A particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal. When

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:

Accepted Answer

1. **Concept:** In projectile motion, the horizontal component of velocity remains constant (ignoring air resistance), while the vertical component reverses its direction (but retains the same magnitude) when the particle returns to the same level [NCERT Class 11, Physics Part I, Sec 4.9 Projectile Motion].
2. **Velocity Components:**
   - Initial Velocity: $\vec{v}_i = v\cos	heta \hat{i} + v\sin	heta \hat{j}$
   - Final Velocity (on landing): $\vec{v}_f = v\cos	heta \hat{i} - v\sin	heta \hat{j}$
3. **Change in Momentum ($\Delta \vec{p}$):**
   $$ \Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) $$
   $$ \Delta \vec{p} = m[(v\cos	heta \hat{i} - v\sin	heta \hat{j}) - (v\cos	heta \hat{i} + v\sin	heta \hat{j})] $$
   $$ \Delta \vec{p} = -2mv\sin	heta \hat{j} $$
4. **Magnitude:**
   $$ |\Delta \vec{p}| = 2mv\sin	heta $$
   Given $	heta = 45^\circ$:
   $$ |\Delta \vec{p}| = 2mv \sin(45^\circ) = 2mv \left(\frac{1}{\sqrt{2}}ight) = \sqrt{2}mv $$

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