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Question

A ball of mass 'm', moving with a velocity 'v' along X-axis, strikes another ball of mass '2m' kept at rest. The first ball comes to rest after collision and the other breaks into two equal pieces. One of the pieces starts moving along Y-axis with a speed v1. What will be the velocity of the other piece?


A

v2+v21

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B

2v2+v21

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C

v2v21

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D

v2+2v21

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Solution

The correct option is A

v2+v21


Can you recall,

vCOM=m1v1+m2v2m1+m2

Considering both balls as one system Initially the velocity of COM of system

vCOM=mv^im+2m=v3^i

After collision,

Let the velocity of other piece be v2=vx^i+vy^j

As there is no net external force, v will not change and fit ball of mass 'm' comes to rest.

(vCOM)x=m1(v1)x+m2(v2)xm+m1+m2

v3=0+mvk3m

vx=v ......... (1)

Similarly,

(vCOM)y=m1(v1)y+m2(v2)ym+m1+m2

0 = mv1+mvy(2m+m)vy=v1

Therefore, v2=v^iv1^j

|v2|=v2+v21


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