A body of mass 2.9 , kg is suspended from a s | vedclass

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(C) Let $M = 2.9 \, kg$ be the mass of the block and $m = 0.1 \, kg$ be the mass of the bullet. Let $v_0 = 150 \, m/s$ be the velocity of the bullet.B

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let $M = 2.9 \, kg$ be the mass of the block and $m = 0.1 \, kg$ be the mass of the bullet. Let $v_0 = 150 \, m/s$ be the velocity of the bullet.By the principle of conservation of linear momentum during the inelastic collision:$m v_0 = (M + m) V$$0.1 	imes 150 = (2.9 + 0.1) V$$15 = 3 V \Rightarrow V = 5 \, m/s$Now,the combined system swings to a height $h$. By the principle of conservation of mechanical energy:$\frac{1}{2} (M + m) V^2 = (M + m) g h$$h = \frac{V^2}{2g} = \frac{5^2}{2 	imes 10} = \frac{25}{20} = 1.25 \, m$From the geometry of the pendulum,the height $h$ is given by $h = l(1 - \cos 	heta)$,where $l = 2.5 \, m$ is the length of the string.$1.25 = 2.5 (1 - \cos 	heta)$$1 - \cos 	heta = \frac{1.25}{2.5} = 0.5$$\cos 	heta = 1 - 0.5 = 0.5$$	heta = \cos^{-1}(0.5) = 60^{\circ}$

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