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Standard VII

Physics

Law of Conservation of Energy

A perfectly h...

Question

A perfectly hard billiard ball of kinetic energy $E_{k}$ collides with another similar ball at rest. After the collision the kinetic energy of the ball becomes $E_{k}^{'}$ . Then

E_{k}^{'} = E_{k}

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E_{k}^{'} > E_{k}

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E_{k}^{'} < E_{k}

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E_{k}^{'} = E_{k}^{2}

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Solution

The correct option is B $E_{k}^{'} < E_{k}$
Even if the collision is elastic there will be some redistribution of K.E. depending on configurations of ball resulting in reduction of K.E. of first ball and increase in K.E. of second ball.

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A perfectly hard billiard ball of kinetic energy Ek collides with another similar ball at rest. After the collision the kinetic energy of the ball becomes E′k. Then

The correct option is B E′k<EkEven if the collision is elastic there will be some redistribution of K.E. depending on configurations of ball resulting in reduction of K.E. of first ball and increase in K.E. of second ball.

A perfectly hard billiard ball of kinetic energy $E_{k}$ collides with another similar ball at rest. After the collision the kinetic energy of the ball becomes $E_{k}^{'}$ . Then

The correct option is B $E_{k}^{'} < E_{k}$
Even if the collision is elastic there will be some redistribution of K.E. depending on configurations of ball resulting in reduction of K.E. of first ball and increase in K.E. of second ball.