1. Conservation of Momentum
* Before the collision: The bullet has momentum (m₁v₁) and the pendulum is at rest (m₂v₂ = 0).
* After the collision: The bullet and pendulum move together as one unit (m₁ + m₂) with a common velocity (v').
The conservation of momentum equation is:
m₁v₁ + m₂v₂ = (m₁ + m₂)v'
2. Solving for the common velocity (v')
* m₁ = 0.012 kg (mass of bullet)
* v₁ = 380 m/s (initial velocity of bullet)
* m₂ = 6 kg (mass of pendulum)
* v₂ = 0 m/s (initial velocity of pendulum)
Substitute the values into the momentum equation and solve for v':
(0.012 kg)(380 m/s) + (6 kg)(0 m/s) = (0.012 kg + 6 kg)v'
v' ≈ 0.76 m/s
3. Conservation of Energy
* Immediately after the collision: The system has kinetic energy (1/2(m₁ + m₂)v'²).
* At the highest point: The system has potential energy (m₁ + m₂)gh, where h is the vertical height it rises.
The conservation of energy equation is:
1/2(m₁ + m₂)v'² = (m₁ + m₂)gh
4. Solving for the vertical height (h)
* v' ≈ 0.76 m/s (calculated above)
* g = 9.8 m/s² (acceleration due to gravity)
Substitute the values into the energy equation and solve for h:
1/2(0.012 kg + 6 kg)(0.76 m/s)² = (0.012 kg + 6 kg)(9.8 m/s²)h
h ≈ 0.029 m
Therefore, the ballistic pendulum rises approximately 0.029 meters (or 2.9 centimeters) vertically.
