Question

The blocks A and B shown in fig. Have masses $M_A=5kg$ and $M_B=4kg$. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is

• A. $\frac{\sqrt{3}}{2}\sqrt{g}$
• B. $\frac{\sqrt{3}}{4}\sqrt{g}$
• C. $\frac{\sqrt{g}}{2\sqrt{3}}$
• D. $\frac{\sqrt{g}}{2}$

Answer 1

The correct option is C

$\frac{\sqrt{g}}{2\sqrt{3}}$

If A moves down the incline by 1 m, B shal move up by $\frac{1}{2}m$.

If the speed of B is v, then the speed of A will be 2v.

From conservation of energy,

Gain in KE = loss in PE

$\frac{1}{2}{m}_A(2v{)}^2+\frac{1}{2}{m}_B{v}^2=m_{A}g\times \frac{3}{5}-m_{B}g\times \frac{1}{2}$

Solving, we get

$v=\frac{\sqrt{g}}{2\sqrt{3}}$