Question

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\(\frac{x}{4}\). The value of x is_____.

Answer 1

Correct Answer: 375

To solve the problem, we analyze the kinetic energy before and after the body hits the ground. The velocity just before hitting the ground (v₁) and just after rebounding (v₂) are related by the given ratio: \( \frac{v_2}{v_1} = \frac{1}{4} \). Since kinetic energy (KE) is given by \( \text{KE} = \frac{1}{2}mv^2 \), we have:

• Initial kinetic energy (just before impact): \( \text{KE}_1 = \frac{1}{2}m(v_1)^2 \)
• Final kinetic energy (just after rebound): \( \text{KE}_2 = \frac{1}{2}m(v_2)^2 \)

Substituting \( v_2 = \frac{v_1}{4} \) gives:
\( \text{KE}_2 = \frac{1}{2}m\left(\frac{v_1}{4}\right)^2 = \frac{1}{32}m(v_1)^2 \)
Now, calculate the percentage loss in kinetic energy:
\( \text{Percentage Loss} = \left(\frac{\text{KE}_1 - \text{KE}_2}{\text{KE}_1}\right) \times 100 \% = \left(1 - \frac{1}{16}\right) \times 100 \% = \frac{15}{16} \times 100 \% = 93.75\% \)
Given the problem statement is \(\frac{x}{4} = 93.75\), we solve for \(x\):
\( x = 93.75 \times 4 = 375 \)
Thus, the value of x is 375, which falls within the specified range 375,375.