Question

A block of mass 0.992 kg rests on a frictionless horizontal surface. The block is attached to an ideal spring. Calibration shows that a force of 0.75 N is required to compress the spring 0.25 cm. A 8.0-g rifle bullet is fired and embeds itself in the block, compressing the spring 15.0 cm before rebounding. (a) What was the speed of the block just after impact? (b) What was the initial bullet speed?

Answer 1

Spring constant, k = F/x = 0.75/(0.25 x 10^-2) = 300 N/m

Using law of conservation of energy,

0.5 kx^2 = 0.5 (m + M) v^2

V = sqrt[(300 x 0.15^2)/(0.008 + 9.992)] = 2.59 m/s

B)

Using law of conservation of momentum,

u = (0.008 + 0.992) x 2.59/0.008 = 323.75 m/s

Comment in case any doubt please rate my answer....