Jee Mechanics

The initial energy of the system is the kinetic energy of the small block $m$ , $E_0 = \frac 12mv^2$ . After the block $m$ hits the irregular shaped block $M$ , part of the initial kinetic energy is converted to moving the two blocks $m+M$ with a velocity $v_1$ , $E_v$ . Another part is converted to potential energy for block $m$ to move vertically up block $M$ , $E_p$ . The minimum $v$ is the one that block $m$ reaches the highest point of block $M$ and stops or when

$\begin{aligned} E_0 & = E_v + E_p \\ \frac 12 mv^2 & = \frac 12 (m+M){\color{#3D99F6}v_1}^2 + mgh & \small \color{#3D99F6} \text{By conservation of momentum:} \\ \frac 12 mv^2 & = \frac 12 (m+M){\color{#3D99F6}\left(\frac {mv}{m+M}\right)}^2 + mgh & \small \color{#3D99F6} mv=(m+M)v_1 \implies v_1 = \frac {mv}{m+M} \\ \frac 12 mv^2 & = \frac 12 \times \frac {m^2v^2}{m+M} + mgh & \small \color{#3D99F6} \text{Multiply both sides by }\frac 2m \\ v^2 & = \frac {mv^2}{m+M} + 2gh & \small \color{#3D99F6} \text{Rearrange} \\ \implies v & = \sqrt{\frac {2gh(m+M)}M} \\ & = \sqrt{\frac {2\cdot 10 \cdot 12 (8+12)}{12}} \\ & = \boxed{20} \end{aligned}$

Let initial velocity of small block be v and final velocity of system be v'

Applying conservation if momentum

mv=(M+m)v'. (1)

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Applying conservation of energy

$\frac{1}{2}$ mv²= $\frac{1}{2}$ (M+m)v'²+mgh. (2)

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Solving (1) and (2) by putting values, we get

v=20m/s

The expression for velocity being $sqrt(\frac{2*(m+M)*g*h}{M}$ ))