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(D) Let $m_1 = 2 \ g$ and $u_1 = 2 \ ms^{-1}$ be the mass and initial velocity of the first ball.Let $m_2 = 8 \ g$ and $u_2 = 0 \ ms^{-1}$ be the mass

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(D) Let $m_1 = 2 \ g$ and $u_1 = 2 \ ms^{-1}$ be the mass and initial velocity of the first ball.Let $m_2 = 8 \ g$ and $u_2 = 0 \ ms^{-1}$ be the mass and initial velocity of the second ball.After collision,the first ball comes to rest,so $v_1 = 0 \ ms^{-1}$.Using the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.Substituting the values: $(2 \ g)(2 \ ms^{-1}) + (8 \ g)(0) = (2 \ g)(0) + (8 \ g)(v_2)$.$4 = 8 v_2 \implies v_2 = 0.5 \ ms^{-1}$.The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.Substituting the values: $e = \frac{0.5 - 0}{2 - 0} = \frac{0.5}{2} = 0.25$.

$A$ ball of mass $2 \ g$ moving with a velocity of $2 \ ms^{-1}$ collides with another ball of mass $8 \ g$ which is at rest and comes to rest after collision. Then the coefficient of restitution is

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