\(\Delta K\) is the same in both the frames.
To determine the change in kinetic energy (\(\Delta K\)) of the ball in both frames, we need to analyze the motion of the ball in both the Earth frame (frame \(S\)) and the moving frame (frame \(S'\)).
- Frame \(S\) (Earth's Frame):
- The ball is dropped from rest, so its initial velocity \(u_0 = 0\).
- Using the equation of motion, the final velocity when it hits the ground is:
\(v_f = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity. - The change in kinetic energy is given by:
\(\Delta K_S = \frac{1}{2}m(v_f^2 - u_0^2) = \frac{1}{2}m(2gh) = mgh\).
- Frame \(S'\) (Moving Frame):
- The initial velocity of the ball is \(-u\) (downwards).
- Applying the relative motion concept and accounting for the frame's upward motion, the velocity of the ball relative to the moving frame just before hitting the ground is:
\(v_f' = v_f - u = \sqrt{2gh} - u\). - The change in kinetic energy in frame \(S'\) is:
\(\Delta K_{S'} = \frac{1}{2}m((v_f')^2 - (-u)^2)\)
= \(\frac{1}{2}m((\sqrt{2gh} - u)^2 - u^2)\). - After simplifying, the change \(\Delta K_{S'}\) comes out to be \(mgh\), which matches \(\Delta K_S\).
Thus, the change in kinetic energy of the ball is the same in both frames \(S\) and \(S'\). Hence, the correct answer is:
\(\Delta K\) is the same in both the frames.
Important Questions from Kinematic equations for uniformly accelerated motion
- A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately
A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of
Which of the following best describes the relationship between distance, time, and speed when a body is NOT accelerating?
If the distance travelled by a body in the $n^{th}$ second is given by $(7 + 5n)$ m, then find the initial velocity and acceleration of the body respectively.
- A particle is released from height $S$ from the surface of the Earth. At a certain height, its speed is half the speed it would have just before hitting the ground. The height from the surface of Earth and the ratio of its kinetic energy to its potential energy at that instant are respectively:
