Question

A particle of mass m is projected with velocity v making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

• A. mv $ \sqrt 2 $
• B. zero
• C. 2 mv
• D. mv / $ \sqrt 2 $

Answer 1

The Correct Option is A

To determine the change in momentum of the particle, we need to consider the initial and final momentum of the projectile. The particle is projected at an angle of 45^\circ from the horizontal with an initial velocity v.

1. The initial velocity components are:
   • Horizontal component: v_x = v \cdot \cos(45^\circ) = \frac{v}{\sqrt{2}}
   • Vertical component: v_y = v \cdot \sin(45^\circ) = \frac{v}{\sqrt{2}}
2. At the highest point in the projectile's motion, the vertical component of velocity becomes zero, but considering the entire flight until the particle hits the ground:
   • The final horizontal component of velocity remains the same as it is not affected by gravity: v_{fx} = \frac{v}{\sqrt{2}}
   • The final vertical component is equal in magnitude and opposite in direction to the initial vertical component as the projectile lands back on the ground: v_{fy} = -\frac{v}{\sqrt{2}}
3. Momentum is given by the product of mass and velocity:
   • Initial momentum: \mathbf{p_i} = m \cdot \mathbf{v_i} = m \left(\frac{v}{\sqrt{2}}\hat{i} + \frac{v}{\sqrt{2}}\hat{j}\right)
   • Final momentum: \mathbf{p_f} = m \cdot \mathbf{v_f} = m \left( \frac{v}{\sqrt{2}}\hat{i} - \frac{v}{\sqrt{2}}\hat{j} \right)
4. The change in momentum is given by the vector difference between the final and initial momentum: \Delta \mathbf{p} = \mathbf{p_f} - \mathbf{p_i}
5. Calculate the change in momentum:
   • \Delta \mathbf{p} = m \left( \frac{v}{\sqrt{2}}\hat{i} - \frac{v}{\sqrt{2}}\hat{j} \right) - m \left(\frac{v}{\sqrt{2}}\hat{i} + \frac{v}{\sqrt{2}}\hat{j}\right)
   • \Delta \mathbf{p} = m \left( 0\hat{i} - 2\frac{v}{\sqrt{2}}\hat{j} \right)
   • \Delta \mathbf{p} = -\sqrt{2} mv \hat{j}
   • The magnitude of the change in momentum is calculated as: |\Delta \mathbf{p}| = mv \sqrt{2}

Therefore, the magnitude of the change in momentum when the particle lands on the level ground is mv \sqrt{2}.