A train is moving along the tracks at a const | vedclass

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(A, B, C) $1$. With respect to the girl, the initial velocity of the ball is $0$ and final velocity is $v$. The kinetic energy is $E_k = \frac{1}{2}mv

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Work done by the train is $mvu$

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(A, B, C) $1$. With respect to the girl, the initial velocity of the ball is $0$ and final velocity is $v$. The kinetic energy is $E_k = \frac{1}{2}mv^2$. Thus, the work done by the girl is $W = \Delta E_k = \frac{1}{2}mv^2$. So, option $A$ and $B$ are correct.$2$. With respect to the ground, the initial velocity of the ball is $u$ and the final velocity is $v+u$. The work done by the train is the change in kinetic energy of the ball as seen from the ground minus the work done by the girl: $W_{	ext{train}} = \Delta E_{k, 	ext{ground}} - W_{	ext{girl}} = [\frac{1}{2}m(v+u)^2 - \frac{1}{2}mu^2] - \frac{1}{2}mv^2 = \frac{1}{2}m(v^2 + u^2 + 2vu - u^2) - \frac{1}{2}mv^2 = mvu$. So, option $C$ is correct.$3$. The gain in kinetic energy as measured by a person on the ground is $\Delta E_{k, 	ext{ground}} = \frac{1}{2}m(v+u)^2 - \frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mvu$. Thus, option $D$ is incorrect.

$A$ train is moving along the tracks at a constant speed $u$. $A$ girl on the train throws a ball of mass $m$ straight ahead along the direction of motion of the train with speed $v$ with respect to herself. Then

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