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

$P_i=Nm\cup \hat{i}{\overrightarrow{P}}_f=-Nmv\hat{i}$
$N$ is Number of balls strikes with wall
$N=100$
$\Delta \overrightarrow{P}={\overrightarrow{P}}_f-{\overrightarrow{P}}_i=-2Nmv\hat{i}$
$=-200Nm\hat{\hat{i}}$
${\overrightarrow{F}}_{\text{Total}}=\frac{\Delta \overrightarrow{P}}{\Delta t}=-\frac{200mvt}{t}$
$|\overrightarrow{F}|=\frac{200mv}{t}$

Answer 2

The momentum change for each ball is given by:

\[ \Delta p = p_f - p_i = -Nm \hat{i} - Nm \hat{i} = -2Nm \hat{i} \]

Where:

• \( N = 100 \) (number of balls),
• \( \hat{i} \) is the unit vector in the direction of motion.

The total force exerted by the balls on the wall is:

\[ F_{\text{total}} = \frac{\Delta p}{\Delta t} = \frac{-200Nm}{t} = \frac{200mv}{t} \]

Thus, the total force is \( \frac{200mv}{t} \).