A man of mass m is standing on a cart of mass | vedclass

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(D) Let the velocity of the man be $v_m$ and the velocity of the cart be $v_c$ with respect to the ground after the jump.By the law of conservation of

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$\frac{mu^2}{3}$

Vedclass · Answer

(D) Let the velocity of the man be $v_m$ and the velocity of the cart be $v_c$ with respect to the ground after the jump.By the law of conservation of linear momentum, the initial momentum is zero, so $m v_m + 2m v_c = 0$, which implies $v_m = -2v_c$.The relative velocity of the man with respect to the cart is $u = v_m - v_c$.Substituting $v_m = -2v_c$, we get $u = -2v_c - v_c = -3v_c$, so $v_c = -u/3$.Then $v_m = -2(-u/3) = 2u/3$.The work done by internal forces is equal to the change in kinetic energy of the system: $W = \Delta K = K_f - K_i$.Since the system was initially at rest, $K_i = 0$.$K_f = \frac{1}{2} m v_m^2 + \frac{1}{2} (2m) v_c^2 = \frac{1}{2} m (2u/3)^2 + m (-u/3)^2$.$K_f = \frac{1}{2} m (4u^2/9) + m (u^2/9) = \frac{2mu^2}{9} + \frac{mu^2}{9} = \frac{3mu^2}{9} = \frac{mu^2}{3}$.Thus, the work done is $\frac{mu^2}{3}$.

$A$ man of mass $m$ is standing on a cart of mass $2m$. Initially, the cart is at rest. The man jumps horizontally with a relative velocity $u$ with respect to the cart. The work done by the internal forces of the man during the process of jumping will be:

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