A bomb falling freely bursts after 10 sec. (g=10ms-2) into two fragments of masses in the ratio of 2:1. The velocity of heavier fragment immediately after the explosion is 200 ms-1 vertically downwards. The velocity of the lighter fragment immediately after the explosion is

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50ms-1 upward

75ms-1 downward

100ms-1 upward

400ms-1 upward

answer is C.

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Detailed Solution

u=gt =10 × 10 m/s = 100 m/s ( downward)Conserving momentum , mv + 2m ×200 = 3m ×100 ⇒v = - 100 m/s (upward )

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