Question

A 8.00 g bullet traveling at 515 m/s embeds itself in a 1.57 kg wooden block at rest on a frictionless surface. The block is attached to a spring with k = 82.0 N/m.

Part A) Find the period.

Part B) Find the amplitude of the subsequent simple harmonic motion.

Part C) Find the total energy of the bullet + block + spring system before the bullet enters the block.

Part D) Find the total energy of the bullet + block +spring system after the bullet enters the block.

Answer 1

Part A
T = 2π √( m / k )
where
T = period = ?
m = total mass = (1.57 kg) + (0.008 kg) = (1.578 kg)
k = spring constant = 82 N/m
so
T = 2π √( (1.578 kg) / (82 N/m) )
T = 0.8716 s

Part B
The speed of the combined mass just after impact is:
(0.008 kg)(515 m/s) + (1.57 kg)(0 m/s) = (1.578 kg)×V
V = 2.6109 m/s
The kinetic energy just after impact is:
KE = (1.578 kg)×(2.6109 m/s)²/2
KE = 5.378 J
The amplitude of the motion is:
(5.378 J) = (82 N/m)×A²/2
A = 0.362 m

Part C
Before impact, the only energy in the system is the kinetic energy of the bullet.
Eb = (0.008 kg)×(515 m/s)²/2
Eb = 1060.9 J

Part D

Energy After collision = 1060.9+5.378 = 1066.278 J