A 200 g arrow is fired and becomes embedded in a 1.8 kg wooden block that is initially at rest and suspended by a string of length 0.50 m. The arrow and the block displace an angle of 60° to the vertical after the impact. Find the initial velocity, v, of the arrow before it hits the block.

M₁ = 0.2 kg

m₂ = 1.8 kg

v₁ = ? m/s

v₂ = 0 m/s

g = 9.81 m/s

1: conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₁₂ = m₁v₁

2: conservation of energy after the hit

0.5 (m₁ + m₂) v₁₂² = (m₁ + m₂) gh = (m₁ + m₂) g * 0.5 * cos60° ⇒ v₁₂ = √ (g cos60°)

1 + 2:

m₁v₁ = (m₁ + m₂) * √ (g cos60°) ⇒ v₁ = (m₁ + m₂) / m₁ * √ (g cos60°)