A particle of mass m is projected with velocity making an angle of `45^(@)` with the horizontal When the particle lands on the level ground the magnitude of the change in its momentum will be .

Options

A. 2 mv
B. `(mv)/(sqrt(2))`
C. `mvsqrt(2)`
D. zero

Answer

To solve the problem step by step, we will analyze the motion of the particle projected at an angle of 45 degrees and calculate the change in momentum when it lands back on the ground.

### Step 1: Understand the Initial Conditions
The particle is projected with an initial velocity \( u \) at an angle of \( 45^\circ \) with the horizontal. The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos 45^\circ = \frac{u}{\sqrt{2}} \)
- Vertical component: \( u_y = u \sin 45^\circ = \frac{u}{\sqrt{2}} \)

### Step 2: Analyze the Final Conditions
When the particle lands on the ground, it will have a final velocity \( v \) which will also be at an angle of \( 45^\circ \) with the horizontal. The final velocity can similarly be resolved:
- Horizontal component: \( v_x = v \cos 45^\circ = \frac{v}{\sqrt{2}} \)
- Vertical component: \( v_y = v \sin 45^\circ = \frac{v}{\sqrt{2}} \)

### Step 3: Determine the Relationship Between Initial and Final Velocities
In projectile motion, the magnitude of the initial velocity \( u \) and the final velocity \( v \) when the particle lands are equal, i.e., \( u = v \).

### Step 4: Calculate Change in Momentum
The momentum of the particle is given by \( p = mv \), where \( m \) is the mass of the particle.

- Initial momentum in the horizontal direction:
\[ p_{ix} = m u_x = m \left(\frac{u}{\sqrt{2}}\right) = \frac{mu}{\sqrt{2}} \]
- Final momentum in the horizontal direction:
\[ p_{fx} = m v_x = m \left(\frac{v}{\sqrt{2}}\right) = \frac{mv}{\sqrt{2}} = \frac{mu}{\sqrt{2}} \quad (\text{since } u = v) \]
- Change in momentum in the horizontal direction:
\[ \Delta p_x = p_{fx} - p_{ix} = \frac{mu}{\sqrt{2}} - \frac{mu}{\sqrt{2}} = 0 \]

- Initial momentum in the vertical direction:
\[ p_{iy} = m u_y = m \left(\frac{u}{\sqrt{2}}\right) = \frac{mu}{\sqrt{2}} \]
- Final momentum in the vertical direction (downward is negative):
\[ p_{fy} = m v_y = m \left(-\frac{v}{\sqrt{2}}\right) = -\frac{mv}{\sqrt{2}} = -\frac{mu}{\sqrt{2}} \quad (\text{since } u = v) \]
- Change in momentum in the vertical direction:
\[ \Delta p_y = p_{fy} - p_{iy} = -\frac{mu}{\sqrt{2}} - \frac{mu}{\sqrt{2}} = -\frac{2mu}{\sqrt{2}} = -\frac{2mu}{\sqrt{2}} \]

### Step 5: Calculate the Magnitude of Change in Momentum
The magnitude of the change in momentum is given by:
\[ |\Delta p| = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} = \sqrt{0^2 + \left(-\frac{2mu}{\sqrt{2}}\right)^2} = \frac{2mu}{\sqrt{2}} = \sqrt{2} mu \]

### Final Answer
Thus, the magnitude of the change in momentum when the particle lands on the level ground is:
\[ |\Delta p| = \sqrt{2} mu \]