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1.	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.

2.	From a rifle of mass `4kg`, a bullet of mass `50g` is fired with an initial velocity of `35m//s`. Calculate the initial recoil velocity of the rifle.
Answer» Mass of the rifle, `m_(1) = 4 kg` Mass of the bullet, `m_(2) = 50g = 0.05 kg ` Recoil velocity of the rifle `= v_(1)` Bullet is fired with an initial velocity, `v_(2) = `35 m/s Initially, the rifle is at rest. Thus, its initial velocity, v = 0 Total initial momentum of the rifle and bullet system `= (m_(1) + m_(2))v = 0 ` Total momentum of the rifle and bullet system after firing: `= m_(1)v_(1) + m_(2)v_(2) = 4(v_(1)) + 0.05xx35 = 4v_(1) + 1.75 ` According to the law of conservation of momentum: Total momentum after the firing = Total momentum before the firing `4v_(1) + 1.75 = 0 ` ` v_(1) = – 1.75//4 = – 0.4375 m//s` The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s

3.	A motorcar of mass `1200kg` is moving along a straight line with a uniform velocity of `90 km//h`. Its velocity is slowed down to `18 km//h` in `4s` by an unbalanced external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.
Answer» Correct Answer - `5 m s ^(-2); 2400 kg ms ^(-1)`;6000 N 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 `implies5 = 25 + a (4)` ` impliesa = - 5 m//s^(2)` Negative sign indicates that its a retarding motion i.e. velocity is decreasing. Change in momentum = mv – mu = m (v – u) `= 1200 (5 – 25) = - 24000 kg ms^(-1) ` `:.` Force = Mass `xx` Acceleration `= 1200 xx - 5 = - 6000 N ` Acceleration of the motor car `= - 5 m//s^(2)` Change in momentum of the motor car `= - 24000 kg ms^(-1)` Hence, the force required to decrease the velocity is 6000 N. (Negative sign indicates retardation, decrease in momentum and retarding force)

4.	The velocity time graph of a ball of mass `20g` moving along a straight line on a long table is given in (figure) How much force does the table exert on the ball to bring it to rest?
Answer» The initial velocity of the ball is 20 cm `s^(-1)`. Due to the frictional force exerted by the table, the velocity of the ball decreases down to zero in 10 s. Thus, u = 20 cm `s^(–1)`, v = 0 cm `s^(-1)` and t = 10 s. Since the velocity-time graph is a straight line, it is clear that the ball moves with a constant acceleration. The acceleration a is `a=(v-u)/(t)` =0 cm `s^(-1)` -20 cm `s^(-1)//10 s` =-2 cm `s^(-2)`=-0.02 m`s^(-2)` The force exerted on the ball F is , `F=ma=(20//1000) kg xx (-0.02ms^(-2))` `= -0.0004 N`. The negative sign implies that the fricitonal force exerted by the table is opposite to the direction of motion of the ball.

5.	Two object, 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?
Answer» Correct Answer - `0 ms^(-1)` Mass of one of the objects, `m_(1) = 1.5 kg ` Mass of the other object, `m_(2) = 1.5 kg` Velocity of m1 before collision, `u_(1) = 2.5 m//s` Velocity of` m_(2)` , moving in opposite direction before collision, `u_(2) = -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,`= (m_(1) + m_(2)) v=m_(1)u_(1) + m_(2) u_(2) ` Or,` = (1.5 + 1.5) v =1.5xx2.5 + 1.5 xx(-2.5)` [negative sign as moving in opposited direction ] Or, `v=0ms^(-1)`

6.	How much momentum will a dumb-bell of mass `10kg` transfer to the floor if it falls a height of `80 cm`? Take its downward acceleration to be `10 m//s^(2)`.
Answer» Correct Answer - 40 kg 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//s^(2) ` 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: ` v^(2)= u^(2) + 2as` ` v^(2)= 0 + 2 (10) 0.8 ` v= 4 m/s Hence, the momentum with which the dumbbell hits the floor is = mv ` = 10 xx 4 kg ms^(-1)` ` = 40 kg ms^(-1)`

7.	A force of `5N` gives a mass `m_(1)`, an acceleration of `10m//s^(2)`, and a mass `m_(2)`, an acceleration of `20m//s^(2)`. What acceleration would it give if both the masses were tied together?
Answer» From Eq. (9.4) we have `m_(1)=F//a_(1)`, and `m_(2)=F//a_(2)`. Here , `a_(1)=10 ms^(-2)`, `a_(2)=20 ms^(-2)` and `F=5 N`. Thus , `m_(1) 5N//10 ms^(-2)=0.50 kg`: and `m_(2) =5 N//20 ms^(-2)=0.25 kg` If the two masses were tied together , the total mass , m would be m=0.50 kg +0.25 =0.75 kg. The acceleration , a produced in the combined mass by the 5 N force would be , `a=F//m=5N//0.75 kg=6.67 ms^(-2)`.

8.	An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer» Yes, 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 tand the velocity of air. The net force on the drop is zero.

9.	Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer» When the bus 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 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 backwardd. So, it is advised to tie any luggage kept on the roof of a bus with a rope.

10.	A stone of `1kg` is thrown with a velocity of `20 ms^(-1)` across the frozen surface of lake and comes to rest after travelling a distance of `50m`. What is the force of friction between the stone and the ice?
Answer» Correct Answer - `-4 N` 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, `v^(2)-u^(2)=2as` Or, `0-20^(2)=2axx50`, Or , `a=-4 ms^(-2)` Force of friction, F=ma= -4N

11.	A hockey ball of mass `200 g` travelling at `10 m//s` is struck by a hockey stick so as to return it along its original path with a velocity of `5 m//s`. Calculate the change in momentum of the hockey ball by the force applied by the hockey stick.
Answer» Correct Answer - 3 kg m `s^(-1)` Mass of the hockey ball, m= 200 g = 0.2 kg Hockey ball travels with velocity, `v_(1)= 10 m//s` Initial momentum `= mv_(1)` Hockey ball travels in the opposite direction with velocity, `v_(2)=` -5 m/s Final momentum = `mv_(2)` Change in momentum `= mv(1)- mv_(2)= 0.2 [10 - (-5)] = 0.2 (15) = 3 kg ms^(-1)` Hence, the change in momentum of the hockey ball is 3 kg `m s^(-1) `.

12.	Which of the following has more interia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train (c) a five-repees 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 for 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, 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, inertia of the five rupee coin is greater than that of the one-rupee coin.