"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
No comments:
Post a Comment