A ball of 0.4 , kg mass and a speed of 3 , m/ | vedclass

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(A) Let $m_1 = 0.4 \, kg$,$u_1 = 3 \, m/s$,$m_2 = 0.6 \, kg$,and $u_2 = 0 \, m/s$.By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 =

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$0.6 \, m/s, 2.4 \, m/s$

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(A) Let $m_1 = 0.4 \, kg$,$u_1 = 3 \, m/s$,$m_2 = 0.6 \, kg$,and $u_2 = 0 \, m/s$.By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.$(0.4 	imes 3) + 0 = 0.4 v_1 + 0.6 v_2 \implies 1.2 = 0.4 v_1 + 0.6 v_2 \implies 2 v_1 + 3 v_2 = 6$ (Equation $1$).For a perfectly elastic collision,the coefficient of restitution $e = 1$,so $v_2 - v_1 = u_1 - u_2$.$v_2 - v_1 = 3 - 0 \implies v_2 - v_1 = 3 \implies v_2 = v_1 + 3$ (Equation $2$).Substitute Equation $2$ into Equation $1$:$2 v_1 + 3(v_1 + 3) = 6 \implies 2 v_1 + 3 v_1 + 9 = 6 \implies 5 v_1 = -3 \implies v_1 = -0.6 \, m/s$.The negative sign indicates the first ball rebounds.Substitute $v_1$ into Equation $2$:$v_2 = -0.6 + 3 = 2.4 \, m/s$.Thus,the speeds are $0.6 \, m/s$ and $2.4 \, m/s$.

$A$ ball of $0.4 \, kg$ mass and a speed of $3 \, m/s$ has a head-on,completely elastic collision with a $0.6 \, kg$ mass initially at rest. Find the speeds of both balls after the collision:

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