A person throws a ball (A) vertically upwards with 40 m/sec. After 4 sec, he throws another identical ball (B) upwards with same speed. If the collision between balls is head-on perfectly elastic, the speed of ball (A) just after collision is

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10 m/sec

20 m/sec

40 m/sec

answer is C.

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Detailed Solution

$v^{2} = u^{2} + 2 gh \Rightarrow 0^{2} = (40^{2}) - 2 (10) (h) \Rightarrow h = 80 m$

where h is maximum height reached by ball A
$v = u - gt \Rightarrow t = 4 \sec t^{'} = \frac{80}{u_{A} + u_{B}} = 2 \sec u_{A}^{'} = {gt}^{'} = 20 m / s u_{B}^{'} = 40 - 2 \times 10 = 20 m / s$
Thus the speed of each ball gets interchanged (for equal mass & elastic collision)

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