Question

A 20-g bullet flying horizontally becomes embedded in a 4-kg wooden block suspended from the ceiling. What was the speed of the bullet if the block swung 5 cm up from its initial level above ground?

Answer 1

When the block swung its : kinetic energy that it gained during collision gets converted completely into gravitational potential energy at height of 5cm or 0.05 m , so applying conservation of energy after bullet get embedded in block

$\frac{mv^2}{2 } = mgh$

m = combined mass of block and bullet

g = 9.8 m/s²

h = 0.05 m

v = speed of bullet and block combined just after collision

$\frac{v^2}{2 } = gh$

$v= \sqrt{2gh}$

$v= \sqrt{2*9.8*0.05}$

$v\approx 1 m/s$

In horizontal direction there is no external force on system during collision so now applying conservation of momentum in horizontal direction during collision

momentum of system before collison = mass of bullet * speed of bullet = 0.02 * v_b

momentum of system after collision = mv = (4+0.02) * 1 = 4.02 kg m/s

equating both the momentums

0.02 * v_b = 4.02

v_b = 201 m/s