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 level ground, the magnitude of change in its momentum will be:

Accepted Answer

1. **Identify Velocity Components:**
   - Initial velocity vector: $\vec{v}_i = v\cos	heta \hat{i} + v\sin	heta \hat{j}$.
   - Final velocity vector (on landing): Due to gravity acting vertically downwards, the horizontal component remains constant, while the vertical component reverses direction (magnitude remains same on level ground).
   - $\vec{v}_f = v\cos	heta \hat{i} - v\sin	heta \hat{j}$ .
2. **Calculate Change in Momentum:**
   - Change in momentum $\Delta\vec{p} = m\vec{v}_f - m\vec{v}_i$.
   - $\Delta\vec{p} = m(v\cos	heta \hat{i} - v\sin	heta \hat{j}) - m(v\cos	heta \hat{i} + v\sin	heta \hat{j})$.
   - $\Delta\vec{p} = -2mv\sin	heta \hat{j}$.
3. **Magnitude of Change:**
   - $|\Delta\vec{p}| = 2mv\sin	heta$.
4. **Substitute Given Values:**
   - 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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