A 5 kg block of lead (c = 128 J/(kg °C)) is at rest on a spring with a constant of 150 N/m. A 2 kg block of lead is thrown on top of the 5 kg block from a height of 2.40m above the block and an initial speed of 5 m/s, and they stick together as they move downwards. Assume there is no air resistance involved, and that both blocks are at a temperature of 25 °C before colliding.

a) How far is the spring compressed from its natural length before the collision happens, in m.
b) How fast is the 2 kg block of lead moving just before colliding with the 5 kg block, in m/s?
c) How fast are the two blocks of lead moving together just before they begin to compress the spring further, in m/s?
d) What is the change in temperature of the lead blocks right after colliding, in °C? Assume all lost energy went into this process.

According to Hooke's law, the spring force is equal to the spring stiffness times the displacement from its natural length. Mechanical energy is conserved during the fall of the block and during the compression of the spring. During the collision, momentum is conserved, and some of the mechanical energy is converted to heat. Using these principles, the questions can be solved.

a) Hooke's law:

Mg = kx

(5 kg) (9.8 m/s²) = (150 N/m) x

x = 0.327 m

b) Conservation of energy.

KE = PE₀ + KE₀

½ mv² = mgh + ½ mv₀²

v² = 2gh + v₀²

v² = 2 (9.8 m/s²) (2.40 m) + (5 m/s)²

v = 8.49 m/s

c) Momentum is conserved.

mu = (m + M) v

(2 kg) (8.49 m/s) = (2 kg + 5 kg) v

v = 2.43 m/s

d) The heat generated is equal to the difference in kinetic energy.

q = KE − KE₀

(m + M) C ΔT = ½ mu² − ½ (m + M) v²

(2 kg + 5 kg) (128 J/kg/°C) ΔT = ½ (2 kg) (8.49 m/s)² − ½ (2 kg + 5 kg) (2.43 m/s)²

ΔT = 0.0574 °C