Question

A block of mass M = 5.60 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5860 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 660 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

Answer 1

m? = mass of the block = 5.60 kg
u? = initial velocity of the block = 0.00 m/s
m? = mass of the bullet = 0.00920 kg
u? = initial speed of the bullet = 660 m/s
k = spring constant = 5860 N/m

Use conservation of momentum to find the speed of the combined mass just after the collision.

m?×u? + m?×u? = (m? + m?)×v
so
v = (m?×u? + m?×u?) / (m? + m?)

v = [(5.6 kg)×(0 m/s) + (0.0092 kg)×(660 m/s)] / [(5.6 kg) + (0.0092 kg)]
v = 1.0825 m/s

The kinetic energy of the combined mass just after the collision is:
KE = (m? + m?) × v² / 2
KE = [(5.6 kg) + (0.0092 kg) × (1.0825 m/s)² / 2
KE = 3.286 J

That must also be the potential energy in the spring at maximum compression.
The maximum compression is also the amplitude.
PE = k × A² / 2
(3.286) = (5860 N/m) × A² / 2
A = 0.0334 m