A abampere (aA) The unit of electric current in the CGSeniu system, defined as that current that, if flowing through two parallel conductors of negligible cross section and infinite length, placed 1 cm apart in vacuo, would produce on each conductor a force of 1 dyne per centimeter of length. 1 abampere = 1 abcoulomb/s = r statampere (where c = speed of light in cm/s) = 10 ampere. aberration Imperfect image formation due to geometric imperfections in the optical elements of a system ablation 1 . The wasting of glacier ice by any process (calving, melting, evaporation, etc.). 2. The shedding of molten material from the outer sur- face of a meteorite or tektite during its flight through the atmosphere. absolute age The age of a natural substance, of a fossil or living organism, or of an artifact, obtained by means of an absolute dating method. See absolute dating method. absolute density Density in kg/m' or, more commonly, in g/cm\ both at STP. Cf. density, relative density abso
Force and Laws of Motion class 9 NCERT physics MCQ & SAQ
Force and Laws of Motion class 9 NCERT solution:
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-rupees coin and a one-rupee coin?
Answer
Inertia is the measure of the mass of the body. The greater is 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 of the same size. Hence, inertia of the stone is greater than that of a rubber ball.
(b) Mass of a train is more than the mass of a bicycle. Hence, the inertia of the train is greater than that of the bicycle.
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, the inertia of the five rupee coin is greater than that of the one-rupee coin.
2. 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.
Answer
The velocity of football changes four times.
First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth, when the goalkeeper kicks the football towards a player of his own team.
Agent supplying the force:
→ First case – First player
→ Second case – Second player
→ Third case – Goalkeeper
→ Fourth case – Goalkeeper
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer
Some leaves of a tree get detached when we shake its branches vigorously because branches come in motion while the leaves tend to remain at rest due to inertia of rest.
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?
Answer
In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain the inertia 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 because when the bus accelerates, the inertia of the rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.
Page No: 126
1. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer
A horse pushes the ground in the backward direction. According to Newton's third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.
2. Explain why it is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer
When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton's third law of motion. As a result of the backward force, the stability of the fireman
decreases. Hence, it is difficult for him to remain stable while holding the hose.
3. 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.
Answer
Mass of the rifle, m1= 4 kg
Mass of the bullet, m2= 50g= 0.05 kg
Recoil velocity of the rifle= v1
Bullet is fired with an initial velocity, v2= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m1+m2)v= 0
Total momentum of the rifle and bullet system after firing:
= m1v1 + m2v2 = 0.05 × 35 = 4v1 + 1.75
According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing 4v1 + 1.75= 0
v1 = -1.75/4 = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
Page No: 127
4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms−1 and 1 ms−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms−1. Determine the velocity of the second object.
Answer
Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other 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
Therefore, m1v1 + m2v2 = m1v3 + m2v4
2(0.1) + 1(0.2) = 1.67(0.1) + v4(0.2)
0.4 = 0.167 + 0.2v4
v4= 1.165 m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
Page No: 128
Exercises
1. 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.
Answer
Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A raindrop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the velocity of air. The net force on the drop is zero.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer
When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer
When a bust starts suddenly, the lower part of the luggage kept on the roof being in contact with the bus begins to move forward with the speed of the bus, but the upper part tends to remain at rest due to inertia of rest. Therefore, the upper part is left behind and hence luggage falls backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.
4. 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
The ball slows down and comes to rest due to opposing forces of air resistance and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the motion is correct.
5. 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 tons (Hint: 1 metric ton = 1000 kg).
Answer
Initial velocity, u = 0 Distance traveled, s = 400 m
Time taken, t = 20 s
We know, s = ut + ½ at2
Or, 400 = 0 + ½ a (20)2
Or, a = 2 ms–2
Now, m = 7 metric ton = 7000 kg, a = 2 ms–2
Or, F = ma = 7000 × 2 = 14000 N Ans.
6. 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?
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance covered by the stone, s = 50 m
Since, v2 - u2 = 2as,
Or, 0 - 202 = 2a × 50,
Or, a = –4 ms-2
Force of friction, F = ma = – 4N
7. 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.
Answer
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F − Ff = 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, Fa = 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 = Fa/M = 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.
8. 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?
Answer
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 x Acceleration = 1500 x (−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.
9. What is the momentum of an object of mass m, moving with a velocity v?
Answer
(d) mv
Mass of the object = m
Velocity = v
Momentum = Mass x Velocity
Momentum = mv
10. 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?
Answer
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.
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?
Mass of one of the objects, m1 = 1.5 kg Mass of the other object, m2 = 1.5 kg
Velocity of m1 before collision, u1 = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, u2 = −2.5 m/s
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 × 2.5 +1.5 × (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms–1
Page No: 129
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.
Answer
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.
13. 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 of 5 m s−1. Calculate the change of momentum that occurs in the motion of the hockey ball by the force applied by the hockey stick.
Answer
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 ms−1
Hence, the change in momentum of the hockey ball is 3 kg ms−1.
14. 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.
Answer
Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)a = -150/0.03 = -5000 m/s2
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.
15. 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.
Answer
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 ms−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
⇒ v = 10/6
= 5/3
The total momentum after a collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg ms−1
Total momentum just after the impact = (m1 + m2)v = 6 × 5/3 = 10 kg ms-1
Hence, velocity of the combined object after collision = 5/3 ms-1
16. 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.
Answer
Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time taken by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms−1
Final momentum = mv = 100 × 8 = 800 kg ms−1
Force exerted on the object, F = (mv - mu)/ t
= m (v-u)/t
= 800 - 500
= 300/6
= 50 N
Initial momentum of the object is 500 kg ms−1.
Final momentum of the object is 800 kg ms−1.
Force exerted on the object is 50 N.
17. Akhtar, Kiran and Rahul were riding in a motor car that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul, while giving an entirely new explanation, said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong.
The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of the large velocity of the motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insects with motor cars, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momentum are also the same. Only the signs of changes in momentum are opposite, i.e., change in momentum of the two occur in opposite directions, though magnitude of change in momentum of each is the same.
18. How much momentum will a dumbbell 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 s−2.
Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v2 = u2 + 2as
⇒ v2 = 0 + 2 (10) 0.8
⇒ v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv = 10 × 4 = 40 kg ms−1
Page No: 130
Additional Exercises
1. The following is the distance-time table of an object in motion:
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b)What do you infer about the forces acting on the object?
Answer
(a) There is an unequal change of distance in an equal interval of time.
Thus, the given object is having a non-uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.
(b) The object is in accelerated condition. According to Newton's second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say unbalanced force is acting on the object.
2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s−2. With what force does each person push the motorcar?
(Assume that all persons push the motorcar with the same muscular effort)
Mass of the motor car = 1200 kg
Only two people manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person,
a = 0.2 m/s2
Let the force applied by the third person be F.
From Newton’s second law of motion:
Force = Mass × Acceleration
F = 1200 × 0.2 = 240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motor car.
3. A hammer of mass 500g, moving at 50 ms−1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Answer
Mass of the hammer, m = 500 g = 0.5 kg
Initial velocity of the hammer, u = 50 m/s
Time taken by the nail to stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, f = m(v-u)/t
= 0.5(0-50)/0.01
= -2500 N
The hammer strikes the nail with a force of −2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.
4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Answer
Mass of the motor car, m = 1200 kg
Initial velocity of the motor car, u = 90 km/h = 25 m/s
Final velocity of the motor car, v = 18 km/h = 5 m/s
Time taken, t = 4 s
According to the first equation of motion:
v = u + at
⇒ 5 = 25 + a (4)
⇒ a = − 5 m/s2
Negative sign indicates that it is a retarding motion i.e. velocity is decreasing.
Change in momentum = mv − mu = m (v−u)
= 1200 (5 − 25) = −24000 kg m s−1
∵ Force = Mass × Acceleration
= 1200 × −5 = −6000 N
Acceleration of the motor car = −5 m/s2
Change in momentum of the motor car = −24000 kg m s−1
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)
Force and Laws of Motion class 9 NCERT physics MCQ:
Question: 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.
a) Three times
b) Five times
c) Four times
d) Six times
Answer: Three times
Question: There will be a change in the speed or in the direction of motion of a body when it is acted upon by
a) An Unbalanced force
b) Balanced Force
c) Uniform force
d) Zero Force
Answer: An Unbalanced force
Question: A hammer of mass 500 g, moving at 50m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. The force of the nail on the hammer is
a) 2500N
b) 1000N
c) 3500N
d) 500N
Answer: 2500N
Question: The action and reaction forces referred to in the third law
a) Must act on different objects
b) Must act on the same object
c) May act on different objects
d) Need not be equal in magnitude but act in the same direction
Answer: Must act on different objects
Question: An object will continue to accelerate until the
a) Resultant force on it is zero
b) The velocity changes direction
c) Resultant force on it is increased continuously
d) Resultant force begins to decrease
Answer: Resultant force on it is zero
Question: Qualitative definition of Force is given by
a) Newton's first law of motion
b) Newton's Second law of motion
c) Newton's third law of motion
d) Newton's law of gravitation
Answer: Newton's first law of motion
Question: The object shown below moves with constant velocity. Two forces are acting on the object. Considering negligible friction , the resultant force will be
a) 3 N leftwards
b) 7N rightwards
c) 17 N leftwards
d) 10 N leftwards
Answer: 3 N leftwards
Question: Type of inertia that tends to resist the change in case of an -Athlete often jumps before taking a long jump
a) Inertia of motion
b) Inertia of rest
c) Inertia of direction
d) Uniformly accelerated motion
Answer: Inertia of motion
Question: A passenger in a moving train tosses a coin which falls
a) Behind him
b) In front of him
c) Falls outside the train
d) Will not move
Answer: Behind him
Question: 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
a) -1m/s
b) 1m/s
c) 2m/s
d) -2m/s
Answer: -1m/s
More Questions...................
Question: Rocket works on the principle of conservation of
a) momentum
b) Mass
c) Energy
d) velocity
Answer: momentum
Question: An object of mass 2 kg is sliding with a constant velocity of 4 m/ s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is
a) 0 N
b) 2N
c) 32N
d) 8N
Answer: 0 N
Question: Inertia is a measure of
a) Mass
b) velocity
c) Force
d) acceleration
Answer: Mass
Question: Formula to find the Force is
a) F= ma
b) F= m/a
c) F= a/m
d) a=Fm
Answer: F= ma
Question: Force required in accelerating a 2 kg mass at 5 m/s2 and a 4 kg mass at 2 m/s2
a) 2kg mass at 5m/s2 is greater than 4 kg mass at 2 m/s2
b) Same in both the cases
c) 2kg mass at 5m/s2 is less than 4 kg mass at 2 m/s2
d) Zero in both the cases
Answer: 2kg mass at 5m/s2 is greater than 4 kg mass at 2 m/s2
Question.. If a bus starts suddenly, the passengers in the bus will tend to fall
(a) In the direction opposite to the direction of motion of the bus.
(b) In the same direction as the direction of motion of the bus.
(c) Sideways.
(d) None of the above.
Answer : A
Question. Which of the following is the unit of linear momentum?
(a) kg m/s
(b) kg ms
(c) kg ms−2
(d) kg
Answer : B
Question. A plate, a ball and a child all have the same mass. The one having more inertia is the
(a) plate
(b) ball
(c) child
(d) All have equal inertia
Answer : D
Question. Inertia is a property of a body by virtue of which the body is
(a) Unable to change by itself the state of rest.
(b) Unable to change by itself the state of uniform motion in a straight line.
(c) Unable to change by itself the direction of motion.
(d) Unable to change by itself the state of rest or uniform motion in a straight line.
Answer : D
Question. A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goalkeeper to:
(a) Exert large force on the ball
(b) Increases the force exerted by the ball on hands
(c) Increase the rate of change of momentum
(d) Decrease the rate of change of momentum
Answer : D
Question. The acceleration of an object is
(a) inversely proportional to its mass
(b) directly proportional to the applied force
(c) resisted by inertia
(d) all of the above
Answer : D
Question.: An object will continue to accelerate until
(a) The resultant force begins to decrease.
(b) The resultant force on it is zero.
(c) The velocity changes direction.
(d) The resultant force on it is increased continuously.
Answer : B
Question. The S.I. unit of force is
(a) Newton-meter
(b) Newton
(c) Newton per second
(d) Newton per square meter
Answer : B
Question.. Rate of change of momentum is equal to
(a) Acceleration
(b) Work done
(c) Force
(d) Impulse
Answer : C
Question. The action and reaction forces referred to in the third law
(a) Must act on the same object.
(b) May act on different objects.
(c) Must act on different objects.
(d) Need not be equal in magnitude but must have the same direction.
Answer : C
Question. If an object moves with a uniform velocity we can conclude that
(a) there is no force acting on the body
(b) no unbalanced force acts on it
(c) an unbalanced force acts on the body
(d) it has uniform acceleration
Answer : B
Question. When a force is exerted on an object, it can change its:
(a) State
(b) Position
(c) Shape
(d) All the above
Answer : D
Question . Rocket works on the principle of conservation of
(a) velocity
(b) Mass
(c) Linear Momentum
(d) Energy
Answer : C
Question. The seat belts are provided in the cars so that if the car stops suddenly due to an emergency braking, the persons sitting on the front seats are not thrown forward violently and saved from getting injured. Can you guess the law due to which a person falls in forward direction on the sudden stopping of the car?
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) Newton’s law of gravitation
Answer : A
Question. Two equal masses m each moving in the opposite direction with the same speed v collide and stick to each other. The velocity of the combined mass is
(a) v
(b) 2v
(c) v/2
(d) zero
Answer : D
Question. When unbalanced forces act on a body, the body:
(a) Must move with uniform velocity
(b) Must remain at rest
(c) Must experience acceleration
(d) Must move in a curved path
Answer : C
Question. A passenger in a moving train tosses a coin which falls behind him. Observing this statement, what can you say about the motion of the train?
(a) Accelerated
(b) Retarded
(c) Along circular tracks
(d) Uniform
Answer : A
Question. An object of mass 20 kg is moving with a velocity of 10 m/s. Its momentum will be:
(a) 2000 kg.m/s
(b) 20 kg.m/s
(c) 2 kg.m/s
(d) 200 kg.m/s
Answer : D
Question. When a 12 N force acts on 3 kg mass for a second, the change in velocity is (in m/s)
(a) 36
(b) 4
(c) 2
(d) 18
Answer : B
Question. The acceleration of an object is
(a) inversely proportional to its mass
(b) directly proportional to the applied force
(c) resisted by inertia
(d) all of the above
Answer : D
Question. If a bus starts suddenly, the passengers in the bus will tend to fall
(a) In the direction opposite to the direction of motion of the bus.
(b) In the same direction as the direction of motion of the bus.
(c) Sideways.
(d) None of the above.
Answer : A
Question. An object of mass 6 kg is sliding on a horizontal surface, with a uniform velocity of 8 m/ s . Assuming force of friction offered by the surface to be zero, The force required to maintain the motion of object with the same uniform velocity is
(a) 0 Newton
(b) 2 Newtons
(c) 8 Newtons
(d) 32 Newtons
Answer : D
1. A plate, a ball and a child all have the same mass. The one having more inertia is the
(a) plate
(b) ball
(c) child
(d) All have equal inertia
► (d) All have equal inertia
2. The inertia of an object tends to cause the object
(a) to increase its speed
(b) to decrease its speed
(c) to resist any change in the state of rest or of motion
(d) to decelerate due to friction
► (c) to resist any change in the state of rest or of motion
3. The rate of change of momentum with respect to time is measured in
(a) kg ms-2
(b) kg ms-1
(c) kg m
(d) kg
► (a) kg ms-2
4. When unbalanced forces act on a body, the body:
(a) Must move with uniform velocity
(b) Must remain at rest
(c) Must experience acceleration
(d) Must move in a curved path
► (c) Must experience acceleration
5. A man throws a ball weighing 200 g vertically upwards with a speed of 10m/s. Its momentum at the highest point of its flight will be:
(a) 2 kg. m/s
(b) 2000 kg.m/s
(c) Insufficient data to find the momentum.
(d) zero
► (d) zero
6. Quantitative expression of force is given by:
(a) Newton’s second law of motion.
(b) Newton’s third law of motion.
(c) Newton’s first law of motion.
(d) Newton’s law of gravitation.
► (a) Newton’s second law of motion.
7. Find the time taken by a body of mass 16 kg to come to rest from a uniform velocity of magnitude 10 m/s, when a force of 4N is applied continuously
(a) 20 s
(b) 30 s
(c) 40 s
(d) 50 s
► (c) 40 s
8. The S.I. unit of force is
(a) Newton-meter
(b) Newton
(c) Newton per second
(d) Newton per square meter
► (b) Newton
9. A force of ‘P* N acts on a particle so as to accelerate it from rest to a velocity 'v' m/s. The force 'P’ is then replaced by ‘Q’ N which decelerates it to rest.
(a) P may be equal to Q
(b) P must be equal to Q
(c) P must be unequal to Q
(d) none of these
► (a) P may be equal to Q
10. If the force acting on the body is zero. Its momentum is:
(a) Zero
(b) Constant
(c) Infinite
(d) None of the above
►(b) Constant
11. When a force is exerted on an object, it can change its:
(a) State
(b) Position
(c) Shape
(d) All the above
► (d) All the above
12. What force can change the velocity of a body of mass 1kg from 20 m/s to 30m/s in 2 seconds?
(a) 10 N
(b) 15 N
(c) 5 N
(d) 25 N
► (c) 5 N
13.One way that you can recognize that a force is acting on an object:
(a) is to note any change in the object's state of motion.
(b) is to determine its mass at different locations.
(c) is to measure the instantaneous velocity of a moving object.
(d) A and C
► (a) is to note any change in the object's state of motion.
14. a 1-kg object is lying on the ground. An unbalanced force of magnitude 1 N is applied to the object. Which of these options explains the motion of the object as a result of the acting force?
(a) The object will accelerate in the direction of the applied force.
(b) The object will accelerate in a direction perpendicular to the applied force.
(c) The object will decelerate in the direction of the applied force.
(d) The object will remain at rest
► (a) The object will accelerate in the direction of the applied force.
15. When a 12 N force acts on 3 kg mass for a second, the change in velocity is (in m/s)
(a) 36
(b) 4
(c) 2
(d) 18
► (b) 4
16. A passenger in a moving train tosses a coin which falls behind him. It means that the motion of the train is
(a) accelerated
(b) uniform
(c) retarded
(d) along circular tracks
► (a) accelerated
17. Force is defined as
(a) change in momentum
(b) rate of change of momentum
(c) the quantity that opposes inertia
(d) the quantity that keeps the velocity constant
► (b) rate of change of momentum
18. A force is defined as a
(a) Fall
(b) Pull
(c) Push or Pull
(d) Push
► (c) Push or Pull
19. What is the momentum of a body of mass 2m and velocity v/2?
(a) mv/4
(b) mv
(c) 2mv
(d) mv/2
► (b) mv
20. Two equal masses m each moving in the opposite direction with the same speed v collide and stick to each other. The velocity of the combined mass is
(a) v
(b) 2v
(c) v/2
(d) zero
► (d) zero
21. In high jump competition the athlete is made to fall on a cushioned bed to:
(a) To decrease his momentum fast.
(b) Make him stop quickly.
(c) Increase the time to stop.
(d) Make him sleep comfortably.
► (c) Increase the time to stop.
22. Momentum of a body of mass 0.5 kg moving with a speed of 10 m/s is
(a) 2.5 kg.m/s
(b) 5 kg.m/s
(c) 0.5 kg.m/s
(d) 50 kg.m/s
► (b) 5 kg.m/s
23. An object of mass of 2 kg is sliding with a velocity of 4 ms-1 on a frictional horizontal surface. The retarding force necessary to stop the object in 1 second is
(a) 2 N
(b) 4 N
(c) 8 N
(d) 0 N
► (c) 8 N
24. The acceleration of an object is
(a) inversely proportional to its mass
(b) directly proportional to the applied force
(c) resisted by inertia
(d) all of the above
► (d) all of the above
25. Which of the following has the largest momentum?
(a) A cat running down the street
(b) A pickup truck traveling down the highway.
(c) A large truck parked in a parking lot.
(d) A car parked in a parking lot.
► (b) A pickup truck traveling down the highway.
26. According to Newton’s third law of motion, action and reaction
(a) always act on the same body
(b) always act on different bodies in opposite directions
(c) have the same magnitude and direction
(d) act on either body at normal to each other
► (b) always act on different bodies in opposite directions
27. Which one of the following statements is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing
(b) Its velocity always changes
(c) It always goes away from earth
(d) A force is always acting on it
► (c) It always goes away from earth
28. Change in momentum when a car weighing 700kg changes its speed from 100m/s to 200 m/s is:
(a) 14000 kg.m/s
(b) 10500000 kg.m/s
(c) 21000000 kg.m/s
(d) 70000 kg.m/s
► (d) 70000 kg.m/s
29. When balanced forces act on a body, the body:
(a) Must remain in its state of rest
(b) Must continue moving with uniform velocity, if already in motion
(c) Must experience some acceleration
(d) Both (A) and (B)
► (d) Both (A) and (B)
30. If an object moves with a uniform velocity we can conclude that
(a) there is no force acting on the body
(b) no unbalanced force acts on it
(c) an unbalanced force acts on the body
(d) it has uniform acceleration
► (b) no unbalanced force acts on it
31. The acceleration of an object is inversely proportional to:
(a) Force.
(b) Momentum.
(c) Mass.
(d) Velocity.
► (c) Mass.
32. Find the time for which a force of 1kgwt acts on a body of mass 1 kg moving with a uniform speed of 4m/s to stop the body.
(a) 0.8 s
(b) 0.2 s
(c) 0.6 s
(d) 0.4 s
► (d) 0.4 s
33. __________ is a force, which always opposes the motion of one body over the other body in contact with it
(a) Gravitational force
(b) Reaction force
(c) Normal force
(d) Frictional force
► (d) Frictional force
34. An object of mass 20 kg is moving with a velocity of 10 m/s. Its momentum will be:
(a) 2000 kg.m/s
(b) 20 kg.m/s
(c) 2 kg.m/s
(d) 200 kg.m/s
► (d) 200 kg.m/s
35. When the driver of a fast moving car suddenly applies brakes, the passengers in the car:
(a) fall backward
(b) fall forward
(c) are not affected
(d) none of the above
► (b) fall forward
36. A force of magnitude' F' acts on a body of mass 'm'. The acceleration of the body depends upon:
(a) Volume of body
(b) Density of body
(c) Area of body
(d) Mass of body
► (d) Mass of body
Force and Laws of Motion class 9 NCERT physics SAQ:
Q.1. A boy of mass 50 kg running at 5 m s-1 jumps on to a 20 kg trolley traveling in the same direction at 1.5 m s-1. Find their common velocity. [CBSE 2016]
Ans. Here mass of boy m1 = 50 kg, initial velocity of boy u1 = 5 m s-1, mass of trolley m2 = 20 kg and initial velocity of trolley u2 = 1.5 m s-1.
If their common velocity be v, then from law of conservation of momentum, we have (m1 + m2)v = m1u1 + m2u2
⇒
Q.2. Do action and reaction act on the same body or on different bodies? Explain your answer with the help of an example. How are they related in magnitude and direction? Write the total momentum of the gun and the bullet before firing. [CBSE 2016]
Ans. Action and reaction forces act on two different bodies. For example, when a person swims in water, he applies a backward push of action on water. In turn, the water exerts a forward reaction force on the swimmer due to which he swims forward.
The magnitude of action and reaction forces are exactly equal but their directions are mutually opposite.
Total momentum of the gun and the bullet system before firing is zero.
Recoil: Total Momentum = 0
Q.3. The velocity-time graph of a car of 1000 kg mass is given alongside. From the graph answer the following: [CBSE 2010, 2016]
(а) When is the maximum force acting on the car? Why?
(b) What is the retarding force?
(c) For how long is there no force acting?
Ans. (a) In the v-t graph, the OA part represents uniformly accelerated motion with an acceleration
∴ Force in this region is maximum having a value
F = ma = 1000 x 3.75 = 3750 N
(b) In the v-t graph the part BC represents uniformly retarded motion with an acceleration.
∴ Retarding force F = ma = 1000 x (-7.5) = - 7500 N
(c) The region AB of the graph represents uniform motion with a constant velocity of 15 m s-1. Hence there is no force acting on the car during the part AB from 4 s to 8 s.
Q.4. A bullet of mass 10 g is fired with a rifle. The bullet takes 0.004 s to move through the barrel and leaves it with a velocity of 400 m s-1. Calculate the force exerted on the bullet by the rifle. [CBSE 2016]
Ans. Here mass of bullet m = 10 g = 0.01 kg, initial velocity of bullet u = 0. final velocity v = 400 m s-1 and time t = 0.004 s.
∴
∴ Force exerted on the bullet by the rifle.
F = ma = 0.01 x (1 x 105) = 1 x 103 N = 1000 N
Q.5. If the time taken to bring a ball to rest from a certain velocity v is reduced to half, what will be the changes in values of: [CBSE 2012, 2013, 2016]
(a) initial and final momentum
(b) change of momentum
(c) rate of change of momentum.
Ans. (a) Initial momentum remains unchanged at pi = mv.
Final momentum remains unchanged at Pf = 0.
(b) Change of momentum of ball remains unchanged at
(c) Initially rate of change of momentum but on reducing time to half, the new rate of change of momentum will be
Q.6. Name the physical quantities whose units are given below:
(i) kg m s-2 (ii) kg m s-1 (iii) N m2 kg-2 (iv) N (v) m s-2 (vi) m s-1 [CBSE 2015,2016]
Ans. The physical quantities having the given units are: (i) Force, (ii) momentum, (iii) gravitational constant G, (iv) force, (v) acceleration and (vi) speed or velocity.
Q.7. Define inertia. What determines the inertia of a body? Between a football and a rock of the same size which will have more inertia? [CBSE 2016]
Ans. Inertia of an object is its natural tendency to maintain its state of rest or of uniform linear motion.
Mass of an object is a measure of its inertia.
A rock of the same size has more inertia than a football because the mass of rock is much more.
Q.8. What's Newton's third law? [CBSE 2011, 2012, 2015, 2016]
Ans. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
Q.9. State the relation between the momentum of a body and the force acting on it. [CBSE 2016]
Ans. Rate of change of momentum of a body is equal to the external unbalanced force acting on it.
Q.10. What is the momentum of a body of mass 5 kg moving with a velocity of 0.20 m s-1? [CBSE 2016]
Ans. Momentum p = mass (m) x velocity (v) = 5 x 0.20 = 1.0 kg m s-1.
Q.11. A boy of mass 50 kg running 5 m/s jumps on to a 20 kg trolley traveling in the same direction at 1.5 m/s. Find their common velocity. [CBSE 2016]
Ans. Mass of boy = m1 = 50 kg
Mass of trolley = m2 = 20 kg
Initial velocity of boy = u1 = 5 ms-1
Initial velocity of trolley = u2 = 1.5 ms-1
Final velocity = v
= 4 ms-1
Q.12. Define SI unit of force. A force of 2N acting on a body changes its velocity uniformly from 2 m/s to 5 m/s in 10 s. Calculate the mass of the body. [CBSE 2016]
Ans. The SI unit of force is Newton (N). Force applied by a body is equal to 1 N if it accelerates a unit mass by 1 m/s2 in its direction.
Given: F = 2N, t = - 10 s, u = 2 ms-1,
v = 5 ms-1.
Since,
or
Q.13. A runner presses the ground with his feet before he starts his run. Identify action and reaction in this situation. [CBSE 2015]
Ans. The force applied by the runner on the ground is an action force. Reaction force of the ground acts on the runner and pushes him forward.
Q.14. Out of the four physical quantities associated with the motion of an object viz force, velocity, acceleration and momentum which one remains constant for all bodies large or small, undergoing a free fall? [CBSE 2015]
Ans. During free fall, the body accelerates. Thus velocity and momentum increase at every point of motion.
Acceleration remains constant. Thus force also remains constant (as F = ma).
Q.15. How are action-reaction forces related in magnitude and direction? [CBSE 2015]
Ans. They have the same magnitude but different directions.
Q.16. A stone released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. [CBSE 2015]
Ans. Given u = 0, s = 19.6 m, g = 9.8 m s-2.
Q.17. A stone released from the top of a tower of height 19.6 in. Calculate its final velocity just before touching the ground. [CBSE 2015]
(Take g = 9.8 m/s2)
Ans. Initial velocity, u = 0
Height of fall, h = 19.6 m
g = 9.8 ms-2
Final velocity,
Q.18. Car A of mass 1500 kg traveling at 25 m/s collides with another car B of mass 1000 kg traveling at 15 m/s in the same direction. After collision, the velocity of car A becomes 20 m/s. Calculate the velocity of car B after collision. [CBSE 2015]
Ans. Masses: ma = 1500 kg; mb = 1000 kg
Q.19. State the meaning of balanced force. [CBSE 2011, 2012, 2015]
Ans. Forces are said to be a balanced force if their resultant force is zero.
Q.20. When an athlete comes running from a distance, he is able to jump longer. Why? [CBSE2015]
Ans. Athlete gains momentum which helps him in taking a longer jump.
Q.21. A heavy person experiences more sideways push when a moving vehicle turns suddenly. Why? [CBSE 2015]
Ans. A heavy person experiences more sideways push when a moving vehicle turns suddenly because his inertia is more.
Q.22. Give the following reasons:
(a) A footballer kicks a ball, which rolls on the ground and after covering some distance comes to rest.
(b) Only the carrom coin at the bottom of a pile is removed when a fast moving striker hits it. [CBSE 2015]
Ans. (a) When football is rolling on the ground, a force of friction acts on it due to the ground in a direction opposite to its motion. As a result, the motion of football gets slowed down and after covering some distance it comes to rest.
(b) The carrom coin at the bottom of a pile comes in a state of motion due to force exerted by the striker on it. However, other coins of the pile remain intact due to their inertia of rest.
Q.23. A person is prone to more serious injuries when falling from a certain height on a hard concrete floor than on a sandy surface. Explain why. [CBSE 2011, 2012, 2015]
Ans. When a person falls from a height on a hard concrete floor, he immediately comes to a resting position. It means change in momentum is taking place in an extremely short time and consequently, force exerted by the floor on the person to destroy its momentum is extremely large. Hence, chances of more injuries. When a person falls on a sandy surface, the surface gets compressed downward and it increases the time of fall. As a result, the same change in momentum force exerted by sandy surfaces on the person is less and chances of him being hurt are less.
Q.1. (a) State the reason why a bullet of small mass fired from a gun kills a person.
(b) A bullet of mass 4 g fired with a velocity of 50 m s-1 enters a wall up to a depth of 10 cm. Calculate the average resistance offered by the wall.
(c) How will the depth of penetration into the wall change if a bullet of mass 5 g strikes against it with a velocity of 40 m s-1? Give reason to justify your answer. [CBSE 2016]
Ans. (a) Although the mass of a bullet is small, its momentum is very large on account of its extremely large velocity. When a bullet hits a person this large momentum is transferred to the person within a very short time and thus produces a large force and may kill the person.
(b) Here mass of bullet m = 4 g = 0.004 kg, initial velocity u = 50 m s-1, final velocity v = 0 and distance covered in wall s = 10 cm = 0.1 m.
As per equation v2 - u2 = 2 as, the acceleration
Average resistance force offered by wall F = ma = 0.004 x (- 12500) = - 50 N
(c) Depth of penetration will decrease because initial velocity of bullet is less but magnitude of retardation due to the wall remains unchanged.
Q.2. If you are trying to push a heavy box on a horizontal surface, list various forces on the box. State the condition under which this box will start sliding on the surface. How will the magnitude of applied force required to move the box change if:
(i) The weight of the box is increased,
(ii) the surface on which the box is placed is made more rough? [CBSE 2016]
Ans. Various forces acting on the box are:
(i) Its weight ‘W' acting vertically downward, (ii) reaction force 'R’ due to horizontal, surface, (iii) force of push 'F’ and (iv) frictional force 'f’. Forces have been shown in Fig. 9.13.
The box will start sliding on the surface only if the pushing force F is greater than the frictional force f i.e.,F>f.
(i) If the weight of the box is increased, the magnitude of applied force must increase.
(ii) If the surface is made more rough, the magnitude of applied force must increase.
Q.3. (i) Define momentum. State its SI unit.
(ii) An object of mass 50 kg is accelerated uniformly from a velocity of 4 m/s to 8 m/s in 8 s. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object. [CBSE 2015]
Ans. The product of mass and velocity of a body is called momentum.
SI unit: kg ms-1
mass, m = 50 kg, velocities,
u = 4 ms-1 and v = 8 ms-1, time, t = 8 s.
Initial momentum, p1 = mu = 50 x 4
= 200 kg m/s
Final momentum, p2 = mv = 50 x 8
= 400 kg ms-1
Force and Laws of Motion class 9 NCERT physics long question:
(a) A cricket player lowers his hands while catching the ball.
(b) The vehicles are fitted with shockers.
(c) A karate player breaks the pile of tiles or bricks with a single blow.
(d) In a high jump athletic event, the athletes are allowed to fall either on a sand bed or cushioned bed.
(e) In a moving car, the drivers and other passengers are advised to wear seat belts.
(f) China and glassware are packed with soft materials.
(g) Athletes are advised to come to a stop slowly after finishing a fast race.
Answer:
(a) if a player does not lower his hands while catching the ball, the time to stop the ball is very small. So a large force has to be applied to reduce the velocity of the ball to zero or to change the momentum of the ball. When a player lowers his hands, the time to stop the ball is increased and hence less force has to be applied to cause the same change in the momentum of the ball. Therefore, the hands of the player are not injured.
(b) The vehicles are fitted with shockers (i.e., springs)
The floor of a vehicle is cónnected to the lower part of the vehicle by springs or shocks. When the vehicle moves over a rough road, the force due to jerks is transmitted to the floor of the vehicle through the shockers. The shockers increase the time of transmission of the force of the jerk to reach the floor of the vehicle. Hence less force is experienced by the passengers in the vehicles.
(c) A karate player can break a pile of tiles with a single blow of his hand because he strikes the pile of tiles with his hand very fast, during which the entire linear momentum of the fast-moving hand is reduced to zero in a very short interval of time. This exerts a very large force on the pile of tiles which is sufficient to break them, by a single blow of his hand.
(d) In a high jump athletic event., the athletes are allowed to fall either on a sand bed or cushioned bed: This is done to increase the time of athletes fall to stop after making the high jump, which decreases the rate of change of linear momentum and decreases the impact.
(e) In a moving car, the drivers and other passengers are advised to wear seat belts: When brakes are applied suddenly, the passengers of the car fall forward due to the inertia of motion. The seat belt worn by passengers of the car prevents them from falling forward suddenly. This enables the entire linear momentum of the passengers to reduce to zero over a long interval of time, hence it prevents injury.
(f) China and glassware are packed with soft material: China and glassware are wrapped in paper before packing to avoid breakage while transporting. During transportation, there may be collisions due to ta jerks of the packed wares. Soft material like paper slows down their rate of change of linear momentum. As a result, the impact is reduced and items are not broken.
(g) Athletes are advised to come to stop slowly after finishing a fast race: By doing so, he decreases the rate of change of linear momentum by increasing the time interval and hence, reducing the impact, which reduces injury.
Explain
(a) How do we swim?
(b) Why does a gun recoil?
(c) It is difficult to walk on sand or ice.
(d) The motion of the rocket.
(e) Why does a fireman struggle to hold a hose-pipe?
(f) Rowing of a boat.
(g) When a man jumps out from a boat, the boat moves backward.
(h) Walking of a person.
Answer:
(a) While swimming, a swimmer pushes the water backward with his hands (i.e., he applies force in the backward direction, which is known as action.) The reaction offered by the water to the swimmer pushes him forward.
(b) Recoiling of a gun: When a bullet is fired from a gun, it exerts a forward force on the bullet and the bullet exerts an equal (in magnitude) and opposite (in direction) force on the gun. Due to the high mass of the gun, it moves a little distance backward and gives a backward jerk to the shoulder of the gunman.
(c) It is difficult to walk on sand or ice: When our feet press the sandy ground in the backward direction, the sand gets pushed away and as a result, we get only a small reaction (forward) from the sandy ground making it difficult to walk.
(d) Rocket propulsion Before firing the rocket, the total linear momentum of the system is zero because the rocket is in the state of rest. When it is fired, chemical fuels inside the rocket are burnt and the hot gasses are passed through a nozzle with great speed. According to the law of conservation of linear momentum, the total linear momentum after firing must be equal to zero. As the hot gasses gain linear momentum to the rear on leaving the rocket, the rocket acquires equal linear momentum in the upward direction i.e., opposite direction.
(c) A fireman has to make a great effort to hold a hose-pipe to throw a stream of water on the fire to extinguish it. This is because the stream of water rushing through the hose-pipe in the forward direction with a large speed exerts a large force on the hose-pipe in the backward direction which is known as the reaction force. This reaction force tends to move the hose-pipe in the backward direction. Therefore, a fireman struggles to hold the hose-pipe strongly to keep it at rest.
(f) Rowing of a boat: The boatman during the rowing of a boat pushes the water backward with oars (this is an action of the boatman). According to the third law of motion, water exerts an equal (in magnitude) and opposite (in direction) push on the boat which moves it forward (this is a reaction by water). Thus, the harder the boatman pushes back the water with oars (i.e., greater is the action), greater is the reaction force exerted by water, and faster the boat moves in the forward direction.
(g) When a man jumps out from a boat, the boat moves backward: When the passengers start jumping out of a rowing boat, they push the boat backward with their feet. The boat exerts an equal (in magnitude) and opposite (in direction) force on passengers in the forward direction which enables them to move forward.
(h) Walking of a person: When a person walks on the ground, he pushes the ground with his foot in the backward direction by pressing the ground. This push is known as the action, According to Newton’s third law of motion, an equal and opposite reaction acts on the foot of the person by the ground. This reaction (force) of the ground on the person pushes him forward.
During a cricket match, a new player Ayush injured his hand while catching the ball. His friend Rudra advised him to catch the ball by lowering his hands backward. When Ayush got another chance to catch the ball, he successfully caught the ball without injuring his hands.
Answer the following questions:
(a) A cricket player lowers his hands while catching the ball. Explain why.
(b) Write down the values shown by Rudra.
(a) If a player does not lower his hands while catching the ball, the time to stop the ball will be very small. So a large force has to be applied to reduce the velocity of the ball to zero or to change the momentum of the ball. When a player lowers his hands, the time to stop the ball is increased and hence less force has to be applied to cause the same change in momentum of the ball. Therefore, the hands of the player are not injured.
(b) Rudra is a good person as he helped his friend Ayush. He is a knowledgeable person.
During servicing his bike Aman advised the mechanic to oil the shockers for its proper functioning.
Answer the following questions:
(a) The vehicles are fitted with shockers. Explain why.
(b) Write down the values shown by Aman.
(a) The floor of a vehicle is connected to the lower part of the vehicle by springs or shocks. When a vehicle moves over a rough road, the force due to jerks is transmitted to the floor of the vehicle through the shockers. The shockers increase the time of transmission of the force of the jerk to reach the floor of the vehicle. Hence less force is experienced by the passengers in the vehicles.
(b) Aman is an intelligent and careful person.
Ranjan advised his son Ayush to wear a seat belt while driving the car.
Answer the following questions
(a) While driving the car, the drivers and other passengers are advised to wear seat belts. Explain why.
(b) Write down the values shown by Ranjan.
Answer:
(a) When brakes are applied suddenly, the passengers of the car fall forward due to the inertia of motion. The seat belt worn by passengers of the car prevents them from falling forward suddenly. This enables the entire linear momentum of the passengers to reduce to zero over a long interval of time, hence it prevents injury.
(b) Ranjan is a noble, intelligent, and careful person.
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