A spring-block system is resting on a frictio | vedclass

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(C) Let $m_1 = 1.0 \ kg$ and $m_2 = 2.0 \ kg$. Initial velocity of $m_1$ is $u_1 = 2.0 \ m \ s^{-1}$ and $m_2$ is $u_2 = 0$. Let $v_1$ and $v_2$ be th

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$2.09$

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(C) Let $m_1 = 1.0 \ kg$ and $m_2 = 2.0 \ kg$. Initial velocity of $m_1$ is $u_1 = 2.0 \ m \ s^{-1}$ and $m_2$ is $u_2 = 0$. Let $v_1$ and $v_2$ be their velocities after the elastic collision.Using conservation of linear momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2 \implies 1(2) = 1(v_1) + 2(v_2) \implies v_1 + 2v_2 = 2$ . . . . . $(1)$Using the coefficient of restitution for an elastic collision $(e=1)$: $v_2 - v_1 = e(u_1 - u_2) = 1(2 - 0) = 2$ . . . . . $(2)$Solving equations $(1)$ and $(2)$,we get $v_2 = 4/3 \ m \ s^{-1}$ and $v_1 = -2/3 \ m \ s^{-1}$.The $2.0 \ kg$ block attached to the spring undergoes simple harmonic motion. The time period is $T = 2\pi \sqrt{m_2/k} = 2\pi \sqrt{2/2} = 2\pi \ s$.The spring returns to its unstretched position after half the time period,i.e.,$\Delta t = T/2 = \pi \ s$.During this time,the $1.0 \ kg$ block moves with constant velocity $v_1 = -2/3 \ m \ s^{-1}$ (away from the spring). Its displacement is $\Delta x_1 = v_1 \Delta t = (-2/3) 	imes \pi = -2\pi/3 \ m$.The $2.0 \ kg$ block moves from $x=0$ to $x_{max}$ and back to $x=0$. Its displacement is $\Delta x_2 = 0$.The distance between the blocks is $|\Delta x_1 - \Delta x_2| = |-2\pi/3 - 0| = 2\pi/3 \approx 2(3.14)/3 = 6.28/3 \approx 2.09 \ m$.

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