A simple pendulum of length L = frac{10}{3} t | vedclass

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(A) $1$. At the highest point of the trajectory,the vertical velocity of the bullet is $0$. The horizontal velocity is $v_x = u \cos 	heta = 50 \cos

Vedclass · Accepted Answer

$\cos^{-1}(0.8)$

Vedclass · Answer

(A) $1$. At the highest point of the trajectory,the vertical velocity of the bullet is $0$. The horizontal velocity is $v_x = u \cos 	heta = 50 \cos 	heta$.$2$. By conservation of linear momentum during the collision: $m v_x = (m + 3m) V$,where $V$ is the velocity of the combined mass immediately after collision. Thus,$V = \frac{50 \cos 	heta}{4} = 12.5 \cos 	heta$.$3$. By conservation of mechanical energy for the pendulum after collision: $\frac{1}{2} (4m) V^2 = (4m) g h$,where $h$ is the vertical height reached. For an angle of $120^{\circ}$ from the vertical,$h = L(1 - \cos 120^{\circ}) = L(1 - (-0.5)) = 1.5 L$.$4$. Substituting $L = \frac{10}{3} 	ext{ m}$,we get $h = 1.5 	imes \frac{10}{3} = 5 	ext{ m}$.$5$. Using $\frac{1}{2} V^2 = g h$,we have $V^2 = 2 	imes 10 	imes 5 = 100$,so $V = 10 	ext{ ms}^{-1}$.$6$. Equating $12.5 \cos 	heta = 10$,we get $\cos 	heta = \frac{10}{12.5} = 0.8$. Thus,$	heta = \cos^{-1}(0.8)$.

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