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:

Options

A. `(2 v^(2))/(5g)`
B. `(v^(2))/(g)`
C. `(2 v^(2))/(7g)`
D. `(v^(2))/(2g)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the system
We have a wedge of mass \( M = 4m \) on a frictionless plane and a particle of mass \( m \) approaching the wedge with speed \( v \). The goal is to find the maximum height \( h \) that the particle climbs on the wedge.

### Step 2: Apply conservation of momentum
Before the collision, the total momentum of the system is given by the momentum of the particle:
\[ P_{\text{initial}} = mv \]
After the collision, the wedge and the particle move together with a common velocity \( v_0 \). The total mass after the collision is \( m + 4m = 5m \). Therefore, the final momentum is:
\[ P_{\text{final}} = (m + 4m)v_0 = 5mv_0 \]
By conservation of momentum:
\[ mv = 5mv_0 \]
From this, we can solve for \( v_0 \):
\[ v_0 = \frac{v}{5} \]

### Step 3: Apply conservation of energy
Initially, the particle has kinetic energy and no potential energy:
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 \]
At the maximum height \( h \), the particle has potential energy and kinetic energy. The potential energy is given by:
\[ PE_{\text{final}} = mgh \]
The kinetic energy of the wedge and the particle after the collision is:
\[ KE_{\text{final}} = \frac{1}{2}(m + 4m)v_0^2 = \frac{1}{2}(5m)\left(\frac{v}{5}\right)^2 = \frac{1}{2}(5m)\frac{v^2}{25} = \frac{mv^2}{10} \]
By conservation of energy:
\[ KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}} \]
Substituting the values we found:
\[ \frac{1}{2}mv^2 = mgh + \frac{mv^2}{10} \]

### Step 4: Solve for height \( h \)
Rearranging the equation gives:
\[ \frac{1}{2}mv^2 - \frac{mv^2}{10} = mgh \]
Factoring out \( m \):
\[ m\left(\frac{1}{2}v^2 - \frac{1}{10}v^2\right) = mgh \]
Simplifying the left side:
\[ \frac{5}{10}v^2 - \frac{1}{10}v^2 = \frac{4}{10}v^2 = \frac{2}{5}v^2 \]
Thus, we have:
\[ \frac{2}{5}v^2 = gh \]
Solving for \( h \):
\[ h = \frac{2v^2}{5g} \]

### Final Answer
The maximum height climbed by the particle on the wedge is:
\[ h = \frac{2v^2}{5g} \]