Kinematics- KEAM (Engineering) 2009 Questions on One dimensional Motion

Today we will discuss some multiple choice questions on one dimensional motion which appeared in Kerala Engineering Entrance 2009 question paper:

(1) A ball is thrown up vertically with speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is

(a) u

(b) 2u

(c) u – gt

(d) √(u² – gt)

(e) gt

If you know the concept of relative velocity correctly, this question will be quite simple for you.

The relative velocity of A with respect to B is given by

V_AB = V_A – V_B where V_A and V_B are respectively the velocities of A and B (with respect to the common frame of reference chosen).

The velocity of A and B at time t are respectively u – gt and – gt, taking upward quantities positive and downward quantities negative.

The speed of A relative to B is u – gt – (– gt) = u [Option (a)].

(2) A body is falling freely under gravity. The distances covered by the body in the first, second and third minutes of its motion are in the ratio

(a) 1 : 4 : 9

(b) 1 : 2 : 3

(c) 1 : 3 : 5

(d) 1 : 5 : 6

(e) 1 : 5 : 13

Since you require the ratio of distances, it is enough to consider the times in seconds (instead of minutes). Strictly, it must be mentioned that the body starts from rest. In the case of such a freely falling body, the distance covered is directly proportional to t² since s = ut + ½ gt² with usual notations where u = 0

The distances covered by the body in one second, two seconds and three seconds are in the ratio 1 : 4 : 9.

Therefore, the distances covered by the body in the first, second and third seconds are in the ratio 1 : (4 – 1) : (9 – 4) which is 1 : 3 : 5

(3) A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of

(a) ²²/₃ cm

(b) ⁴⁰/₃ cm

(c) ²⁰/₃ cm

(d) ²²/₅ cm

(e) ²⁶/₅ cm

We have v² = u² + 2as where u is the initial velocity, v is the final velocity after suffering a displacement s and a is the acceleration.

Therefore we have

u²/4 = u² + 2a × 0.4

Or, – 3u²/4 = 2a × 0.4 …………..(i)

If the bullet comes to rest after penetrating a further distance of s₁ we have

0 = u²/4 + 2a × s₁

Or, – u²/4 = 2a × s₁ …………….(ii)

Dividing Eq(i) by Eq(ii) we get

3 = 0.4/s₁ from which s₁ = 0.4/3 m = ⁴⁰/₃ cm.

[The above question can be worked out using the work energy principle as well].

You will find similar useful multiple choice questions on kinematics (with solution) here.