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PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

The blocks $A$ and $B$ shown in the figure have masses $M_{A} = 5 kg$ and $M_{B} = 4 kg$ . the system is released from rest. The speed of $B$ after $A$ has travelled a distance $1 m$ along the incline is $(g = 10 m s^{- 2})$

Options:

Answer:

Solution:

If $A$ moves down the incline by $1 m$ , $B$ shall move up by $\frac{1}{2} m$ . If the speed of $B$ is $v$ then the speed of $A$ will be $2 v$ .

From the conservation of energy: Gain in K.E. = loss in P.E.

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

Solving we get, $v = \frac{1}{2} \sqrt{\frac{g}{3}} = \sqrt{\frac{5}{6}}$

Subject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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