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A particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal from level ground. When the particle lands on the level ground, the magnitude of the change in its momentum will be -
A
$\sqrt{2}\,mv$
B
zero
C
$2mv$
D
$\dfrac{mv}{\sqrt{2}}$
Detailed Solution
Only the vertical component of velocity changes during projectile motion, so the change in momentum is in the vertical direction only.
Horizontal component of velocity: $v_x = v\cos 45^\circ$. It stays constant throughout the flight, so $\Delta p_x = 0$.
Vertical component at the time of projection: $v_{iy} = +v\sin 45^\circ$ (upward).
Vertical component when the particle lands on the same level: $v_{fy} = -v\sin 45^\circ$ (downward).
Initial vertical momentum: $p_{iy} = mv\sin 45^\circ$.
Final vertical momentum: $p_{fy} = -mv\sin 45^\circ$.
Change in momentum: $\Delta \vec{p} = (p_{fy} - p_{iy})\,\hat{j} = (-mv\sin 45^\circ - mv\sin 45^\circ)\,\hat{j} = -2mv\sin 45^\circ\,\hat{j}$
$|\Delta \vec{p}| = 2mv \times \dfrac{1}{\sqrt{2}} = \sqrt{2}\,mv$
The negative sign only shows that the change in momentum is directed vertically downward (along $-y$).
So the magnitude of the change in momentum is $\sqrt{2}\,mv$.
Horizontal component of velocity: $v_x = v\cos 45^\circ$. It stays constant throughout the flight, so $\Delta p_x = 0$.
Vertical component at the time of projection: $v_{iy} = +v\sin 45^\circ$ (upward).
Vertical component when the particle lands on the same level: $v_{fy} = -v\sin 45^\circ$ (downward).
Initial vertical momentum: $p_{iy} = mv\sin 45^\circ$.
Final vertical momentum: $p_{fy} = -mv\sin 45^\circ$.
Change in momentum: $\Delta \vec{p} = (p_{fy} - p_{iy})\,\hat{j} = (-mv\sin 45^\circ - mv\sin 45^\circ)\,\hat{j} = -2mv\sin 45^\circ\,\hat{j}$
$|\Delta \vec{p}| = 2mv \times \dfrac{1}{\sqrt{2}} = \sqrt{2}\,mv$
The negative sign only shows that the change in momentum is directed vertically downward (along $-y$).
So the magnitude of the change in momentum is $\sqrt{2}\,mv$.