JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 10)
100 balls each of mass $$\mathrm{m}$$ moving with speed $$v$$ simultaneously strike a wall normally and reflected back with same speed, in time $$\mathrm{t ~s}$$. The total force exerted by the balls on the wall is
$$\frac{200 m v}{t}$$
$$\frac{100 m v}{t}$$
$$\frac{m v}{100 t}$$
$$200 m v t$$
Explanation
When the balls strike the wall, the change in momentum of each ball is given by:
$$\Delta p = mv - (-mv) = 2mv$$
Since there are 100 balls, the total change in momentum of all the balls is $$\Delta P = 2m(100v) = 200mv.$$ The time taken for all the balls to strike the wall is $\mathrm{t}$ seconds. Therefore, the average force exerted on the wall is given by:
$$F = \frac{\Delta P}{\mathrm{t}} = \frac{2m(100v)}{\mathrm{t}} = \frac{200mv}{\mathrm{t}}$$
Therefore, the total force exerted by the balls on the wall is $$\boxed{F = \frac{200mv}{\mathrm{t}}}$$
$$\Delta p = mv - (-mv) = 2mv$$
Since there are 100 balls, the total change in momentum of all the balls is $$\Delta P = 2m(100v) = 200mv.$$ The time taken for all the balls to strike the wall is $\mathrm{t}$ seconds. Therefore, the average force exerted on the wall is given by:
$$F = \frac{\Delta P}{\mathrm{t}} = \frac{2m(100v)}{\mathrm{t}} = \frac{200mv}{\mathrm{t}}$$
Therefore, the total force exerted by the balls on the wall is $$\boxed{F = \frac{200mv}{\mathrm{t}}}$$
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