A 15 g bullet is fired horizontally into a block of wood with mass of 2.5kg and embedded in the block. Init?

m₁ = mass of the bullet = 0.015 kg

m₂ = mass of the block = 2.5 kg

h = height gained by the combined masses = 0.15 m

g = acceleration by gravity = 9.8 m/s²

The potential energy of the total mass at a height of 15 cm is:

PE = (m₁ + m₂) × g × h

That energy must be the same as the kinetic energy of the total mass just after impact.

KE = (m₁ + m₂) × v² / 2

so

(m₁ + m₂) × v² / 2 = (m₁ + m₂) × g × h

v² / 2 = g × h

v² = 2 × g × h

v = √(2 × g × h)

That is the velocity of the combined mass (block + bullet) just after impact.

That velocity is caused by the inelastic collision of the bullet with the block

m₁×v₁ + m₂×v₂ = (m₁ + m₂) × √(2 × g × h)

where

v₁ = velocity of the bullet before impact

v₂ = velocity of the block before impact

As the velocity of the block before impact is 0 m/s, the formula simplifies to

m₁×v₁ = (m₁ + m₂) × √(2 × g × h)

so

v₁ = (m₁ + m₂) × √(2 × g × h) / m₁

Now fill in the numbers.

v₁ = ((0.015 kg) + (2.5 kg)) × √(2 × (9.8 m/s²) × (0.15 m)) / (0.015 kg)

v₁ = (2.515 kg) × √(2.94 m²/s²) / (0.015 kg)

v₁ = (167.7) × (1.714 m/s)

v₁ = 287.5 m/s

Convert to m/k/s system.

m of bullet: .015 kg

m of block: 2.5kg

h=.15m

Find GPE.

GPE=(.015+2.5)(9.81)(.15)

GPE=3.7 J

FIND velocity using KE.

3.7J=(.5)(.015+2.5)v^2

3.7J=1.25kg(v^2)

3.7/1.25=2.96

Take square root of both sides

v=1.72 m/s