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:

We are told that a wedge of mass $$M$$ rests on a perfectly smooth horizontal table and that a small particle of mass $$m$$ slides up the wedge without any friction anywhere. Numerically, the wedge is four times heavier than the particle, so $$M = 4m$$. The particle is launched horizontally towards the wedge with speed $$v$$.

Because the horizontal table is friction-free, no external horizontal force acts on the two-body system “particle + wedge”. Therefore the total horizontal linear momentum of the system will remain constant throughout the motion. This is our first key principle:

(i) Conservation of horizontal momentum: $$P_{\text{initial}} = P_{\text{final}}$$.

Secondly, all surfaces are smooth, so no mechanical energy is dissipated as heat. The only potential energy that changes is the gravitational potential energy of the particle as it climbs the wedge. Therefore the total mechanical energy of the system is also conserved:

(ii) Conservation of mechanical energy: $$E_{\text{initial}} = E_{\text{final}}$$.

We now apply these two principles step by step.

Initial state (just before contact)
The wedge is at rest, so its horizontal velocity is $$0$$. The particle of mass $$m$$ moves towards the wedge with speed $$v$$. Hence

Initial horizontal momentum: $$P_{\text{initial}} = m\,v.$$

Initial kinetic energy: $$E_{\text{kin,\,initial}} = \frac12\,m\,v^{2}.$$

The initial gravitational potential energy of the particle can be taken as zero (reference level at the foot of the wedge).

Final state (highest point reached by the particle on the wedge)
At the highest point the particle is momentarily at rest relative to the wedge. Because both bodies are still free to slide on the table, the particle and the wedge must, at that instant, move together with some common horizontal speed, say $$u$$, relative to the ground. Therefore

Final horizontal momentum: $$P_{\text{final}} = (M + m)\,u.$$

Final kinetic energy: $$E_{\text{kin,\,final}} = \frac12\,(M + m)\,u^{2}.$$

Let the particle have climbed a vertical height $$h$$ on the wedge. Its gain in gravitational potential energy is then

$$E_{\text{pot,\,gain}} = m\,g\,h.$$

We now write the conservation equations explicitly.

Step 1 : Linear momentum conservation

$$m\,v = (M + m)\,u.$$

We solve this for $$u$$:

$$u = \frac{m\,v}{M + m}.$$

Given $$M = 4m$$, we substitute:

$$u = \frac{m\,v}{4m + m} = \frac{m\,v}{5m} = \frac{v}{5}.$$

Step 2 : Mechanical energy conservation

Initial total energy = Final total energy:

$$\frac12\,m\,v^{2} = \frac12\,(M + m)\,u^{2} + m\,g\,h.$$

First we insert $$M + m = 4m + m = 5m$$ and $$u = \dfrac{v}{5}$$:

$$\frac12\,m\,v^{2} = \frac12\,(5m)\left(\frac{v}{5}\right)^{2} + m\,g\,h.$$

We now simplify term by term. Start with the kinetic term on the right:

$$\left(\frac{v}{5}\right)^{2} = \frac{v^{2}}{25},$$

so

$$\frac12\,(5m)\left(\frac{v^{2}}{25}\right) = \frac12 \times 5m \times \frac{v^{2}}{25} = \frac{5m\,v^{2}}{50} = \frac{m\,v^{2}}{10}.$$

Our energy equation is now

$$\frac12\,m\,v^{2} = \frac{m\,v^{2}}{10} + m\,g\,h.$$

To clear the fractions, multiply every term by $$\dfrac{2}{m}$$ (this also cancels the common factor $$m$$):

$$v^{2} = \frac{2}{m}\times\frac{m\,v^{2}}{10} + 2\,g\,h \;\;\Longrightarrow\;\; v^{2} = \frac{2\,v^{2}}{10} + 2\,g\,h.$$

Simplify the first term on the right:

$$\frac{2\,v^{2}}{10} = \frac{v^{2}}{5}.$$

Hence

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

Subtract $$\dfrac{v^{2}}{5}$$ from both sides:

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

Write the left side with a common denominator:

$$\frac{5v^{2}}{5} - \frac{v^{2}}{5} = \frac{4v^{2}}{5}.$$

Thus

$$\frac{4v^{2}}{5} = 2\,g\,h.$$

Finally, solve for $$h$$ by dividing both sides by $$2g$$:

$$h = \frac{\frac{4v^{2}}{5}}{2g} = \frac{4v^{2}}{10g} = \frac{2v^{2}}{5g}.$$

We have reached an explicit expression for the maximum height climbed by the particle. Comparing with the options given, this corresponds to Option C.

Hence, the correct answer is Option C.