Two blocks A and B of mass m and 2m respectiv | vedclass

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(A) Since there is no external force on the system,the linear momentum is conserved. Initially,the system is at rest,so the total momentum is $0$. Thu

Vedclass · Accepted Answer

$x\sqrt{\frac{3k}{2m}}$

Vedclass · Answer

(A) Since there is no external force on the system,the linear momentum is conserved. Initially,the system is at rest,so the total momentum is $0$. Thus,$mv_1 = 2mv_2$,which gives $v_1 = 2v_2$.By the law of conservation of energy,the potential energy stored in the spring is converted into the kinetic energy of the blocks: $\frac{1}{2}kx^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2$.Substituting $v_1 = 2v_2$ into the energy equation: $\frac{1}{2}kx^2 = \frac{1}{2}m(2v_2)^2 + mv_2^2 = \frac{1}{2}m(4v_2^2) + mv_2^2 = 3mv_2^2$.Solving for $v_2$: $v_2^2 = \frac{kx^2}{6m} \Rightarrow v_2 = x\sqrt{\frac{k}{6m}}$.Then $v_1 = 2v_2 = 2x\sqrt{\frac{k}{6m}} = x\sqrt{\frac{4k}{6m}} = x\sqrt{\frac{2k}{3m}}$.The relative velocity $v_{rel} = v_1 + v_2 = 2x\sqrt{\frac{k}{6m}} + x\sqrt{\frac{k}{6m}} = 3x\sqrt{\frac{k}{6m}} = x\sqrt{\frac{9k}{6m}} = x\sqrt{\frac{3k}{2m}}$.

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