A body is dropped on ground from a height ' h1 ' and after hitting the ground, it rebounds to a height ' h2 '. If the ratio of velocities of the body just before and after hitting ground is 4 , then percentage loss in kinetic energy of the body is x/4 The value of x is _______.

JEE Mains - Kinetic Energy Loss on Rebound

Question: A body is dropped from a height \( h_1 \) and, after hitting the ground, rebounds to a height \( h_2 \). If the ratio of the velocities of the body just before and after hitting the ground is 4, then the percentage loss in kinetic energy of the body is given by \(\frac{x}{4}\%\). Determine the value of \( x \).
[6-Apr-2023 Shift 2]

Select the correct answer:

(A) \(300\)

(B) \(350\)

(C) \(375\)

(D) \(400\)

Solution

Let \( v_1 \) be the speed just before impact and \( v_2 \) be the speed just after the rebound. Given: \[ \frac{v_1}{v_2} = 4 \quad \Longrightarrow \quad v_1 = 4v_2. \]

The kinetic energy before collision is: \[ KE_{\text{before}} = \frac{1}{2} m v_1^2 = \frac{1}{2} m (4v_2)^2 = \frac{1}{2} m (16v_2^2) = 8 m v_2^2. \]

The kinetic energy after collision is: \[ KE_{\text{after}} = \frac{1}{2} m v_2^2. \]

Thus, the loss in kinetic energy is: \[ \Delta KE = KE_{\text{before}} - KE_{\text{after}} = 8mv_2^2 - \frac{1}{2}mv_2^2 = \frac{15}{2} m v_2^2. \]

The percentage loss in kinetic energy is: \[ \text{Percentage loss} = \frac{\Delta KE}{KE_{\text{before}}} \times 100\% = \frac{\frac{15}{2} m v_2^2}{8 m v_2^2} \times 100\% = \frac{15}{16} \times 100\% \approx 93.75\%. \]

According to the problem, this percentage loss is expressed as \(\frac{x}{4}\%\). Thus, \[ \frac{x}{4} = 93.75 \quad \Longrightarrow \quad x = 93.75 \times 4 = 375. \]

Therefore, the value of \( x \) is 375.