A body is dropped on ground from a height 'h1 and after hitting t… — Solution | JEE Physics

NVAEasyJEE 2023Kinetic Energy & Work-Energy Theorem

A body is dropped on ground from a height 'h $_1$ and after hitting the ground, it rebounds to a height 'h $_2$ . If the ratio of velocities of the body just before and after hitting the ground is $4$ , then percentage loss in kinetic energy of the body is $\frac{x}{4}$ . The value of $x$ is

Answer

Correct answer:375

Step-by-step solution

Standard Method

Given: Let $u$ and $v$ be the speeds just before and after the body strikes the ground. Given $\frac{u}{v} = 4$ .

Find: The value of $x$ if percentage loss in kinetic energy is $\frac{x}{4}$ .

The loss in kinetic energy is

\Delta K.E. = \frac{1}{2}mu^2 - \frac{1}{2}mv^2

So,

\Delta K.E. = \frac{1}{2}m\left(u^2 - v^2\right)

Substitute the speed ratio

Using $u = 4v$ ,

\Delta K.E. = \frac{1}{2}m\left((4v)^2 - v^2\right)

= \frac{1}{2}m\left(16v^2 - v^2\right)

= \frac{1}{2}m\cdot 15v^2

Convert to percentage loss

Initial kinetic energy just before collision is

K_i = \frac{1}{2}mu^2 = \frac{1}{2}m(4v)^2 = \frac{1}{2}m\cdot 16v^2

Loss in kinetic energy is therefore

\frac{\Delta K.E.}{K_i} = \frac{15}{16}

Hence percentage loss is

\frac{15}{16}\times 100 = 93.75\%

Now,

\frac{x}{4} = 93.75 = \frac{375}{4}

Therefore, the value of $x$ is $375$ .

Common mistakes

Using the ratio $\frac{u}{v} = 4$ as if it means $u^2 - v^2 = 4$ is incorrect. The kinetic energy depends on the square of speed, so first write $u = 4v$ and then square it.
Finding the energy loss correctly but dividing by the final kinetic energy instead of the initial kinetic energy gives a wrong percentage. Percentage loss must be calculated with respect to the initial kinetic energy before impact.
Stopping at $93\%$ due to rough rounding is incorrect here. Since the question writes the percentage loss as $\frac{x}{4}$ , the exact value is $93.75\% = \frac{375}{4}$ .

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