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Byju's Answer

Standard XI

Physics

Law of Conservation of Mechanical Energy

The blocks A ...

Question

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

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

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$\frac{\sqrt{3}}{4} \sqrt{g}$

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

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

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Solution

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

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

Solving, we get

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

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Similar questions

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

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