A 1 ,kg ball moving at 12 ,ms^{-1} collides w | vedclass

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(B) Given: $m_1 = 1 \,kg$, $u_1 = 12 \,ms^{-1}$, $m_2 = 2 \,kg$, $u_2 = -24 \,ms^{-1}$ (negative sign indicates opposite direction).Coefficient of res

Vedclass · Accepted Answer

$-28 \,ms^{-1}, -4 \,ms^{-1}$

Vedclass · Answer

(B) Given: $m_1 = 1 \,kg$, $u_1 = 12 \,ms^{-1}$, $m_2 = 2 \,kg$, $u_2 = -24 \,ms^{-1}$ (negative sign indicates opposite direction).Coefficient of restitution $e = 2/3$.The formula for the coefficient of restitution is $e = \frac{v_2 - v_1}{u_1 - u_2}$.Substituting the values: $\frac{2}{3} = \frac{v_2 - v_1}{12 - (-24)} = \frac{v_2 - v_1}{36}$.Thus, $v_2 - v_1 = 24$ ... $(i)$.By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.$(1)(12) + (2)(-24) = (1)v_1 + (2)v_2$.$12 - 48 = v_1 + 2v_2 \Rightarrow v_1 + 2v_2 = -36$ ... (ii).From $(i)$, $v_2 = v_1 + 24$. Substituting into (ii):$v_1 + 2(v_1 + 24) = -36$.$v_1 + 2v_1 + 48 = -36 \Rightarrow 3v_1 = -84 \Rightarrow v_1 = -28 \,ms^{-1}$.Then $v_2 = -28 + 24 = -4 \,ms^{-1}$.The velocities are $-28 \,ms^{-1}$ and $-4 \,ms^{-1}$.

$A$ $1 \,kg$ ball moving at $12 \,ms^{-1}$ collides with a $2 \,kg$ ball moving in the opposite direction at $24 \,ms^{-1}$. If the coefficient of restitution is $2/3$, then their velocities after the collision are

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