“Maturity is
often more absurd than youth and very frequently is most unjust to youth.”
– Thomas
A. Edison
Today’s post covers a few multiple choice
practice questions related to one dimensional kinematics meant for AP Physics C
aspirants.
(1) Water drops from a leaking overhead tank
falls on the ground 5 m below, at regular intervals, the 11th drop just beginning to fall when the first
drop strikes the ground. What will be the height of the 9th drop
when the first drop strikes the ground? (Acceleration due to gravity = 10 ms–2)
(a) 4.9 m
(b) 4.8 m
(c) 4.6 m
(d) 4.4 m
(e) 4.2 m
The time t
taken by a drop to fall through 5 m is given by the relevant equation of
motion,
5
= 0 + ½ gt2
Since g
= 10 ms–2 the above equation gives t = 1 second
When the first drop strikes the ground after
falling freely for one second, the 9th drop has fallen freely for one-fifth of one second, as is evident
from the adjoining figure (remembering that the drops fall at regular intervals
of time). Therefore, the 9th drop has fallen through a distance s given by
s = 0 + ½ g(0.2)2 = (½)×10×0.04 = 0.2 m
Therefore the height
of the 9th
drop when the first drop strikes the ground is 5 m – 0.2 m = 4.8 m.
(2) An object projected vertically upwards with initial velocity u attains maximum height in 5 s. The
ratio of the distance traveled by the object in the 1st second and the
6th second is (Acceleration due to
gravity = 10 ms–2)
(a) 6 : 1
(b) 8 : 1
(c) 9 : 1
(d) 10: 1
(e) 11 : 1
The velocity of projection (u)
of the object is given by the equation of uniformly accelerated linear motion,
0 = u – gt
[We have used the equation, v = u +
at with usual notations].
Substituting for g and t we have
0 = u – 10×5
Therefore u = 50
ms–1
The distance s1 traveled by the object in
the 1st second is given by
s1 = (50×1) – (½ ×10×12)
= 45 m
[We have used the equation, s = ut
+ ½ at2 with usual
notations]
At the end of 5 seconds the object is at the highest point
of its trajectory where its velocity is zero.
Therefore, the distance s6 traveled by the object during the next
one second (6th second) is given by
s6 = 0 + ½ ×10×12 = 5 m.
[We have used the
equation of motion, s = ut + ½ gt2 with usual notations]
The ratio of the distance
traveled by the object in the 1st second and the 6th
second is
s1 : s6 = 45 : 5 = 9 : 1
(3) A ball projected vertically upwards with initial
velocity u reaches maximum height h in t
sec. What is the total time taken by the ball (from the instant of projection)
to reach a height h/4 while
returning?
(a) 1.75t
(b) 1.65t
(c) 1.5t
(d) 1.4t
(e) 1.3t
From the equation of motion, v2 = u2
+ 2as with usual notations, we have
for the upward journey
0
= u2 – 2gh
Therefore h = u2/2g
From the equation of motion, v = u + at with usual notations, we have for the
upward journey
0
= u – gt so
that u = gt
Substituting for u in the
expression for h we have
h = gt2/2 …………. (i)
We now use the equation of motion, s
= ut + ½ gt2 for the
free fall of the ball from the maximum height h to the height h/4.
Since distance of fall is 3h/4)
we have
3h/4 = 0 + ½ gt12 where t1 is the time of fall from
the maximum height h to the height h/4.
Substituting for h from
equation (i) we have
gt2/8
= gt12/2
This gives t1 =
t/2
The total time time taken by the ball (from the instant of projection) to
reach a height h/4 while returning is
t + t1 = t + (t/2) = 3t/2 = 1.5 t.
(4) The velocity-time graph of an object moving along the x-direction is
shown in the figure. What is the displacement of the object when it moves with
the maximum acceleration?
(a) 12 m
(b) 8 m
(c) 6 m
(d) 4.8 m
(e) 4.2 m
The object has maximum acceleration from 10 sec to 12 sec since the slope
of the velocity-time graph is maximum during this interval. The displacement is
given by the area under the velocity-time graph for the interval from 10 sec to
12 sec which is 8 m.
[The displacement of the object when it moves with the maximum acceleration can be calculated
using the equation, s = ut + ½ gt2
as well:
As is evident from the velocity-time graph, the maximum acceleration a is given by
a = Change of velocity/Time = (5 – 3)/(12
– 10) = 1 ms–2
Since the initial velocity u =
3 ms–1 and the time
interval t = 2 s, we have
s = (3×2) + (½)×1×22
= 8 m].
(5) A particle moving
along the x-axis is initially at the origin with velocity 2 ms–1. If the acceleration
a of the particle is given by a = 6t, the position of the particle after 4
seconds is
(a) 24 m
(b) 48 m
(c) 72 m
(d) 96 m
(e) 120 m
The velocity v of the particle is given by
v
= ∫adt = ∫6t dt = 3t2 + C where C is the constant of integration which we can find from the initial
condition.
Initially (at time t = 0) the particle has velocity 2 ms–1. Therefore from the above expression
for velocity we have
C = 2 ms–1
Thus the expression for velocity becomes
v = 3t2 + 2
The position x
of the particle is given by
x = ∫v
dt = ∫(3t2 + 2)dt = t3 + 2t + C’ where C’ is the constant of integration in this case.
Initially (at t
= 0) since the particle is at the origin (where x = 0) we obtain (from the above expression for v)
C’ = 0
Therefore, the expression for the position x becomes
x = t3 + 2t
The position x’
at t = 4 seconds is given by
x’ = 43 + (2×4) = 72 m
You can find a few
more useful questions (with solution) in this section here.
