A ball of 0.5 kg colided with wall at `30^(@)` and bounced back elastically.The speed of ball was `12m//s`. The contact remained for 1s. What is the force applied by wall on ball?

Options

A. `12sqrt(3)N`
B. `sqrt(3)N`
C. `6sqrt(3)N`
D. `3sqrt(3)N`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
The ball collides with the wall at an angle of \(30^\circ\) and bounces back elastically. The mass of the ball is \(0.5 \, \text{kg}\) and its speed before the collision is \(12 \, \text{m/s}\). The contact time with the wall is \(1 \, \text{s}\).

### Step 2: Determine the initial and final momentum components
Since the collision is elastic, we need to analyze the momentum in the direction perpendicular to the wall (horizontal direction).

- The initial horizontal momentum (\(P_i\)) can be calculated using:
\[ P_i = m \cdot v \cdot \cos(\theta) \]
where \(m = 0.5 \, \text{kg}\), \(v = 12 \, \text{m/s}\), and \(\theta = 30^\circ\).

### Step 3: Calculate the initial horizontal momentum
Using the cosine of \(30^\circ\):
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Now substituting the values:
\[ P_i = 0.5 \cdot 12 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \, \text{kg m/s} \]

### Step 4: Determine the final momentum
After the collision, the ball bounces back elastically, meaning it retains the same speed but in the opposite direction. Therefore, the final horizontal momentum (\(P_f\)) is:
\[ P_f = -m \cdot v \cdot \cos(\theta) = -3\sqrt{3} \, \text{kg m/s} \]

### Step 5: Calculate the change in momentum
The change in momentum (\(\Delta P\)) is given by:
\[ \Delta P = P_f - P_i = -3\sqrt{3} - 3\sqrt{3} = -6\sqrt{3} \, \text{kg m/s} \]

### Step 6: Calculate the force
Using the formula for force:
\[ F = \frac{\Delta P}{\Delta t} \]
Substituting the values:
\[ F = \frac{-6\sqrt{3}}{1} = -6\sqrt{3} \, \text{N} \]
The negative sign indicates that the force is in the opposite direction of the initial momentum (which is expected as the wall exerts a force in the opposite direction).

### Final Answer
The magnitude of the force applied by the wall on the ball is:
\[ F = 6\sqrt{3} \, \text{N} \]