A 2 ~kg ball moving at 24 ~ms^{-1} undergoes | vedclass

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(A) Given: $m_1 = 2 ~kg$,$u_1 = 24 ~ms^{-1}$,$m_2 = 4 ~kg$,$u_2 = -48 ~ms^{-1}$ (opposite direction),and $e = 2/3$. Using the formula for final veloci

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$-56, -8$

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(A) Given: $m_1 = 2 ~kg$,$u_1 = 24 ~ms^{-1}$,$m_2 = 4 ~kg$,$u_2 = -48 ~ms^{-1}$ (opposite direction),and $e = 2/3$. Using the formula for final velocities after a one-dimensional collision: $v_1^{\prime} = \frac{m_1 u_1 + m_2 u_2 + e m_2 (u_2 - u_1)}{m_1 + m_2}$ $v_1^{\prime} = \frac{(2)(24) + (4)(-48) + (2/3)(4)(-48 - 24)}{2 + 4}$ $v_1^{\prime} = \frac{48 - 192 + (8/3)(-72)}{6} = \frac{-144 - 192}{6} = \frac{-336}{6} = -56 ~ms^{-1}$. Using the conservation of momentum or the restitution equation $v_2^{\prime} - v_1^{\prime} = e(u_1 - u_2)$: $v_2^{\prime} - (-56) = (2/3)(24 - (-48))$ $v_2^{\prime} + 56 = (2/3)(72) = 48$ $v_2^{\prime} = 48 - 56 = -8 ~ms^{-1}$.

$A$ $2 ~kg$ ball moving at $24 ~ms^{-1}$ undergoes an inelastic head-on collision with a $4 ~kg$ ball moving in the opposite direction at $48 ~ms^{-1}$. If the coefficient of restitution is $2/3$,what are their velocities in $ms^{-1}$ after the impact?

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