Question

A wedge of mass `M = 2m` rests on a smooth horizontal plane. A small block of mass m rests over it at left end A as shown in figure. A sharp impulse is applied on the block, due to which it starts moving to the right with velocity `v_(0) = 6 ms^(-1)`. At highest point of its trajectory, the block collides with a particle of same mass m moving vertically downwards with velocity `v = 2 ms^(-1)` and gets stuck with it. If the combined body lands at the end point A of body of mass M, calculate length `l`. Neglect friction `(g = 10 ms^(-2))`.

Answer 1

Correct Answer - `40 cm`
`COLM rArr mv_(0) + 0 = (m + 2m)v_(1) rArr v_(1) = (v_(0)/(3))`
After collision at highest point
`v_(x) = 1 m//s (= (mv_(1))/(2m))`
`v_(y) = 1m//s (= (m xx 2)/(2m))`
`COME rArr (1)/(2) mv_(0)^(2) = (1)/(2) m(v_(1)^(2) + v_(2)^(2)) + (1)/(2) (2m)v_(1)^(2) rArr v_(2) = sqrt(24) m//s`
Max height attained `= (v_(2)^(2))/(2g) = 1.2 m`
For the block `v_(x) = 1 m//s`
while for the wedge it has `v_(x) = 2m//s`
`(v_(X_("wedge")) - v_(X_("wedge")))t = t` & `(ut +(1)/(2)at^(2))` block `= 1.2`
`rArr t = 0.4 sec` and `l = (2 - 1)t = 0.4 m = 40 cm`