Question

A bullet with a mass ?b=12.7 g is fired into a block of wood at velocity ?b=261 m/s. The block is attached to a spring that has a spring constant ? of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block.

Answer 1

given

mass of bullet mb = 12.7*10^-3 kg

speed of bullet vb = 261 m/s

k = 205 N/m

compressed length of the spring x = 0.35 m

from the conservation of energy

spring potential energy = bullet and block kinetic energy

1/2kx^2 = 1/2(mb+M)v^2

from the conservation of moemntum

initial momentum of bullet = final momentum of bullet and block

mb*vb = (mb+M)v...(1)

1/2kx^2 = 1/2*mb*vb*v

v = 205*0.35^2/12.7*10^-3*261

v = 7.57 m/s

from equation 1

(mb+M) = mb*vb/v = 12.7*10^-3*261/7.57

mb+M = 437.87*10^-3

M = 437.87*10^-3-12.7*10^-3

M = 0.425 kg