Physics-
General
Easy
Question
The ball A collides elastically with an identical ball B with a speed v. The ball B touches another identical ball C. The velocity of C, after the impact, is equal to
- v
- v/2
- v/4
- None of these
The correct answer is: v
Since the ball B & C touch each other, B&C act as a combined mass (let D) whose mass is 2m. Therefore,
conserving the momentum
we write
mv = mv1 + 2mv2
Þ v = v1 + 2v2 . . . (i)
Newton’s impact formula yields
v1 – v2 = -ev
putting e = 1 we obtain
-v1 + v2 = +v. . . (ii)
From (I) & (ii), 3v2 = 2v Þ v2 = 
v
\ The velocity of C is equal to the velocity of B as these balls move with same velocity just after the collision.
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