Question

Figure shows two uniform wedge of mass M each placed on frictionless surface. A block of mass m $=\frac{M}{2}$ is released from rest from wedge A from height H. All surfaces are smooth and block can climb or drop from wedge smoothly. Then

• A. once descended block will climb over wedge A one more time
• B. maximum velocity attained by wedge B is $\frac{4}{3}\sqrt{\frac{gH}{3}}$
• C. final speed of approach of wedge A and block is zero
• D. final speed attained by block is $\sqrt{\frac{gH}{3}}$

Answer 1

The correct option is B maximum velocity attained by wedge B is $\frac{4}{3}\sqrt{\frac{gH}{3}}$


Conservation of momentum $\Rightarrow \frac{M}{2}{V}_1=M{V}_{A1}$

Conservation of energy $\Rightarrow \frac{M}{2}gH=\frac{1}{2}\frac{M}{2}{V}_1^2+\frac{1}{2}M{V}_A{1}^2$

Solving we get $\Rightarrow V_1=2\sqrt{\frac{gH}{3}}$ and $V_{A1}=\sqrt{\frac{gH}{3}}$



Conservation of momentum $\Rightarrow \frac{M}{2}{V}_1=M{V}_{B1}-\frac{M}{2}{V}_2\dots \left(1\right)$
Since there is no energy loss, velocity of approach = velocity of seperation.
$\Rightarrow V_1=V_2+V_{B1}\dots \dots \dots \left(2\right)$

Solving (1) and (2)

$V_2=\frac{V_1}{3}=\frac{2}{3}\sqrt{\frac{gH}{3}}$

and $V_{B1}=\frac{4}{3}\sqrt{\frac{gH}{3}}$
Final speed of approach of wedge A and block is coming negative that is body are getting separated.