Two balls X(2 kg) and Y(4 kg) approach each | vedclass

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(C) Given: $m_1 = 2 \ kg$,$m_2 = 4 \ kg$,$u_1 = 10 \ ms^{-1}$,$u_2 = -10 \ ms^{-1}$.In a perfectly elastic collision,both linear momentum and kinetic

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$-\frac{50}{3} \ ms^{-1}, \frac{10}{3} \ ms^{-1}$

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(C) Given: $m_1 = 2 \ kg$,$m_2 = 4 \ kg$,$u_1 = 10 \ ms^{-1}$,$u_2 = -10 \ ms^{-1}$.In a perfectly elastic collision,both linear momentum and kinetic energy are conserved.By conservation of linear momentum:$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$2(10) + 4(-10) = 2v_1 + 4v_2$$20 - 40 = 2v_1 + 4v_2$$-20 = 2v_1 + 4v_2 \Rightarrow v_1 + 2v_2 = -10 \quad \dots (i)$For a perfectly elastic collision,the coefficient of restitution $e = 1$,so the velocity of separation equals the velocity of approach:$v_2 - v_1 = u_1 - u_2$$v_2 - v_1 = 10 - (-10) = 20 \quad \dots (ii)$Adding equations $(i)$ and $(ii)$:$(v_1 + 2v_2) + (v_2 - v_1) = -10 + 20$$3v_2 = 10 \Rightarrow v_2 = \frac{10}{3} \ ms^{-1}$Substituting $v_2$ into equation $(ii)$:$\frac{10}{3} - v_1 = 20$$v_1 = \frac{10}{3} - 20 = \frac{10 - 60}{3} = -\frac{50}{3} \ ms^{-1}$Thus,the final velocities are $v_1 = -\frac{50}{3} \ ms^{-1}$ and $v_2 = \frac{10}{3} \ ms^{-1}$.

Two balls $X(2 \ kg)$ and $Y(4 \ kg)$ approach each other with equal speeds of $10 \ ms^{-1}$. If the collision is perfectly elastic,then the new velocities of balls $X$ and $Y$ are respectively

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