A wooden block of mass `10 gm` is dropped from the top of a cliff `100 m` high. Simultaneously a bullet of same mass is fired from the foot of the cliff vertically upwards with a velocity of `100 ms^(-1)`. If the bullet after collision gets embedded in the block, the common velocity of the bullet and the block immediately after collision is `(g=10 ms^(-2))`.

Options

A. `40 ms^(-1) downward`
B. `40ms^(-1) upward`
C. `80ms^(-1) upward`
D. zero

Answer

`(2)`

Explanation

Time after which collision takes place ` t = h/u` before collision, Initial velocity of the wooden block `(u_(1)), u_(1) = gt `
Initial velocity of the bullet `(u_(2)) = u -gt `
`Rightarrow m_(1)u_(1) - m_(2)u_(2) = (m_(1) + m_(2))v`