"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

*μ*_{s}*mg*where*μ*_{s}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

*a*_{max}, we have*ma*

_{max}=

*μ*

_{s}

*mg*

Therefore,

*a*_{max}=*μ*_{s}*g*= 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 ‘v_{1}’. An object of mass*M*/4 is dropped vertically into the box and the box continues to move forward with velocity v_{2}. After some time another object of mass 3*M*/4 is dropped vertically into the box and then the box moves forward with velocity v_{3}. Which one among the following is correct?
(a) v

_{1}= v_{2}= v_{3}
(b) v

_{2}= v_{1}/4
(c) v

_{2}= 3 v_{1}/4
(d) v

_{3}= v_{1}/4
(e) v

_{3}= v_{1}/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*M*v_{1}of the box and we have*M*v

_{1}= [

*M +*(

*M*/4) +

*(3*

*M*/4)]v

_{3}

Therefore v

_{3}= v_{1}/2
Therefore, the correct option is (e).

[We have

*M*v_{1}= [*M +*(*M*/4)]v_{2}and this yields v_{2}= 4v_{1}/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)

*m*^{2}*v*^{2}/[2*gd*(*M + m*)^{2}]
(b)

*mv*^{2}/[2*gd*(*M + m*)]
(c)

*m*^{2}/[2(*M + m*)^{2}]
(d) (

*M + m*)*v*^{2}/2*gdM*
(e)

*2**m*^{2}*v*^{2}/[*gd*(*M + m*)^{2}]
The
common velocity (of the bullet and wooden block) after the impact (

*v*_{f}) is given by the law of conservation of momentum:*mv*+

*M*×0 = (

*M + m*)

*v*

_{f}

Therefore,

*v*_{f}=*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*)*v*_{f}^{2}=*μ*_{k}*Rd*where*μ*_{k}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

*v*_{f }and*R*,*we have*
½
(

*M + m*)*[**mv/*(*M + m*)]^{2}*= μ*_{k}(*M + m*)*gd*
This gives

*μ*_{k}=*m*^{2}*v*^{2}/[2*gd*(*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
(40

*g*–*F*)/40 =*g/*4*where**F*is the upthrust on the balloon.
Or,

*g*– (*F*/40) =*g/*4 from which*F =*30*g*
If the total mass
of the balloon while ascending is

*x*we have
(30

*g*–*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

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