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Byju's Answer

Standard XII

Physics

Relation between Work Done and PE

Two blocks ...

Question

Two blocks $A$ and $B$ of mass $m$ and $2 m$ respectively are moving towards a massive (mass $>> 2 m$ ) cliff with velocities $2 v$ and $v$ respectively. The cliff moves with a velocity $v$ . If the coefficient of restitution of collision at the surface of the cliff is $e = \frac{1}{2}$ . Find the work done by the cliff during collision.

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Solution

$v_{A} = 2 v^i, v_{B} = v^i, v_{c l} = - v^i v_{B / c l} = 2 v^i e = \frac{1}{2} = - \frac{v_{f B / c l}}{v_{B / c l}} \Rightarrow v_{f B / c l} = - \frac{1}{2} v_{B / c l} \Rightarrow v_{f B / c l} = - v^i \Rightarrow v_{f B} = v_{f B / c l} + v_{c l} = - v^i - v^i = - 2 v^i W o r k d o n e b y t h e c l i f f c h a n g e s t h e k i n e t i c e n e r g y o f b o d y B . W = \frac{1}{2} 2 m {v_{f B}}^{2} - \frac{1}{2} 2 m {v_{B}}^{2} = m (4 v^{2} - v^{2}) = 3 m v^{2}$

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vA=2v^i,vB=v^i,vcl=−v^ivB/cl=2v^ie=12=−vfB/clvB/cl⇒vfB/cl=−12vB/cl⇒vfB/cl=−v^i⇒vfB=vfB/cl+vcl=−v^i−v^i=−2v^iWorkdonebythecliffchangesthekineticenergyofbodyB.W=122mvfB2−122mvB2=m(4v2−v2)=3mv2