Question

A block of mass m is released from the top of a wedge of mass M as shown in figure. Find the displacement of wedges on the horizontal ground when the block reaches the bottom of the wedges. Neglect friction everywhere.

Answer 1

Correct Answer - A::C
Here, the system is wedge + block. Net force on the system in horizontal direction (x-direction) is zero, therefore, the centre of mass of the system will not move in x-direction so we can apply,
`x_Rm_R=x_Lm_L`
Let x be the displacement of wedge. Then,
`x_L="displacement of wedge towards left"=x`
`m_L="mass of wedge moving towards left" =M`
`x_R="displacement of block with respect to ground towards right" =h cot theta-x`
and `m_R="mass of block moving towards right"=m`
Substituting in Eq. (i), we get
`m(h cot theta-x)=xM`
`x=(mh cot theta)/(M+m)`