Question

A 13.1-g bullet is fired into a block of wood at 245 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

PE at full compression = Initial KE system

½*k*x² = ½* (M+m)*v²
½*(205)*(0.35)² = ½*(M + 0.0131)*v²
12.55625 = ½*(M + 0.0131)*v² ……………eq1

Using conservation of momentum between the bullet and the block
0.0131*(245) = (M + 0.0131)*v
3.2095 = (M + 0.0131)*v
v = 3.2095/(M + 0.0131) ………………….eq2

solving eq1 and eq2

12.55625 = ½*(M + 0.0131)*{ 3.2095/(M + 0.0131)}²

12.55625 = ½*(M + 0.0131)*{ 3.2095/(M + 0.0131)}²

25.1125 *(M + 0.0131)² = 10.30089025*M² + 0.13494162275

25.1125*M² + 0.004309556 + 0.6579475*M = 10.30089025*M² + 0.13494162275

14.8116M² + 0.6579475M – 0.130632066

M = 0.0743 kg

M = 74.3 g