Question

A wedge of mass $M = 4m$ lies on a frictionless plane. A particle of mass $ m $ approaches the wedge with speed $ v $ . There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:
A) $ \dfrac{{2{v^2}}}{{7g}} $
B) $ \dfrac{{{v^2}}}{g} $
C) $ \dfrac{{2{v^2}}}{{5g}} $
D) $ \dfrac{{{v^2}}}{{2g}} $

Answer 1

Hint : In this solution, we will use the law of conservation of momentum to determine the velocity of the block when it starts climbing the ramp. At the top of the ramp, the block will only have potential energy due to its height and at the bottom, it will have only kinetic energy.

Formula used: In this solution, we will use the following formula:
-Potential energy due to height: $ U = mgh $ where $ m $ is the mass of the block, $ g $ is the gravitational acceleration, and $ h $ is its height above the ground.
- Kinetic energy due to velocity: $ K = \dfrac{1}{2}m{v^2} $ where $ v $ is the velocity of the block.

Complete step by step answer
When the block has not started climbing the ramp, only the block will have linear momentum. When it starts climbing the ramp, the momentum will be partly transferred to the ramp and the ramp and the block will start moving with the same velocity.
So, from the law of conservation of momentum, we can write
 $ mv = (m + M)v' $
As $M = 4m$, we can calculate the new velocity as
 $ v' = \dfrac{v}{5} $
Now that we know the combined velocity of the ramp and the block, we can use the law of conservation of energy to determine the maximum height of the block on the ramp. As mentioned in the hint, at the top of the ramp, the block will only have potential energy due to its height as well as kinetic energy associated with the movement of the ramp and the block. And at the bottom, it will have only kinetic energy. So, we can write the law of conservation of energy as
 $ \dfrac{1}{2}m{v^2} = \dfrac{1}{2}(m + M)v{'^2} + mgh $
Substituting $M = 4m$ and $ v' = \dfrac{v}{5} $ , we can determine $ h $ as
 $ h = \dfrac{{2{v^2}}}{{5g}} $
Hence option (C) is the correct choice.

Note
We must realize the fact that after rising, the block and the ramp will move with the same velocity. While this might seem counterintuitive in a practical scenario, there is no friction in this case, so there is no reason for the ramp to slow down, and hence the block and the ramp will have the same velocity at the maximum height.