p = mv
where:
Linear momentum is a vector quantity that describes the motion of an object considering both its mass and velocity. The direction of momentum is the same as the direction of velocity.
A 2 kg ball moving at 5 m/s to the right has a momentum of:
p = (2 kg)(5 m/s) = 10 kg⋅m/s to the right
J = F∆t = ∆p = m∆v
where:
A 0.145 kg baseball moving at 40 m/s is caught in 0.05 s. Calculate the average force applied to stop the ball.
Solution:
∆p = m∆v = (0.145 kg)(0 - 40 m/s) = -5.8 kg⋅m/s
F = ∆p/∆t = -5.8/-0.05 = 116 N
When no external forces act on a system, the total linear momentum remains constant.
p₁ + p₂ (before) = p₁ + p₂ (after)
1. Elastic Collisions
Momentum and Kinetic Energy are both conserved
m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f
½m₁v₁i² + ½m₂v₂i² = ½m₁v₁f² + ½m₂v₂f²
2. Inelastic Collisions
Only Momentum is conserved
m₁v₁i + m₂v₂i = (m₁ + m₂)vf
A 3 kg object moving at 4 m/s collides and sticks to a 2 kg object at rest. Find the final velocity.
Solution:
(3 kg)(4 m/s) + (2 kg)(0 m/s) = (5 kg)vf
12 = 5vf
vf = 2.4 m/s
Conservation of momentum in x-direction:
m₁v₁xi + m₂v₂xi = m₁v₁xf + m₂v₂xf
Conservation of momentum in y-direction:
m₁v₁yi + m₂v₂yi = m₁v₁yf + m₂v₂yf
xcm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
vcm = (m₁v₁ + m₂v₂)/(m₁ + m₂)
F = v(dm/dt) + m(dv/dt)
where:
A rocket ejects mass at 1000 m/s relative to the rocket. If it ejects 2 kg/s of mass, what thrust is produced?
Solution:
F = v(dm/dt)
F = (1000 m/s)(-2 kg/s) = 2000 N
Common Mistakes to Avoid:
Percentage error in momentum:
δp/p = √[(δm/m)² + (δv/v)²]
AP Physics C Exam Tips:
