2000 small balls, each weighing 1 g, strike one square cm of area per second with a velocity of 100 m/s normal to the surface and rebounds with the same velocity. The pressure on the surface is

Options

A. `2xx10^(3)N//m^(2)`
B. `2xx10^(5)N//m^(2)`
C. `4xx10^(3)N//m^(2)`
D. `4xx10^(6)N//m^(2)`

Answer

To solve the problem of finding the pressure exerted on a surface by small balls striking it, we can follow these steps:

### Step 1: Understand the Given Information
We have:
- Number of balls (n) = 2000
- Mass of each ball (m) = 1 g = \(1 \times 10^{-3}\) kg
- Velocity of the balls (v) = 100 m/s
- Area (A) = 1 cm² = \(1 \times 10^{-4}\) m²
- Time (t) = 1 second (since the balls strike the area per second)

### Step 2: Calculate the Change in Momentum for One Ball
The change in momentum (Δp) for one ball when it strikes and rebounds is given by:
\[ \Delta p = \text{Final momentum} - \text{Initial momentum} \]
The initial momentum (p_initial) is:
\[ p_{\text{initial}} = mv = (1 \times 10^{-3} \, \text{kg})(100 \, \text{m/s}) = 0.1 \, \text{kg m/s} \]
The final momentum (p_final) is:
\[ p_{\text{final}} = -mv = -(1 \times 10^{-3} \, \text{kg})(100 \, \text{m/s}) = -0.1 \, \text{kg m/s} \]
Thus, the change in momentum is:
\[ \Delta p = -0.1 - 0.1 = -0.2 \, \text{kg m/s} \]
The magnitude of the change in momentum is:
\[ |\Delta p| = 0.2 \, \text{kg m/s} \]

### Step 3: Calculate the Force Exerted by One Ball
The force (F) exerted by one ball can be calculated using the formula:
\[ F = \frac{\Delta p}{t} \]
Substituting the values:
\[ F = \frac{0.2 \, \text{kg m/s}}{1 \, \text{s}} = 0.2 \, \text{N} \]

### Step 4: Calculate the Total Force Exerted by All Balls
The total force (F_total) exerted by all 2000 balls is:
\[ F_{\text{total}} = n \cdot F = 2000 \cdot 0.2 \, \text{N} = 400 \, \text{N} \]

### Step 5: Calculate the Pressure on the Surface
Pressure (P) is defined as force per unit area:
\[ P = \frac{F_{\text{total}}}{A} \]
Substituting the values:
\[ P = \frac{400 \, \text{N}}{1 \times 10^{-4} \, \text{m}^2} = 4 \times 10^{6} \, \text{N/m}^2 \]

### Final Answer
Thus, the pressure on the surface is:
\[ P = 4 \times 10^{6} \, \text{N/m}^2 \]

---