Question

A bullet with a mass ?b=12.3 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

Using law of conservation of energy  

0.5 Mvi^2 + 0.5 kxi^2 = 0.5 Mvf^2 + 0.5 kxf^2 ...... (1)

At maximum compression,

M = k xf^2/vi^2 ...... (2)

Using law of conservation of momentum

m1 u1 + m2 u2 = Mv

M = m1 u1/v ....... (3)

Equating (2) and (3)

Vi = k xf^2/m1 u1 = (205 x 0.35^2)/(0.0123 x 261) = 7.82 m/s ..... (4)

Substituting in (3)

M = m1 u1/vi

m2 = (m1 u1/vi) - m1

m2 = (0.0123 x 261/7.82) - 0.0123 = 0.398 kg = 398 g

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