Question

100 balls each of mass $m$ moving with speed $v$ simultaneously strike a wall normally and reflected back with same speed, in time $t s$ The total force exerted by the balls on the wall is

• A. $\frac{100 m v}{t}$
• B. $200 m v t$
• C. $\frac{200 m v}{t}$
• D. $\frac{m v}{100 t}$

Hint:

The total force exerted by a group of objects striking and bouncing off a surface is proportional to the rate of change of their momentum.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the total force exerted by the balls on the wall.

Concept: When a ball strikes a wall and is reflected back with the same speed, the change in momentum for one ball can be calculated. By using Newton's second law of motion, the force can be calculated.

Step-by-step Explanation:

The mass of each ball is \(m\) and its initial speed is \(v\). After reflecting with the same speed, the final velocity becomes \(-v\) (since the direction reverses).

The initial momentum of one ball is \(mv\) and the final momentum after reflection is \(-mv\).

The change in momentum for one ball is thus:

\(\Delta p = -mv - mv = -2mv\)

Here, the negative sign indicates a reversal in direction.

According to Newton's second law, force is the change in momentum per unit time.

For all 100 balls, the total change in momentum is:

\(\Delta P = 100 \times (-2mv) = -200mv\)

Thus, the total force \(F\) exerted by the 100 balls on the wall is given by:

\(F = \frac{|\Delta P|}{t} = \frac{200mv}{t}\)

This results in the magnitude of force being \(\frac{200mv}{t}\) N.

Conclusion: The correct answer is \(\frac{200mv}{t}\).
Therefore, the correct option is

$\frac{200 m v}{t}$