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(D) In a perfectly elastic collision between two bodies of equal mass where one body is initially at rest, the bodies exchange their velocities.Let th

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$20$

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(D) In a perfectly elastic collision between two bodies of equal mass where one body is initially at rest, the bodies exchange their velocities.Let the mass of the ball be $m$ and the mass of the bob be $m$.Initial velocity of the ball $u_1 = 2 \,ms^{-1}$ and initial velocity of the bob $u_2 = 0 \,ms^{-1}$.After the elastic collision, the ball comes to rest $(v_1 = 0)$ and the bob acquires the velocity of the ball $(v_2 = u_1 = 2 \,ms^{-1})$.The kinetic energy of the bob immediately after the collision is converted into potential energy as it rises to a height $h$.Using the conservation of energy: $\frac{1}{2}mv_2^2 = mgh$.Substituting the values: $\frac{1}{2} 	imes (2)^2 = 10 	imes h$.$2 = 10h$.$h = 0.2 \,m = 20 \,cm$.

$A$ ball moving horizontally with a velocity $2 \,ms^{-1}$ strikes the bob of a seconds pendulum at rest. If the mass of the bob is equal to the mass of the ball and the collision is perfectly elastic, after collision the bob of the pendulum will raise to a height of $(g=10 \,ms^{-2})$ (in $\,cm$)

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