = 10 kg.m.s-1
As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also 10 kg.m.s-1
Now, (m1 + m2) × v = 10kg.m.s-1

The resulting object moves with a velocity of 1.66 meters per second.

### More Question

1. A bullet fired from a gun is more dangerous than an air molecule hitting a person, though both bullet and air molecule are moving with same velocity. Explain.
2. Why are road accidents at high speeds very much worse than accidents at low speeds?.
3. What was the misbelief about the theory of motion before Newtonian motion theory? Why was it overruled?
4. When force acting on a body has an equal and opposite reaction, then why should the body move at all?
5. What can you say about the speed of a moving object if no force is acting on to it.?
6. Is a marble rolling down an inclined plane moving with constant velocity? Explain.
7. Two forces of 5N &22N are acting in a body in the same direction what will be the resultant force& in which direction will it act?
If the two forces in the above example would have been acting in the opposite direction What would be the resultant force& in which direction will it act?
8. If we push the box with a small force, the box does not move, why?
9. What should be the force acting on an object moving with uniform velocity?
10. Give reasons:
(a) Carpet is beaten with a stick to clean it,
(b) Seat belts are provided in the cars to prevent accidents.
(c) Only the carom coin at the bottom of a pile is removed when a fast moving carom coin (or striker) hits it.
(d) Place a water-filled tumbler on a tray.Hold the tray and turn around as fast as you can. We observe that the water spills. Why?
(e) a groove is provided in a saucer for placing the tea cup.

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