JEE Physics 2023 Question with Solution

Given: 100 balls, each of mass $m$, move normally toward the wall with speed $v$ and rebound with the same speed in time $t$.

Find: The total force exerted by the balls on the wall.

The solution uses the rate of change of momentum.

Initial momentum of all the balls together:

$$P_i = 100mv$$

Final momentum of all the balls together is in the opposite direction:

$$P_f = -100mv$$

Hence, change in momentum is:

$$\Delta p = P_f - P_i = -100mv - 100mv = -200mv$$

Therefore, the magnitude of force is:

$$F = \left|\frac{\Delta p}{t}\right| = \frac{200mv}{t}$$

So, the total force exerted by the balls on the wall is $\frac{200mv}{t}$. The correct option is B.

The solution also shows an option label C, but the worked result clearly matches option $\frac{200mv}{t}$, which is option B in the provided options.

Given: Reflection is normal and the speed before and after collision is the same.

Find: Total force on the wall due to all the balls.

For one ball, if motion toward the wall is taken as positive, then:

$$p_i = mv, \qquad p_f = -mv$$

So for one ball:

$$\Delta p = p_f - p_i = -mv - mv = -2mv$$

For 100 balls:

$$\Delta p_{\text{total}} = 100(-2mv) = -200mv$$

Now use:

$$F_{\text{total}} = \frac{\Delta p_{\text{total}}}{t}$$

Hence, magnitude of the force is:

$$|F_{\text{total}}| = \frac{200mv}{t}$$

Therefore, the total force exerted by the balls on the wall is $\frac{200mv}{t}$.