A 14 g bullet is fired into the bob of a ballistic pendulum of mass 1.6 kg. When the bob is at its maximum height, the strings make an angle of 60° with the vertical. The length of the pendulum is 2.3 m. Find the speed of the bullet.
?m/s
Any help is greatly appreciated. Thanks!
0.06133
To find the speed of the bullet, you can use the principle of conservation of momentum and the conservation of mechanical energy.
1. Conservation of momentum:
The momentum of the bullet before impact is equal to the momentum of the bullet and the pendulum after impact.
Momentum of bullet before impact = Momentum of bullet and pendulum after impact
Given:
Mass of the bullet (m1) = 14 g = 0.014 kg
Mass of the pendulum (m2) = 1.6 kg
Velocity of the bullet before impact (u1) = Unknown
Velocity of the bullet and pendulum after impact (v) = 0 (at maximum height)
Using the equation of momentum:
m1 * u1 = (m1 + m2) * v
Substituting the given values:
0.014 kg * u1 = (0.014 kg + 1.6 kg) * 0
Simplifying the equation:
0.014 kg * u1 = 0
This equation tells us that the momentum of the bullet before impact is zero because the bullet gets embedded in the pendulum bob.
2. Conservation of mechanical energy:
The mechanical energy of the system (bullet + pendulum) is conserved throughout the motion. At the maximum height, all of the kinetic energy is converted into potential energy.
Kinetic energy before impact = Potential energy after impact
Given:
Mass of the pendulum (m2) = 1.6 kg
Length of the pendulum (L) = 2.3 m
Angle with the vertical (θ) = 60°
Using the equation of potential energy:
m2 * g * L * (1 - cosθ) = 0.5 * (m1 + m2) * v^2
Substituting the given values:
1.6 kg * 9.8 m/s^2 * 2.3 m * (1 - cos60°) = 0.5 * (0.014 kg + 1.6 kg) * v^2
Simplifying the equation and solving for v:
v^2 = (1.6 kg * 9.8 m/s^2 * 2.3 m * (1 - 0.5)) / (0.014 kg + 1.6 kg)
v^2 = (34.496 m^2/s^2) / 1.614 kg
v^2 ≈ 21.35 m^2/s^2
Taking the square root of both sides:
v ≈ √(21.35 m^2/s^2)
v ≈ 4.62 m/s
Therefore, the speed of the bullet is approximately 4.62 m/s.