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) 2*u*

(c) *u *– *gt*

(d) √(*u*^{2} – *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 *t*^{2} since s = *ut* + ½ *gt*^{2} 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 *v*^{2} = *u*^{2} + 2*as* where *u* is the initial velocity, *v* is the final velocity after suffering a displacement *s* and *a* is the acceleration.

Therefore we have

*u*^{2}/4 = *u*^{2} + 2*a *×* *0.4

Or, – 3*u*^{2}/4 = 2*a *×* *0.4 …………..(i)

If the bullet comes to rest after penetrating a further distance of *s*_{1}* *we have

* *0* = u*^{2}/4 + 2*a *× *s*_{1}_{}

Or, – *u*^{2}/4 = 2*a *× *s*_{1} …………….(ii)

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

3 = 0.4/*s*_{1} from which *s*_{1} = 0.4/3 m = ** ^{40}/_{3} cm**.

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

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