A projectile of mass m is fired with velocity | vedclass

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Question

(A) At point $P$,the velocity components are $v_x = v \cos 45^{\circ}$ and $v_y = v \sin 45^{\circ}$. The momentum is $\vec{P}_i = m(v \cos 45^{\circ}

Vedclass · Accepted Answer

$mv \sqrt{2}$

Vedclass · Answer

(A) At point $P$,the velocity components are $v_x = v \cos 45^{\circ}$ and $v_y = v \sin 45^{\circ}$. The momentum is $\vec{P}_i = m(v \cos 45^{\circ} \hat{i} + v \sin 45^{\circ} \hat{j})$.At point $Q$,which is at the same horizontal level as $P$,the horizontal velocity remains $v_x = v \cos 45^{\circ}$ and the vertical velocity is $v_y' = -v \sin 45^{\circ}$. The momentum is $\vec{P}_f = m(v \cos 45^{\circ} \hat{i} - v \sin 45^{\circ} \hat{j})$.The change in momentum is $\Delta \vec{P} = \vec{P}_f - \vec{P}_i = -2mv \sin 45^{\circ} \hat{j}$.The magnitude of change in momentum is $|\Delta \vec{P}| = 2mv \sin 45^{\circ} = 2mv 	imes \frac{1}{\sqrt{2}} = \sqrt{2} mv$.

$A$ projectile of mass $m$ is fired with velocity $v$ from the point $P$ at an angle $45^{\circ}$ with the horizontal. The magnitude of change in momentum,when it passes through the point $Q$ on the same horizontal line on which $P$ lies,is :

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