Question

Two smooth prisms of similar right-triangular
sections are arranged on a smooth horizontal plane
as shown in figure. The lower prism has a mass n
times the upper prism. The prisms are held in an
initial position as shown and are then released. As
the upper prism touches the horizontal plane, the
distance moved by the lower prism is​

Answer 1

Answer:

The distance moved by the lower prism is​ $x=\frac{(a-b)}{4}$

Explanation:

In an ideal environment there is conservation of momentum. Therefore, the momentum of prism a to the right side is equal to the momentum of prism be on the left side. Since, there is no external force applied from outside.

The momentum of prism $\mathrm{b}=\frac{M(a-b-x)}{t}$ where X is the distance moved by the lower prism momentum of prism $a=\frac{3 M x}{t}$

Hence,  $\frac{M(a-b-x)}{t}=\frac{3 M x}{t}$

Or  a - b - x = 3x

Or $x=\frac{(a-b)}{4}$

Answer 2

Answer:

$\frac{a+b}{1+n\\}$

Explanation: Refer to the attachments below

hope it helps.......