Question

a wedge shaped block A of mass M is at rest on a smooth horizontal surface. A small block B of mass m placed at the top edge of inclined plane of length L as shown in the figure by the time the block reaches the bottom and the wedge A moves a distance of​

Answer 1

Answer:

mLcosθ/(m+M) .

Explanation:

If we take that the center of mass of the wedge to be x(taken from the initial point) and y cm to be the distance which the wedge A is moved.

So,the total center of mass of the whole system will be same through out the process. Where the wedge is of mass M and the block B is of mass m. Hence, (Mx +mLcosθ)/(m+M)   =  {M(x+y) + my}/(m+M).

Which on solving we will get that the distance which his moved by the wedge A is y=mLcosθ/(m+M) .

Answer 2

ans is option 2 .... mark it as a brain list