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(B) Let the velocity of the block of mass $m_1 = 0.1\,kg$ at position $\frac{x}{2}$ be $u.$ When it hits the second block of mass $m_2,$ the first blo

Vedclass · Accepted Answer

$0.6$

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(B) Let the velocity of the block of mass $m_1 = 0.1\,kg$ at position $\frac{x}{2}$ be $u.$ When it hits the second block of mass $m_2,$ the first block comes to rest,meaning $v_1 = 0.$ The second block moves with $v_2 = 3\,m/s.$By conservation of momentum: $m_1 u + m_2(0) = m_1(0) + m_2(3) \Rightarrow 0.1u = 3m_2.$By conservation of energy for the collision: $\frac{1}{2}m_1 u^2 = \frac{1}{2}m_2(3)^2 \Rightarrow 0.1u^2 = 9m_2.$Dividing the two equations: $\frac{0.1u^2}{0.1u} = \frac{9m_2}{3m_2} \Rightarrow u = 3\,m/s.$Now,using energy conservation for the spring-block system from the initial compressed position to the position $\frac{x}{2}$:Total initial energy $E = \frac{1}{2}kx^2.$Energy at position $\frac{x}{2}$ is $E' = \frac{1}{2}k(\frac{x}{2})^2 + \frac{1}{2}m_1 u^2.$Since $E = E',$ we have $\frac{1}{2}kx^2 = \frac{1}{8}kx^2 + \frac{1}{2}(0.1)(3)^2.$$\frac{3}{8}kx^2 = 0.45 \Rightarrow \frac{1}{2}kx^2 = \frac{0.45 	imes 8}{3 	imes 2} = 0.6\,J.$

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