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 $u_2 = 0$.By conservation of linear

Vedclass · Accepted Answer

$2.09$

Vedclass · Answer

(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 $u_2 = 0$.By conservation of linear momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2 \implies 1(2) = 1(v_1) + 2(v_2) \implies v_1 + 2v_2 = 2$  $(i)$For an elastic collision, the coefficient of restitution $e = 1$: $v_2 - v_1 = e(u_1 - u_2) = 1(2 - 0) = 2$  $(ii)$Solving $(i)$ and $(ii)$: Adding them gives $3v_2 = 4 \implies v_2 = \frac{4}{3} \,m \,s^{-1}$. Substituting into $(ii)$, $v_1 = v_2 - 2 = \frac{4}{3} - 2 = -\frac{2}{3} \,m \,s^{-1}$.The $2.0 \,kg$ block attached to the spring undergoes simple harmonic motion. The time taken for the spring to return to its unstretched position is half the time period: $\Delta t = \frac{T}{2} = \pi \sqrt{\frac{m_2}{k}} = \pi \sqrt{\frac{2}{2}} = \pi \,s$.During this time, the $1.0 \,kg$ block moves with constant velocity $v_1 = -\frac{2}{3} \,m \,s^{-1}$ (moving left). The $2.0 \,kg$ block moves to the right, compresses the spring, and returns to the equilibrium position.Displacement of $1.0 \,kg$ block: $s_1 = v_1 \Delta t = -\frac{2}{3} \pi \,m$.Displacement of $2.0 \,kg$ block: $s_2 = 0$ (as it returns to the starting position).The distance between the blocks is $|s_2 - s_1| = |0 - (- \frac{2}{3} \pi)| = \frac{2}{3} \pi = \frac{2 	imes 3.14159}{3} \approx 2.094 \,m$. Rounded to two decimal places, the distance is $2.09 \,m$.

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