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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Tuesday, October 2, 2012

### AP Physics B & C - Newton’s Laws - Multiple Choice Practice Questions

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend."
– Sir Isaac Newton

Let us discuss a few multiple choice practice questions involving Newton’s laws of motion.
(1) Coefficient of static friction between a horizontal conveyor belt and a 3 kg packet placed on it is 0.2. What is the maximum permissible acceleration of the conveyor belt so that the packet will not slip?
(a) 0.2 ms–2
(b) 0.6 ms–2
(c) 2 ms–2
(d) 3 ms–2
(e) 6 ms–2
It is the frictional force between the packet and the conveyor belt that keeps the packet stationary with respect to the belt when the belt moves. The maximum value of this frictional force is μsmg where μs is the coefficient of static friction, m is the mass of the packet and g is the acceleration due to gravity.
If the maximum permissible acceleration of the belt is amax, we have
mamax = μsmg
Therefore, amax = μsg = 0.2×10 = 2 ms–2
[Note that the mass of the packet is given just to serve the purpose of a distraction]

(2) An empty open box of mass M (Fig.) is sliding along a smooth horizontal surface with constant velocity ‘v1’. An object of mass M/4 is dropped vertically into the box and the box continues to move forward with velocity v2. After some time another object of mass 3M/4 is dropped vertically into the box and then the box moves forward with velocity v3. Which one among the following is correct?
(a) v1 = v2 = v3
(b) v2 =  v1/4
(c) v2 = 3 v1/4
(d) v3 =  v1/4
(e) v3 =  v1/2
The objects dropped vertically into the box will not contribute any horizontal momentum to the box. But they can change the velocity of the box obeying the law of conservation of momentum. The momentum of the system throughout its motion is equal to the initial momentum Mv1 of the box and we have
Mv1 = [M + (M/4) + (3M/4)]v3
Therefore v3 =  v1/2
Therefore, the correct option is (e).
[We have Mv1 = [M + (M/4)]v2 and this yields v2 = 4v1/5. But we don’t have this answer in the given options]

The following questions are specifically meant for AP Physics C aspirants:
(3) A bullet of mass m voving horizontally with speed v strikes a wooden block of mass M which rests on a horizontal surface. After the collision the block and the bullet move together and come to rest after moving through a istanc d. The coefficient of kinetic friction between the block and the horizontal surface is
(a) m2v2/[2gd(M + m)2]
(b) mv2/[2gd(M + m)]
(c) m2/[2(M + m)2]
(d) (M + m)v2/2gdM
(e) 2m2v2/[gd(M + m)2]
The common velocity (of the bullet and wooden block) after the impact (vf) is given by the law of conservation of momentum:
mv + M×0 = (M + m)vf
Therefore, vf = mv/(M + m)
The kinetic energy possessed by the the bullet and wooden block, just after the impact, is utilized in doing work against friction while moving through the distance d.
Therefore we have
½ (M + m)vf2 = μkRd where μk is the coefficient of kinetic friction and R is the normal force which is the weight of the bullet and the wooden block.
Substituting for vf  and R, we have
½ (M + m) [mv/(M + m)]2 = μk(M + m)gd
This gives μk = m2v2/[2gd(M + m)2]
(4) A balloon with its payload has mass 40 kg. It is found to descend with acceleration equal to g/4 where g is the acceleration due to gravity. Keeping the volume of the balloon unchanged, the payload is reduced by M kg. The balloon is then found to ascend with acceleration of the same magnitude g/4. What is the value of M? (Ignore viscous drag force)
(a) 4 kg
(b) 6 kg
(c) 8 kg
(d) 16 kg
(e) 20 kg
If the acceleration due to gravity is g we have
(40g F)/40 = g/4 where F is the upthrust on the balloon.
Or, g – (F/40) = g/4  from which F = 30g
If the total mass of the balloon while ascending is x we have
(30g xg)/x = g/4
Or, (30/x) – 1 = ¼
Or, 30/x = 5/4
Therefore x = 24 kg
The reduction (M) in the mass of the payload is given by
M = 40 kg – 24 kg = 16 kg