The projectile motion of a particle of mass 5 | vedclass

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(D) Given: Mass $m = 5\, g = 5 	imes 10^{-3}\, kg$. Initial velocity $u = 5 \sqrt{2}\, m/s$ at an angle $	heta = 45^{\circ}$.In projectile motion,th

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$5$

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(D) Given: Mass $m = 5\, g = 5 	imes 10^{-3}\, kg$. Initial velocity $u = 5 \sqrt{2}\, m/s$ at an angle $	heta = 45^{\circ}$.In projectile motion,the speed at the same horizontal level is the same,so the magnitude of final velocity $v = u = 5 \sqrt{2}\, m/s$.The initial velocity vector is $\vec{u} = u \cos 45^{\circ} \hat{i} + u \sin 45^{\circ} \hat{j}$.The final velocity vector at point $B$ is $\vec{v} = u \cos 45^{\circ} \hat{i} - u \sin 45^{\circ} \hat{j}$.The change in momentum is $\Delta \vec{P} = m(\vec{v} - \vec{u}) = m(u \cos 45^{\circ} \hat{i} - u \sin 45^{\circ} \hat{j} - (u \cos 45^{\circ} \hat{i} + u \sin 45^{\circ} \hat{j}))$.$\Delta \vec{P} = m(-2u \sin 45^{\circ} \hat{j}) = -2mu \sin 45^{\circ} \hat{j}$.The magnitude is $|\Delta \vec{P}| = 2mu \sin 45^{\circ}$.Substituting the values: $|\Delta \vec{P}| = 2 	imes (5 	imes 10^{-3}) 	imes (5 \sqrt{2}) 	imes \frac{1}{\sqrt{2}} = 2 	imes 5 	imes 10^{-3} 	imes 5 = 50 	imes 10^{-3} = 5 	imes 10^{-2}\, kg \cdot m/s$.Comparing with $x 	imes 10^{-2}$,we get $x = 5$.

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