Question

Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity of A is v0, and the initial velocity of B is 0, find the final velocities of the two blocks and the distance that A slides relative to B.

Answer 1

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$  

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$  which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$