Q.1: 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.
Ans: (a) stone (b) train (c) five rupee coin.
Q.2: In the following example, try to identify the number of times the velocity of 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 velocity of the ball changes three times. First time, the velocity changes when the football player of one team kicks the ball. Second time the velocity changes when another player of the same team kicks the football. Third time the velocity changes when the goalkeeper of the opposite team kicks the football.
The agent supplying the force in each case, have been underlined.
Q.3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: Before shaking the branches the leaves are at rest. When the branches are shaken, they come in motion at once while the leaves tend to remain at rest due to inertia of rest. As a result leaves get detached from the branches and fall down.
Q.4: Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Ans: When a moving bus brakes to a stop, the lower part of our body in contact with bus comes to rest while the upper part of our body tends to keep moving due to inertia of motion. Hence we fall forwards. When the bus accelerates from rest, the lower part of our body comes into motion along with the bus while the upper part of body tends to remain at rest due to inertia of motion and as a result which we fall backwards.
Q.5: If action is always equal to reaction, explain how a horse can pull cart.
Ans: The horse pulls the cart with a force (action) in the forward direction. Since every action has an equal and opposite reaction so, the cart also pulls the horse with an equal force (reaction) in the backward direction. As a result of which the two forces get balanced. But while pulling the cart the horse also pushes the ground with its feet in the backward direction. The reaction of the earth of the earth makes it forward direction along with the cart. This is how the horse applies force and pulls the cart.
Q.7: Explain why it is difficult for a fireman to hold a hose, which ejects large amount of water at a high velocity.
Ans: Water is ejected with a large forward force (action). As we know by Newton’s third law of motion that every action has an equal and opposite reaction so, because of this action fireman experiences a large backward force or reaction. That is why he feels difficulty in holding the hose.
Q.9: From a rifle of mass 4 kg a bullet of mass 50 gm is fired with an initial velocity of 35 ms-1. Calculate the initial recoil velocity of the rifle.
Ans:
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing,
Or, m1v1 + m2v2 = m1u1 + m2u2
Or, 0.05 x 35 + 4v2 = 0 + 0
Or, v2 = – 0.44 ms-1 The negative sign indicates the direction of recoil (backward).
Q.10: Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 ms-1 and 1 ms-1 respectively. They collide and after collision, the first object moves at a velocity of 1.67 ms-1. Determine the velocity of the second object.Ans: By the law of conservation of momentum, m1v1 + m2v2 = m1u1 + m2u2
Or, 0.1 x 1.67 + 0.2 v2 = 0.1 x 2+ 0.23 x 1 Or, v2 = 1.165 ms–1.
It will move in the same direction after collision.
NCERT / CBSE Textbook Exercise Questions Solved
Q.1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with the 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: Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.
Q.2: When a carpet is beaten with a stick, dust comes out. Explain, why?
Ans: When a carpet is beaten with a stick it comes into motion at once. But the dust particles continue to be at rest due to inertia and get detached from the carpet.
Q.3: why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: Due to sudden jerks or due to the bus taking sharp turns on the road, the luggage may fall down from the roof because of its tendency to continue to be either at rest or in motion in the same direction (inertia of motion). To avoid this, it advised to tie the luggage kept on the roof of a bus with a rope.
Q.5: a truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 sec. Find its acceleration. Also find the force acting on it if its mass is 7 metric tones.
Ans: Here, u = 0, s = 400 m, t = 20 s
We know, s = ut + ½ at2 Or, 400 = 0 + ½ a (20)2 Or, a = 2 ms–2
Now, m = 7 MT = 7000 kg, a = 2 ms–2 Or, F = ma = 7000 x 2 = 14000 N Ans.
Q.6: 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?Ans:Here, m = 1 kg, u = 20 ms-1 v = 0, s = 50 m
Since, v2 - u2 = 2as, Or, 0 - 202 = 2a x 50, Or, a = – 4 ms-2
Force of friction, F = ma = – 4N
Q.7: 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;
(b) The acceleration of the train; and
(c) The force of wagon 1 on wagon 2.
Ans: Total mass, m = mass of engine + mass of wagons
Or, m = 8000 + 5 x 2000 = 18000 kg.
(a) The net accelerating force, F = Engine force - Frictional force
Or, F = 40000 - 5000 = 35000 N
(b) The acceleration of the train, a = F ÷ m = 35000 ÷ 18000 = 1.94 ms–2.
(c) The force of wagon 1 on wagon 2
= The net accelerating force - (mass of wagon x acceleration)
= 35000 - (2000 x 1.94) = 31111.2 N Ans.
Q.10: Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at constant velocity. What is the force of friction that will be exerted on the cabinet?
Ans: The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.
Q.11: 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?
Ans: Here, m1 = m2 = 1.5 kg, u1 = 2.5 ms–1 u2 = –2.5 ms–1
Let v be the velocity of the combined object after collision.
By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m1 + m2) v = m1u1 + m2u2
Or, (1.5 + 1.5) v = 1.5 x 2.5 +1.5 x (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms–1 Ans.
Q.12: 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.
Ans: The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.
Q.13: A hockey ball of mass 200 gm 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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans: Change in momentum = m (v - u) = 0.2 (–5 - 10) = –3 kg ms–1.
(The negative sign indicates a change in direction of hockey ball after it is struck by hockey stick. Magnitude of change in momentum = 3 kg ms–1).
Q.16: An Object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6 sec. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.
Ans: Here, m = 100 kg, u = 5 ms–1, v = 8 ms–1, t = 6 sec.
Initial momentum, p1 = mu = 500 kg ms–1
Final momentum, p2 = mv = 800 kg ms–1
The magnitude of the force exerted on the object, F = (p2 - p1) ÷ t = (800 - 500) ÷ 6 = 50 N
Q.17. A constant force acts on an object of mass 5kg for a duration of 2sec. It increases the object velocity from 3ms- 1 to 7ms-1 find the magnitude of force of the applied force. Now if the force was applied for a duration of 5s. What would be the final velocity?
Ans: F = ma = m(v-u)/t = 5(7-3)/2 = 10N
Now, same force 10N applies for 5 sec then, final velocity will be v
So, F = ma = m(v-u)/t Þ 10 = 5 (v-3)/5 Þ v = 13m/s
Q.18. Prove that impulse is equal to change in momentum of the body?
Ans: Impulse of a force is a measure of total effect of the force. It is given by the time for which force acts on the body Þ Impulse = force x time =Ft
By 2nd law of motion, F = Change in momentum/time
Þ F = d P/t Þ F x t = Change in momentum Þ impulse = Change in momentum
Q. 19. A car weighing 2000 kg and moving with a speed of 20m/s is stopped in 10sec on applying brakes. Calculate the retardation and the retarding force?
Ans: Let’s first calculate the retardation = (v-u)/t = (0-20)/10= -2 m/s2
Retarding Force = ma = 2000kg x – 2m/s2 = -4000N
Q.20. The change in momentum of a body in 0.01 sec is 10kg m/s. Find the force acting on this body.
Ans: Force = change in momentum/time taken = 10/0.01 = 1000 N
Q. 21. A railway wagon of mass 2500kg moving a velocity 36 km/hr has a head-on collision with a stationary wagon of mass 3000kg, if two wagons move together after collision calculate their common velocity after collision.
Ans: Using, law of conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
250o x 36 + 3000 x 0 + 2500xv + 300v Þ v = 16.36km/h
Q.22. Two balls of masses 50gm and 100gm are moving along the same line and direction with velocity 3 m/s and 1.5 m/s respectively. They collide and after collision the first ball moves with a velocity 2.5 m/s, determine the velocity of second ball.
Using law of conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
50 x 3 + 100 x 2.5 = 50 x 2.5 + 100xv2 Þ v2 = 2.75m/s
Q.23. A bullet of mass 20gm is fired from a gun of mass 8kg with a velocity of 400 m/s, calculate the recoil velocity of gun
Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
0 + 0 = 0.02 x 400 + 8 x v2 Þ v2 = -1m/s
Q. 24.Calculate the acceleration produced when a force of 350 Newton acts on a body of mass 500kg.
Ans: the acceleration produced = 350/500m/s2
Q. 25. A body of mass 25kg has a momentum of 125kg m/s, calculate velocity of body ?
Ans: We have momentum = mass X velocity p = m v OR, p/m = v v = (125kg m/s)/ 25 kg v = ( 125/25 )m/s v = 5 m/s
Q.26. Why is it easier to stop a tennis ball than a cricket ball moving with same speed ?
Ans: Momentum is product of mass and velocity of object. Here Mass of cricket ball is more than of tennis ball. So the momentum of tennis ball is less than that of cricket ball. That’s why it is easy to stop a tennis ball than a cricket ball moving with same speed.
Q.27. A boy of 40kg runs from left to right at 5m/s & catches a ball of 200g coming from right to left at 10m/s .After taking the catch , find the speed of the boy with the ball in his hand.
Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
40 x 5 + 0.2 x 10 + 40 v1 + 0 Þ v1 = 5.05m/s
Q.28. What is impulse of a Force?
Ans: Impulse of a Force is the product of force and the time for which the force
acts. According to Newton’s 2nd Law,
F = ma or F = m(v-u)/t or Ft = mv – mu
i.e. Impulse of a Force=Change in momentum.
Impulse of a Force is defined as the Change in momentum produced by the force
Q. 29.State the Law of conservation of momentum and prove it for an isolated system.
Ans: The sum of momenta of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. i.e. In an isolated system, the total momentum remains conserved.
Proof: Suppose two objects (two balls A and B, say) of masses m A and m B are traveling in the same direction along a straight line.
Let the velocities uA and uB, respectively and Let uA > uB.
There are no other external unbalanced forces acting on them.
The two balls collide with each other for a time t,
Then, the force on ball B by ball A(action)= the force on ball A(reaction) by ball B
Q.30. A revolver of mass 500 g , fires a bullet of mass 10g with a speed of 100 m/s. Find (i)momentum of the bullet. (ii) Initial momentum of revolver and bullet as a system. (iii) Recoil velocity of the revolver.
Ans: Before the bullet was shot the system of bullet and revolver was at rest.
Mass of the revolver = 500 g = 0.5 Kg Mass of the bullet = 10 g = .01 Kg
Velocity with which the bullet is fired = 100 ms-1
(i) Momentum of the bullet after being shot = .01 X 100 = 1 Ns
(ii) Initial momentum of the system = 0
(iii) Let v1 be the recoil velocity of the revolver.
Final momentum of the system is, momentum of the bullet + momentum of the revolver = 1 + 0.5v1
Applying momentum conservation, 0 = 1 + 0.5v1 Or, v1 = -1/0.5 = -2 ms-1
So, the velocity of the revolver is -2 ms-1, the negative sign indicates that the direction of velocity of the revolver is opposite to the direction of the bullet.
NCERT solution Force and Law of motion for class 9
Guess Questions 2012 Ch-9 FORCE AND LAWS OF MOTION
QUESTION BANK Class 9th Physics (Forces and Laws of motion)
9th Forces and Laws of motion teat paper-1
9th force and law of motion Test paper -2
9th Forces and Laws of motion teat papers unsolved
CBSE PHYSICS Force and law of motion Numerical for class9 unsolved
Solved Numerical Problems in 9th Force and Laws of motion
(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.
Ans: (a) stone (b) train (c) five rupee coin.
Q.2: In the following example, try to identify the number of times the velocity of 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 velocity of the ball changes three times. First time, the velocity changes when the football player of one team kicks the ball. Second time the velocity changes when another player of the same team kicks the football. Third time the velocity changes when the goalkeeper of the opposite team kicks the football.
The agent supplying the force in each case, have been underlined.
Q.3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: Before shaking the branches the leaves are at rest. When the branches are shaken, they come in motion at once while the leaves tend to remain at rest due to inertia of rest. As a result leaves get detached from the branches and fall down.
Q.4: Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Ans: When a moving bus brakes to a stop, the lower part of our body in contact with bus comes to rest while the upper part of our body tends to keep moving due to inertia of motion. Hence we fall forwards. When the bus accelerates from rest, the lower part of our body comes into motion along with the bus while the upper part of body tends to remain at rest due to inertia of motion and as a result which we fall backwards.
Q.5: If action is always equal to reaction, explain how a horse can pull cart.
Ans: The horse pulls the cart with a force (action) in the forward direction. Since every action has an equal and opposite reaction so, the cart also pulls the horse with an equal force (reaction) in the backward direction. As a result of which the two forces get balanced. But while pulling the cart the horse also pushes the ground with its feet in the backward direction. The reaction of the earth of the earth makes it forward direction along with the cart. This is how the horse applies force and pulls the cart.
Q.7: Explain why it is difficult for a fireman to hold a hose, which ejects large amount of water at a high velocity.
Ans: Water is ejected with a large forward force (action). As we know by Newton’s third law of motion that every action has an equal and opposite reaction so, because of this action fireman experiences a large backward force or reaction. That is why he feels difficulty in holding the hose.
Q.9: From a rifle of mass 4 kg a bullet of mass 50 gm is fired with an initial velocity of 35 ms-1. Calculate the initial recoil velocity of the rifle.
Ans:
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing,
Or, m1v1 + m2v2 = m1u1 + m2u2
Or, 0.05 x 35 + 4v2 = 0 + 0
Or, v2 = – 0.44 ms-1 The negative sign indicates the direction of recoil (backward).
Q.10: Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 ms-1 and 1 ms-1 respectively. They collide and after collision, the first object moves at a velocity of 1.67 ms-1. Determine the velocity of the second object.Ans: By the law of conservation of momentum, m1v1 + m2v2 = m1u1 + m2u2
Or, 0.1 x 1.67 + 0.2 v2 = 0.1 x 2+ 0.23 x 1 Or, v2 = 1.165 ms–1.
It will move in the same direction after collision.
NCERT / CBSE Textbook Exercise Questions Solved
Q.1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with the 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: Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.
Q.2: When a carpet is beaten with a stick, dust comes out. Explain, why?
Ans: When a carpet is beaten with a stick it comes into motion at once. But the dust particles continue to be at rest due to inertia and get detached from the carpet.
Q.3: why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: Due to sudden jerks or due to the bus taking sharp turns on the road, the luggage may fall down from the roof because of its tendency to continue to be either at rest or in motion in the same direction (inertia of motion). To avoid this, it advised to tie the luggage kept on the roof of a bus with a rope.
Q.5: a truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 sec. Find its acceleration. Also find the force acting on it if its mass is 7 metric tones.
Ans: Here, u = 0, s = 400 m, t = 20 s
We know, s = ut + ½ at2 Or, 400 = 0 + ½ a (20)2 Or, a = 2 ms–2
Now, m = 7 MT = 7000 kg, a = 2 ms–2 Or, F = ma = 7000 x 2 = 14000 N Ans.
Q.6: 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?Ans:Here, m = 1 kg, u = 20 ms-1 v = 0, s = 50 m
Since, v2 - u2 = 2as, Or, 0 - 202 = 2a x 50, Or, a = – 4 ms-2
Force of friction, F = ma = – 4N
Q.7: 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;
(b) The acceleration of the train; and
(c) The force of wagon 1 on wagon 2.
Ans: Total mass, m = mass of engine + mass of wagons
Or, m = 8000 + 5 x 2000 = 18000 kg.
(a) The net accelerating force, F = Engine force - Frictional force
Or, F = 40000 - 5000 = 35000 N
(b) The acceleration of the train, a = F ÷ m = 35000 ÷ 18000 = 1.94 ms–2.
(c) The force of wagon 1 on wagon 2
= The net accelerating force - (mass of wagon x acceleration)
= 35000 - (2000 x 1.94) = 31111.2 N Ans.
Q.10: Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at constant velocity. What is the force of friction that will be exerted on the cabinet?
Ans: The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.
Q.11: 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?
Ans: Here, m1 = m2 = 1.5 kg, u1 = 2.5 ms–1 u2 = –2.5 ms–1
Let v be the velocity of the combined object after collision.
By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m1 + m2) v = m1u1 + m2u2
Or, (1.5 + 1.5) v = 1.5 x 2.5 +1.5 x (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms–1 Ans.
Q.12: 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.
Ans: The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.
Q.13: A hockey ball of mass 200 gm 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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans: Change in momentum = m (v - u) = 0.2 (–5 - 10) = –3 kg ms–1.
(The negative sign indicates a change in direction of hockey ball after it is struck by hockey stick. Magnitude of change in momentum = 3 kg ms–1).
Q.16: An Object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6 sec. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.
Ans: Here, m = 100 kg, u = 5 ms–1, v = 8 ms–1, t = 6 sec.
Initial momentum, p1 = mu = 500 kg ms–1
Final momentum, p2 = mv = 800 kg ms–1
The magnitude of the force exerted on the object, F = (p2 - p1) ÷ t = (800 - 500) ÷ 6 = 50 N
Q.17. A constant force acts on an object of mass 5kg for a duration of 2sec. It increases the object velocity from 3ms- 1 to 7ms-1 find the magnitude of force of the applied force. Now if the force was applied for a duration of 5s. What would be the final velocity?
Ans: F = ma = m(v-u)/t = 5(7-3)/2 = 10N
Now, same force 10N applies for 5 sec then, final velocity will be v
So, F = ma = m(v-u)/t Þ 10 = 5 (v-3)/5 Þ v = 13m/s
Q.18. Prove that impulse is equal to change in momentum of the body?
Ans: Impulse of a force is a measure of total effect of the force. It is given by the time for which force acts on the body Þ Impulse = force x time =Ft
By 2nd law of motion, F = Change in momentum/time
Þ F = d P/t Þ F x t = Change in momentum Þ impulse = Change in momentum
Q. 19. A car weighing 2000 kg and moving with a speed of 20m/s is stopped in 10sec on applying brakes. Calculate the retardation and the retarding force?
Ans: Let’s first calculate the retardation = (v-u)/t = (0-20)/10= -2 m/s2
Retarding Force = ma = 2000kg x – 2m/s2 = -4000N
Q.20. The change in momentum of a body in 0.01 sec is 10kg m/s. Find the force acting on this body.
Ans: Force = change in momentum/time taken = 10/0.01 = 1000 N
Q. 21. A railway wagon of mass 2500kg moving a velocity 36 km/hr has a head-on collision with a stationary wagon of mass 3000kg, if two wagons move together after collision calculate their common velocity after collision.
Ans: Using, law of conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
250o x 36 + 3000 x 0 + 2500xv + 300v Þ v = 16.36km/h
Q.22. Two balls of masses 50gm and 100gm are moving along the same line and direction with velocity 3 m/s and 1.5 m/s respectively. They collide and after collision the first ball moves with a velocity 2.5 m/s, determine the velocity of second ball.
Using law of conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
50 x 3 + 100 x 2.5 = 50 x 2.5 + 100xv2 Þ v2 = 2.75m/s
Q.23. A bullet of mass 20gm is fired from a gun of mass 8kg with a velocity of 400 m/s, calculate the recoil velocity of gun
Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
0 + 0 = 0.02 x 400 + 8 x v2 Þ v2 = -1m/s
Q. 24.Calculate the acceleration produced when a force of 350 Newton acts on a body of mass 500kg.
Ans: the acceleration produced = 350/500m/s2
Q. 25. A body of mass 25kg has a momentum of 125kg m/s, calculate velocity of body ?
Ans: We have momentum = mass X velocity p = m v OR, p/m = v v = (125kg m/s)/ 25 kg v = ( 125/25 )m/s v = 5 m/s
Q.26. Why is it easier to stop a tennis ball than a cricket ball moving with same speed ?
Ans: Momentum is product of mass and velocity of object. Here Mass of cricket ball is more than of tennis ball. So the momentum of tennis ball is less than that of cricket ball. That’s why it is easy to stop a tennis ball than a cricket ball moving with same speed.
Q.27. A boy of 40kg runs from left to right at 5m/s & catches a ball of 200g coming from right to left at 10m/s .After taking the catch , find the speed of the boy with the ball in his hand.
Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2
40 x 5 + 0.2 x 10 + 40 v1 + 0 Þ v1 = 5.05m/s
Q.28. What is impulse of a Force?
Ans: Impulse of a Force is the product of force and the time for which the force
acts. According to Newton’s 2nd Law,
F = ma or F = m(v-u)/t or Ft = mv – mu
i.e. Impulse of a Force=Change in momentum.
Impulse of a Force is defined as the Change in momentum produced by the force
Q. 29.State the Law of conservation of momentum and prove it for an isolated system.
Ans: The sum of momenta of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. i.e. In an isolated system, the total momentum remains conserved.
Proof: Suppose two objects (two balls A and B, say) of masses m A and m B are traveling in the same direction along a straight line.
Let the velocities uA and uB, respectively and Let uA > uB.
There are no other external unbalanced forces acting on them.
The two balls collide with each other for a time t,
Then, the force on ball B by ball A(action)= the force on ball A(reaction) by ball B
Q.30. A revolver of mass 500 g , fires a bullet of mass 10g with a speed of 100 m/s. Find (i)momentum of the bullet. (ii) Initial momentum of revolver and bullet as a system. (iii) Recoil velocity of the revolver.
Ans: Before the bullet was shot the system of bullet and revolver was at rest.
Mass of the revolver = 500 g = 0.5 Kg Mass of the bullet = 10 g = .01 Kg
Velocity with which the bullet is fired = 100 ms-1
(i) Momentum of the bullet after being shot = .01 X 100 = 1 Ns
(ii) Initial momentum of the system = 0
(iii) Let v1 be the recoil velocity of the revolver.
Final momentum of the system is, momentum of the bullet + momentum of the revolver = 1 + 0.5v1
Applying momentum conservation, 0 = 1 + 0.5v1 Or, v1 = -1/0.5 = -2 ms-1
So, the velocity of the revolver is -2 ms-1, the negative sign indicates that the direction of velocity of the revolver is opposite to the direction of the bullet.
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