A particle of mass m is projected from the gr | vedclass

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Question

(A) Let the first particle be $P_1$ and the second be $P_2$. At the highest point,the velocity of $P_1$ is $v_{x1} = u_0 \cos \alpha$ and $v_{y1} = 0$

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Let the first particle be $P_1$ and the second be $P_2$. At the highest point,the velocity of $P_1$ is $v_{x1} = u_0 \cos \alpha$ and $v_{y1} = 0$.The second particle $P_2$ is thrown vertically upward with speed $u_0$. At the highest point of $P_1$ (height $H = \frac{u_0^2 \sin^2 \alpha}{2g}$),the velocity of $P_2$ is $v_{y2} = \sqrt{u_0^2 - 2gH} = \sqrt{u_0^2 - u_0^2 \sin^2 \alpha} = u_0 \cos \alpha$.Applying conservation of linear momentum in the horizontal direction:$m(u_0 \cos \alpha) + m(0) = (2m)v_x \implies v_x = \frac{u_0 \cos \alpha}{2}$.Applying conservation of linear momentum in the vertical direction:$m(0) + m(u_0 \cos \alpha) = (2m)v_y \implies v_y = \frac{u_0 \cos \alpha}{2}$.The angle $	heta$ with the horizontal is given by $	an 	heta = \frac{v_y}{v_x} = \frac{u_0 \cos \alpha / 2}{u_0 \cos \alpha / 2} = 1$.Therefore,$	heta = 45^{\circ} = \frac{\pi}{4}$.

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