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
A train moving at constant speed is an inertial frame, so the work–energy theorem holds in it as well as in the ground frame. Kinetic energy and work depend on the frame, and the extra energy seen from the ground must be supplied by the train through the girl's feet.
Step 1 — In the train frame the ball goes from rest to speed v: \(K_{\text{train}}=\dfrac{1}{2}mv^{2}\), and this is exactly the work the girl does by her own effort.
Step 2 — In the ground frame the ball goes from u to u + v: \(\Delta K_{\text{ground}}=\dfrac{1}{2}m(u+v)^{2}-\dfrac{1}{2}mu^{2}=\dfrac{1}{2}mv^{2}+muv\).
Step 3 — While the girl pushes the ball forward with force F for a time t (Ft = mv), the floor pushes her forward with the same force, and the train moves a distance ut, so \(W_{\text{train}}=F\,(ut)=\dfrac{mv}{t}\,ut=muv\).
Step 4 — The girl's share plus the train's share equals the gain seen from the ground: mv2/2 + muv.
The kinetic energy of mv2/2 in the train frame, the girl's work of mv2/2 and the train's work of mvu all hold. The gain measured by a person by the track is mv2/2 + muv, larger than mv2/2, so that statement fails.
Kinetic energy of the ball as measured by the girl on the train is mv2/2., Work done by the girl in throwing the ball is mv2/2. and Work done by the train is mvu. options are correct.
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