A 1 kg particle strikes a wall with velocity 1 m/s at an angle of `60^@` with the wall and reflects at the same angle. If it remains in contact with wall for 0.1 s, then the force is

Options

A. `0`
B. `10 sqrt(3)` N
C. `30 sqrt(3) N`
D. `40 sqrt(3) N`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Problem
A 1 kg particle strikes a wall at a velocity of 1 m/s at an angle of 60 degrees and reflects off at the same angle. The contact time with the wall is 0.1 seconds. We need to find the force exerted on the particle during the collision.

### Step 2: Resolve the Initial Velocity
The initial velocity \( \vec{u} \) can be resolved into two components:
- The component parallel to the wall (x-direction):
\[ u_x = u \cos(60^\circ) = 1 \cdot \cos(60^\circ) = 1 \cdot \frac{1}{2} = 0.5 \, \text{m/s} \]
- The component perpendicular to the wall (y-direction):
\[ u_y = u \sin(60^\circ) = 1 \cdot \sin(60^\circ) = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \, \text{m/s} \]

### Step 3: Determine the Final Velocity Components
After the collision, the particle reflects off the wall with the same speed and angle:
- The x-component of the final velocity \( v_x \) remains the same in magnitude but changes direction:
\[ v_x = -u_x = -0.5 \, \text{m/s} \]
- The y-component of the final velocity \( v_y \) remains the same:
\[ v_y = u_y = \frac{\sqrt{3}}{2} \, \text{m/s} \]

### Step 4: Calculate the Change in Momentum
The momentum before and after the collision can be calculated as follows:
- Initial momentum \( \vec{p_i} \):
\[ \vec{p_i} = m \vec{u} = 1 \cdot (0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \]
- Final momentum \( \vec{p_f} \):
\[ \vec{p_f} = m \vec{v} = 1 \cdot (-0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = -0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \]
- Change in momentum \( \Delta \vec{p} \):
\[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} = \left(-0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\right) - \left(0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\right) \]
\[ = -0.5 \hat{i} - 0.5 \hat{i} + \frac{\sqrt{3}}{2} \hat{j} - \frac{\sqrt{3}}{2} \hat{j} = -1 \hat{i} + 0 \hat{j} \]
\[ = -1 \hat{i} \, \text{kg m/s} \]

### Step 5: Calculate the Force
Using the formula for force:
\[ F = \frac{\Delta \vec{p}}{\Delta t} \]
Substituting the values:
\[ F = \frac{-1 \hat{i}}{0.1} = -10 \hat{i} \, \text{N} \]
The magnitude of the force is \( 10 \, \text{N} \) in the negative x-direction.

### Final Answer
The force exerted on the particle is \( 10 \, \text{N} \) in the direction opposite to the initial x-component of the velocity.

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