A 1 kg ball moving with a speed of 20 m/s strikes a hard wall at an angle of `30^(@)` with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for 0.5 seconds, the average force acting on the wall is

Options

A. 96 N
B. 48 N
C. 24 N
D. 40 N

Answer

To solve the problem, we need to calculate the average force acting on the wall when a ball strikes it and is reflected back. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
- Mass of the ball (m) = 1 kg
- Initial speed of the ball (v) = 20 m/s
- Angle of incidence with the wall (θ) = 30 degrees
- Time of contact with the wall (Δt) = 0.5 seconds

### Step 2: Resolve the Initial Velocity into Components
The velocity of the ball can be resolved into two components:
- **Horizontal Component (v_x)**:
\[ v_x = v \cdot \cos(θ) = 20 \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

- **Vertical Component (v_y)**:
\[ v_y = v \cdot \sin(θ) = 20 \cdot \sin(30^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]

### Step 3: Calculate Initial Momentum
The initial momentum vector (P_initial) can be expressed as:
\[ P_{initial} = m \cdot v = (1 \, \text{kg}) \cdot (10\sqrt{3} \, \hat{i} - 10 \, \hat{j}) \, \text{kg m/s} \]
\[ P_{initial} = 10\sqrt{3} \, \hat{i} - 10 \, \hat{j} \, \text{kg m/s} \]

### Step 4: Calculate Final Momentum
After striking the wall, the horizontal component of the velocity reverses direction while the vertical component remains unchanged:
- **Final Horizontal Component (v'_x)**:
\[ v'_x = -20 \cdot \cos(30^\circ) = -10\sqrt{3} \, \text{m/s} \]

- **Final Vertical Component (v'_y)**:
\[ v'_y = 20 \cdot \sin(30^\circ) = 10 \, \text{m/s} \]

Thus, the final momentum vector (P_final) is:
\[ P_{final} = m \cdot v' = (1 \, \text{kg}) \cdot (-10\sqrt{3} \, \hat{i} + 10 \, \hat{j}) \, \text{kg m/s} \]
\[ P_{final} = -10\sqrt{3} \, \hat{i} + 10 \, \hat{j} \, \text{kg m/s} \]

### Step 5: Calculate Change in Momentum (ΔP)
The change in momentum (ΔP) is given by:
\[ \Delta P = P_{final} - P_{initial} \]
Substituting the values:
\[ \Delta P = (-10\sqrt{3} \, \hat{i} + 10 \, \hat{j}) - (10\sqrt{3} \, \hat{i} - 10 \, \hat{j}) \]
\[ \Delta P = -20\sqrt{3} \, \hat{i} + 20 \, \hat{j} \, \text{kg m/s} \]

### Step 6: Calculate the Magnitude of Change in Momentum
The magnitude of ΔP can be calculated as:
\[ |\Delta P| = \sqrt{(-20\sqrt{3})^2 + (20)^2} = \sqrt{1200 + 400} = \sqrt{1600} = 40 \, \text{kg m/s} \]

### Step 7: Calculate Average Force (F_avg)
The average force acting on the wall can be calculated using the formula:
\[ F_{avg} = \frac{\Delta P}{\Delta t} \]
Substituting the values:
\[ F_{avg} = \frac{40 \, \text{kg m/s}}{0.5 \, \text{s}} = 80 \, \text{N} \]

### Final Answer
The average force acting on the wall is **80 N**.

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