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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.

Respuesta :

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°)

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