1. Set up the situation

A wedge of mass $M=4m$ is on a frictionless horizontal plane. A particle of mass $m$ approaches with speed $v$ and climbs the wedge.

Since all संपर्क surfaces are frictionless:

• external horizontal force on the system is zero,
• so horizontal momentum is conserved.

At the maximum height, the particle is instantaneously at rest relative to the wedge, so particle and wedge move together with the same horizontal speed, say $u$.

2. Apply conservation of horizontal momentum

Initially:

• particle momentum $= mv$,
• wedge momentum $=0$.

So initial horizontal momentum is $p_i = mv.$

At maximum height, both move with common horizontal speed $u$: $p_f = (m+4m)u = 5mu.$

Hence, $mv = 5mu$ $u = \frac{v}{5}.$

3. Apply conservation of mechanical energy

Initially, total kinetic energy is only of the particle: $K_i = \frac12 mv^2.$

At maximum height, both masses move horizontally with speed $u=\frac v5$, so total kinetic energy is $K_f = \frac12 (m+4m)u^2 = \frac12 (5m)\left(\frac v5\right)^2.$

Thus, $K_f = \frac12 \cdot 5m \cdot \frac{v^2}{25} = \frac{mv^2}{10}.$

If the particle rises by height $h$, increase in gravitational potential energy is $mgh.$

By energy conservation, $\frac12 mv^2 = \frac{mv^2}{10} + mgh.$

So, $mgh = \frac12 mv^2 - \frac{mv^2}{10}.$

$mgh = mv^2\left(\frac12 - \frac1{10}\right) = mv^2\left(\frac{5-1}{10}\right) = \frac{2mv^2}{5}.$

Therefore, $h = \frac{2v^2}{5g}.$

4. Match with options

$h = \frac{2v^2}{5g}$

So the correct option is:

D: $\displaystyle \frac{2v^2}{5g}$

5. Comparison with stored answer

Stored correct answer: D

This matches our derived answer.