A body of mass m hits a wall with speed v at an angle of `30^o` with normal and reflects at same angle from normal on other side of normal. The magnitude of change in linear momentum of the body is (Consider elastic collision)

Options

A. `frac{8sqrt3mv}{2}`
B. mv
C. `sqrt3mv`
D. Zero

Answer

To solve the problem of finding the magnitude of change in linear momentum of a body of mass \( m \) that hits a wall at an angle of \( 30^\circ \) with the normal and reflects elastically, we can follow these steps:

### Step 1: Understand the Initial Conditions
The body hits the wall with speed \( v \) at an angle of \( 30^\circ \) with respect to the normal. We need to resolve the velocity into its components.

### Step 2: Resolve Initial Momentum
The initial momentum \( \vec{p_i} \) can be resolved into two components:
- The component parallel to the wall (x-direction): \( p_{ix} = mv \cos(30^\circ) \)
- The component perpendicular to the wall (y-direction): \( p_{iy} = -mv \sin(30^\circ) \) (negative because it is directed towards the wall)

Using the values of \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \):
\[ p_{ix} = mv \cdot \frac{\sqrt{3}}{2} = \frac{mv\sqrt{3}}{2} \]
\[ p_{iy} = -mv \cdot \frac{1}{2} = -\frac{mv}{2} \]

Thus, the initial momentum vector is:
\[ \vec{p_i} = \left( \frac{mv\sqrt{3}}{2} \hat{i} - \frac{mv}{2} \hat{j} \right) \]

### Step 3: Resolve Final Momentum
After the collision, the body reflects off the wall at the same angle, so:
- The x-component remains the same: \( p_{fx} = -mv \cos(30^\circ) = -\frac{mv\sqrt{3}}{2} \)
- The y-component changes direction: \( p_{fy} = -(-mv \sin(30^\circ)) = \frac{mv}{2} \)

Thus, the final momentum vector is:
\[ \vec{p_f} = \left( -\frac{mv\sqrt{3}}{2} \hat{i} + \frac{mv}{2} \hat{j} \right) \]

### Step 4: Calculate Change in Momentum
The change in momentum \( \Delta \vec{p} \) is given by:
\[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} \]
Substituting the values:
\[ \Delta \vec{p} = \left( -\frac{mv\sqrt{3}}{2} \hat{i} + \frac{mv}{2} \hat{j} \right) - \left( \frac{mv\sqrt{3}}{2} \hat{i} - \frac{mv}{2} \hat{j} \right) \]
\[ = -\frac{mv\sqrt{3}}{2} \hat{i} + \frac{mv}{2} \hat{j} - \frac{mv\sqrt{3}}{2} \hat{i} + \frac{mv}{2} \hat{j} \]
\[ = -mv\sqrt{3} \hat{i} + mv \hat{j} \]

### Step 5: Find the Magnitude of Change in Momentum
The magnitude of the change in momentum is:
\[ |\Delta \vec{p}| = \sqrt{(-mv\sqrt{3})^2 + (mv)^2} \]
\[ = \sqrt{3m^2v^2 + m^2v^2} = \sqrt{4m^2v^2} = 2mv \]

### Final Answer
The magnitude of the change in linear momentum of the body is \( 2mv \).