Question

In the shown figure, the heavu block of mass `2 kg` rests on the horizontal surface and the lighter block of mass `1 kg` is dropped from a height of `0.9 m`. At the instant the string gets taut, find the upwards speed (in `m//s` ) of the heavy block.

Answer 1

Correct Answer - `2 m//s`
Velocity of lighter block at the instatn the string just gets taut `v = sqrt(2gh) = sqrt(2 xx 10 xx 1.8) = 6 m//s`
Now by impulse `-` momentum theorem, let common speed be `v_(1)` then `(2 + 1) v_(1) = (1) v rArr v_(1) = (v)/(3) = (6)/(3) = 2 m//s`