A bullet of mass $20 \mathrm{~g}$ moving horizontally with a velocity of $500 \mathrm{~m/s}$ strikes

Question

A bullet of mass $20 \mathrm{~g}$ moving horizontally with a velocity of $500 \mathrm{~m/s}$ strikes a wooden block of mass $1.98 \mathrm{~kg}$ which is suspended by a light inextensible string. As a result, the centre of gravity of the block is found to rise a vertical distance of $5 \mathrm{~cm}$. The speed of the bullet after it emerges out horizontally from the block will be (Take $g = 10 \mathrm{~m/s}^2$)

Professor PC Thomas Classes · Answer

(c) : Let the mass of the bullet be $m = 0.02 \mathrm{~kg}$ and its initial speed be $u = 500 \mathrm{~m/s}$. Let the mass of the block be $M = 1.98 \mathrm{~kg}$. The block rises by $h = 0.05 \mathrm{~m}$. By energy conservation, the velocity $V$ of the block immediately after the collision is given by $\frac{1}{2} M V^2 = Mgh \implies V = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 0.05} = 1 \mathrm{~m/s}$. By momentum conservation for the bullet-block system: $mu = MV + mv \implies 0.02 	imes 500 = 1.98 	imes 1 + 0.02 	imes v$. This simplifies to $10 = 1.98 + 0.02v \implies 8.02 = 0.02v$. Thus, $v = 8.02 / 0.02 = 401 \mathrm{~m/s}$.