Intext Questions

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 rupees coin and a one-rupee coin?
Sol.     (a) A stone.
(b) A train
(c) A five rupees coin.Explanation – Inertia is associated with mass. Objects having more mass have more inertia.

Q.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.
Sol.     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 playerSecond case – Second playerThird case – GoalkeeperFourth case – Goalkeeper

Q.3     Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Sol.     The answer of this cause lies behind the Newton’s First Law of Motion. Initially, leaves and tree both are in rest. But when the tree is shaken vigorously, tree comes in motion while leaves have tendency to be in rest. Thus, because of remaining in the position of rest some of the leaves may get detached from a tree if we vigorously shake its branch.

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?
Sol.     In a moving bus, passengers are in motion along with bus. When brakes are applied to stop a moving bus, bus comes in the position of rest. But because of tendency to be in the motion a person falls in forward direction.Similarly, when a bus is accelerated from rest, the tendency to be in rest, a person in the bus falls backwards.

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Q.1     If action is always equal to the reaction, explain how a horse can pull a cart.
Sol.     Horse pushes the ground in backward direction. In reaction to this, the horse moves forward and cart moves along with horse in forward direction.In this case when horse tries to pull the cart in forward direction, cart pulls the horse in backward direction, but because of the unbalanced force applied by the horse, it pulls the cart in forward direction.

Q.2     Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Sol.     When large amount of water is ejected from a hose at a high velocity, according to Newton’s Third Law of Motion, water pushes the hose in backward direction with the same force. Therefore, it is difficult for a fireman to hold a hose in which ejects large amount of water at a high velocity.

Q.3     From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.
Sol.     Given,Mass of rifle (m₁) = 4 kgInitial velocity of rifle (u₁) = 0Mass of bullet (m₂) = 50 g = 50/1000 kg = 0.05 kgInitial velocity of bullet (u₂) = 0Final velocity (v₂) of bullet = 35 m/sFinal velocity [Recoil velocity] of rifle (v₁) = ? We
know that,
m1u1+m2u2=m1v1+m2v2
⇒4kg×0+0.5kg×0=4kg×v1+0.005kg×35ms−1
⇒0=4kg×v1+1.75kgm/s
⇒−4kg×v1=1.75kgm/s
⇒vl=1.75kgm/s−4kg=−0.4375ms≈−0.44m/s
Here negative sign of velocity of rifle shows that rifle moves in the opposite direction of the movement of bullet. Therefore, recoil velocity of rifle is equal to 0.44 m/s.

Q.4     Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.
Sol.      Given,Mass of first object (m₁) = 100 g = 100/1000 kg = 0.1 kgInitial velocity of first object (u₁) = 2 m/sFinal velocity of first object after collision (v₁) = 1.67 m/sMass of second object (m₂) = 200 g = 200/1000 kg = 0.2 kgInitial velocity of second object (u_{2) = 1 m/s}Final velocity of second object after collision (v₂) = ? We know that,

m1u1+m2u2=m1v1+m2v2
⇒0.1kg×2ms−1+0.2kg×1ms−1=0.1kg×1.67ms−1+0.2kg×v2
⇒0.2kgms−1+0.2kgms−1=0.167kgms−1+0.2kg×v2
⇒0.4kgms−1−0.167ms−1=0.2kg×v2
⇒v2=0.233kgms−10.2kg=1.165ms−1
Thus, velocity of the second object after collision = 1.165 ms−1

Exercise

Q.1     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.
Sol.     When a net zero external unbalanced force is applied on the body, it is possible for the object to be travelling with a non-zero velocity. In fact, once an object comes into motion and there is a condition in which its motion is unopposed by any external force; the object will continue to remain in motion. It is necessary that the object moves at a constant velocity and in a particular direction.

Q.2     When a carpet is beaten with a stick, dust comes out of it. Explain.
Sol.     Beating of a carpet with a stick; makes the carpet come in motion suddenly, while dust particles trapped within the pores of carpet have tendency to remain in rest, and in order to maintain the position of rest they come out of carpet. This happens because of the application of Newton’s First Law of Motion which states that any object remains in its state unless any external force is applied over it.

Q.3     Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Sol.     Luggage kept on the roof of a bus has the tendency to maintain its state of rest when bus is in rest and to maintain the state of motion when bus is in motion according to Newton’s First Law of Motion. When bus will come in motion from its state of rest, in order to maintain the position of rest, luggage kept over its roof may fall down. Similarly, when a moving bus will come in the state of rest or there is any sudden change in velocity because of applying of brake, luggage may fall down because of its tendency to remain in the state of motion. This is the cause that it is advised to tie any luggage kept on the roof a bus with a rope so that luggage can be prevented from falling down.

Q.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.
Sol.     (c) There is a force on the ball opposing the motion. Explanation: When ball moves on the ground, the force of friction opposes its movement and after some time ball comes to the state of rest.

Q.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 tonnes (Hint: 1 tonne = 1000 kg.)
Sol.     Given,  Initial velocity of truck (u) = 0 (Since, truck starts from rest) Distance travelled, s = 400 m
Time (t) = 20 Acceleration (a) = ?
We know that,
s=ut+12at2
⇒400m=0×20s+12×a×(20s)2
⇒400m=12×a×400s2
⇒400m=a×200s2
⇒a=400m200s2=2ms−2
Force acting upon truck:
Given mass of truck = 7 ton = 7 X 1000 kg = 7000 kg
We know that, force, P = m x a
Therefore, P = 7000 kg x 2 m s – 2
Or, P = 14000 Newton
Thus, Acceleration = 2 m s – 2 and force acting upon truck in the given condition = 14000 N

Q.6     A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Sol.     Given,
Mass of stone = 1 kg
Initial velocity, u = 20 m/s
Final velocity, v = 0 (as stone comes to rest)
Distance covered, s = 50 m
Force of friction = ?
We know that,
v2=u2+2as
⇒(0)2=(20m/s)2+2a×50m
⇒−400m2s−2=100ma
⇒a=−400m2s−2100m=−4ms−2
Now, we know that, force, F = mass x acceleration
Therefore, F = 1 kg X -4ms-2
Or, F = -4ms-2
Thus, force of friction acting upon stone = -4ms-2. Here negative sign shows that force is being applied in the opposite direction of the movement of the stone.

Q.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.
Sol.     Given,
           force of engine = 40000 NForce of friction = 5000 N
           Mass of engine = 8000 kg
           Total weight of wagons = 5 x 2000 kg = 10000 kg
           (a) The net accelerating force
           = Force exerted by engine – Force of fricition
           = 40000 N – 5000 N = 35000 N
           (b) The acceleration of the trainWe know that, F = mass x acceleration
Or, 35000 N = (mass of engine + mass )
Or a=35000N18000kg=1.94ms−2 of 5 wagons X a
           Or, 35000 N = (8000 kg + 10000 kg) X a
           Or, 35000N = 18000 kg X a
           (c) The force of wagon 1 on wagon 2
           Since, net accelerating force = 35000 N
           Mass of all 5 wagons = 10000 kg
           We know that, F = m x a
           Therefore,
a=35000N10000kg=3.5ms−2
Therefore, acceleration of wagons = 3.5 ms−2
Thus, force wagon 1 on 2 = mass of four wagons × acceleration
Or, F = 4 × 2000 kg × 3.5 ms−2
Or, F = 8000 kg × 3.5 ms−2
Or, F = 28000N
Thus, (a) The net accelerating force = 35000N
(b) The acceleration of train = 1.944 ms−2
(c) The force of wagon 1 on 2 = 280000N 35000N = 10000 kg x a