"Whenever you are confronted with an
opponent, conquer him with love."
– Mahatma Gandhi
Let us discuss a few interesting multiple choice practice questions on one dimensional motion. Here are some questions beneficial for AP Physics B as well as AP Physics C aspirants:
(1) The adjoining
figure shows the velocity time graph of an object. Total displacement suffered
by the object during the interval when it has non-zero acceleration and
retardation is
(a) 80 m
(b) 70 m
(c) 60 m
(d) 40 m
(e) 30 m
The object has non-zero
acceleration and retardation during the time intervals from 5 sec to 15 sec and
from 20 sec to 30 sec. The total area under the velocity time graph during
these intervals gives the required displacement.
Displacement from 5
sec to 15 sec = 30 m
Displacement from
20 sec to 30 sec = 40 m
Therefore, total
displacement = 70 m
(2) Successive
positions (x) of an object (moving
from left to right) at equal time intervals are shown in the following figure:
Which one among the
following position-time graphs best represents the motion of the object?
Change of position
is slowest in the beginning and in the end. The motion is best represented by
graph (c).
[Note that graph (d)
is not the answer since the change of position in the beginning and in the end
is shown as fastest in it]
(3) Which one among
the following velocity-time graphs best represents the motion of the object
mentioned in question no.(2)?
The graph (c) is
the answer.
The following questions are meant for AP Physics C aspirants:
(4) A particle projected
vertically upwards attains the maximum height h in time t. While
returning from the highest point it takes an additional time t1 to fall to the height h.2. If air resistance is negligible, how
is t1 related to t?
(a) t1 = t/2
(b) t1 = √(t/2)
(c) t1 = t/√2
(d) t1 = t/√3
(e) t1 = t/3
For the upward
motion we have
0 – u2 = – 2gh
…………(i)
[We have used the
equation of motion, v2 – u2 = 2as. The sign of the
gravitational acceleration g is negative since it is opposite to the
direction of the velocity of projection u]
Using the equation
of motion, v = u + at we have
0 = u – gt from which u = gt
Substituting this
value of u in Eq (i), we have
g2t2 = 2gh
Therefore, t =√(2h/g)
……..(ii)
For the fall
through h/2 from the highest point we
have
h/2
= 0×t1 + ½ gt12
[We have used the
equation of motion, s = ut + ½ gt2]
Therefore, t1 =√(h/g)
Comparing this with
the value of t given in Eq (ii) we
obtain t1 = t/√2.
(5) A ball projected
vertically up has the same vertical displacement h at times t1
second and t2 second. If
air resistance is negligible, the maximum height reached by the ball is
(a) gt1t2/2
(b) g(t12 + t22)/2
(c) g(t1 + t2)2/4
(d) g(t1 + t2)2/8
(e) 2g(t1 + t2)2
The time taken by
the ball to move from height h to the
top of its trajectory and back to the height h is t2 – t1.
Therefore, the time taken to move from height h to the top of the trajectory is (t2 – t1)/2.
The total time taken for the upward journey (from ground to the top most point) is evidently t1 + t2 – t1)/2 = (t1 + t2)/2.
Time taken for the return journey (from the top most point
to the ground) also is equal to (t1 + t2)/2.
Therefore, the maximum height H (using
the equation, s = ut + ½ gt2) is given by
H = 0 + ½ g [(t1 + t2)/2]2
Or, H = g(t1 + t2)2/8
Questions on kinematics were discussed earlier on
this site. You can access them either by clicking on the label ‘kinematics’
below this post or by trying a search for ‘kinematics’ using
the search box provided on this page.