Question

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. The kinetic energy of the ball as measured by the girl on the train is \(\frac{mv^2}{2}\)
• B. work done by the girl in throwing the ball is \(\frac{mv^2}{2}\).
• C. Work done by the train mvu
• D. The gain is the kinetic energy of the ball as measured by a person standing by the rail track is \(\frac{mv^2}{2}\).

Answer 1

The Correct Option is A, B, C

🚄 Analysis: Girl Throwing a Ball from a Moving Train 

1. Train Frame (Girl’s Reference Frame)

• Initial velocity of the ball: $0$
• Final velocity of the ball: $v$ (relative to train)
• Kinetic Energy: $\text{K.E.} = \frac{1}{2}mv^2$
• Work done by the girl = Change in K.E. = $\frac{1}{2}mv^2$

2. Ground Frame (Stationary Observer)

• Initial velocity of the ball: $u$ (same as train's speed)
• Final velocity of the ball: $u + v$ (vector sum)
• Final K.E.: $\frac{1}{2}m(u + v)^2 = \frac{1}{2}mu^2 + \frac{1}{2}mv^2 + muv$
• Initial K.E.: $\frac{1}{2}mu^2$
• Change in K.E.: $\left[\frac{1}{2}mv^2 + muv\right]$
• This includes:
  • $\frac{1}{2}mv^2$ – work done by the girl
  • $muv$ – work done by the train (since it's moving)

✅ Conclusions

• Option A is correct: Work done by girl = $\frac{1}{2}mv^2$ (in train frame)
• Option B is correct: Ground frame sees increased total K.E.
• Option C is correct: Change in K.E. includes both girl’s and train’s work
• Option D is incorrect: It misses the $muv$ term (train’s contribution)