Q58 / 270 · NDA 2026 (Phase 1) · General Ability
A ball is dropped from rest from height $h$ above the ground in frame $S$ (the Earth's frame). Another frame $S'$ is moving upwards with constant speed $u$ relative to the Earth. In frame $S'$, the ball has initial downward speed $v$. Pertaining to the change in kinetic energy $(\Delta K)$ of the ball from release to just before hitting the ground as measured in frames $S$ and $S'$ separately, which one of the following is correct?
- A$\Delta K$ is the same in both the frames.
- B$\Delta K$ is larger in frame $S$.
- Correct answerC$\Delta K$ is larger in frame $S'$.
- D$\Delta K$ cannot be compared without the information of mass of the ball.
Solution
In frame $S$ (Earth's frame):
Initial velocity of the ball, $v_{i} = 0$
Final velocity of the ball just before hitting the ground, $v_{f} = \sqrt{2gh}$ (downwards)
Change in kinetic energy in frame $S$ is:
$\Delta K_{S} = \dfrac{1}{2} m v_{f}^2 - \dfrac{1}{2} m v_{i}^2 = \dfrac{1}{2} m (2gh) - 0 = mgh$
In frame $S'$ (moving upwards with constant speed $u$):
Initial velocity of the ball, $v'_{i} = u = v$ (downwards)
Final velocity of the ball, $v'_{f} = \sqrt{2gh} + v$ (downwards)
Change in kinetic energy in frame $S'$ is:
$\Delta K_{S'} = \dfrac{1}{2} m (v'_{f})^2 - \dfrac{1}{2} m (v'_{i})^2$
$\Delta K_{S'} = \dfrac{1}{2} m (\sqrt{2gh} + v)^2 - \dfrac{1}{2} m v^2$
$\Delta K_{S'} = \dfrac{1}{2} m (2gh + v^2 + 2v\sqrt{2gh}) - \dfrac{1}{2} m v^2$
$\Delta K_{S'} = mgh + mv\sqrt{2gh}$
Comparing the two expressions, we get $\Delta K_{S'} = \Delta K_{S} + mv\sqrt{2gh}$.
Since $m, v, g, h$ are all positive quantities, $\Delta K_{S'} > \Delta K_{S}$.
Therefore, the change in kinetic energy is larger in frame $S'$.
Answer: $\Delta K$ is larger in frame $S'$.
Initial velocity of the ball, $v_{i} = 0$
Final velocity of the ball just before hitting the ground, $v_{f} = \sqrt{2gh}$ (downwards)
Change in kinetic energy in frame $S$ is:
$\Delta K_{S} = \dfrac{1}{2} m v_{f}^2 - \dfrac{1}{2} m v_{i}^2 = \dfrac{1}{2} m (2gh) - 0 = mgh$
In frame $S'$ (moving upwards with constant speed $u$):
Initial velocity of the ball, $v'_{i} = u = v$ (downwards)
Final velocity of the ball, $v'_{f} = \sqrt{2gh} + v$ (downwards)
Change in kinetic energy in frame $S'$ is:
$\Delta K_{S'} = \dfrac{1}{2} m (v'_{f})^2 - \dfrac{1}{2} m (v'_{i})^2$
$\Delta K_{S'} = \dfrac{1}{2} m (\sqrt{2gh} + v)^2 - \dfrac{1}{2} m v^2$
$\Delta K_{S'} = \dfrac{1}{2} m (2gh + v^2 + 2v\sqrt{2gh}) - \dfrac{1}{2} m v^2$
$\Delta K_{S'} = mgh + mv\sqrt{2gh}$
Comparing the two expressions, we get $\Delta K_{S'} = \Delta K_{S} + mv\sqrt{2gh}$.
Since $m, v, g, h$ are all positive quantities, $\Delta K_{S'} > \Delta K_{S}$.
Therefore, the change in kinetic energy is larger in frame $S'$.
Answer: $\Delta K$ is larger in frame $S'$.
Hence, the correct answer is option: 'C'.