JEE Physics 2024 Question with Solution

Given: mass of the body is $m = 0.1 \, \text{kg}$, height of fall is $10 \, \text{m}$, rebound height is $5 \, \text{m}$, and $g = 9.8 \, \text{m/s}^2$.

Find: the impulse imparted by the ground to the body.

The velocity just before hitting the ground is found from

with $u = 0$, so

This velocity is downward.

The velocity just after rebounding is found from the rebound height:

This velocity is upward.

Impulse is the change in momentum. Taking upward as positive, the initial velocity is $-v_1$ and the final velocity is $+v_2$. Therefore,

The solution working gives the final numerical value $2.39 \, \text{kg m/s}$. The solution states The Correct Option is D, while the answer key says (3) and both options C and D contain the same value $2.39 \, \text{kg m/s}$. Since the solution is the primary source, the answer is marked as D.

An alternate approach shown on the page also concludes the impulse as $2.39 \, \text{kg m/s}$, although it contains an inconsistent intermediate line for rebound speed. The final value still matches the standard calculation above.