A particle of mass m is projected with veloci | vedclass

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(C) The initial velocity of the particle is $\vec{u} = v \cos 	heta \hat{i} + v \sin 	heta \hat{j}$.The final velocity of the particle when it lands

Vedclass · Accepted Answer

$m v \sqrt{2}$

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(C) The initial velocity of the particle is $\vec{u} = v \cos 	heta \hat{i} + v \sin 	heta \hat{j}$.The final velocity of the particle when it lands on the ground is $\vec{v}_f = v \cos 	heta \hat{i} - v \sin 	heta \hat{j}$.The initial momentum is $\vec{p}_i = m \vec{u} = m v \cos 	heta \hat{i} + m v \sin 	heta \hat{j}$.The final momentum is $\vec{p}_f = m \vec{v}_f = m v \cos 	heta \hat{i} - m v \sin 	heta \hat{j}$.The change in momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = (m v \cos 	heta \hat{i} - m v \sin 	heta \hat{j}) - (m v \cos 	heta \hat{i} + m v \sin 	heta \hat{j}) = -2 m v \sin 	heta \hat{j}$.The magnitude of the change in momentum is $|\Delta \vec{p}| = 2 m v \sin 	heta$.Given $	heta = 45^{\circ}$,we have $\sin 45^{\circ} = 1 / \sqrt{2}$.Therefore,$|\Delta \vec{p}| = 2 m v (1 / \sqrt{2}) = \sqrt{2} m v$.

$A$ 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

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