Momentum and collisions

College Physics I. Also known as: phys 101, phys 1401, phys 2425, phys 1, college physics, college physics i, algebra-based physics, physics 1, general physics 1, general physics i, introductory physics i, phys 201

Momentum, impulse, conservation of momentum, and elastic and inelastic collisions.

Momentum and impulse

The momentum of an object is its mass times its velocity: p = m v. It is a vector and has units of kg·m/s. Newton's second law can be written as net force equals the rate of change of momentum. Rearranging gives the impulse-momentum theorem: F Δt = Δp = m Δv. The product of force and the time it acts is the impulse. This explains why an airbag or a bent knee helps: extending the stopping time lowers the average force for the same change in momentum.

Conservation of momentum

If no net external force acts on a system, its total momentum stays the same. Internal forces between the objects in a collision are equal and opposite (Newton's third law), so they change each object's momentum but not the total. You must still keep track of direction: choose a positive direction and give velocities in the opposite direction a negative sign.

Types of collision

  • Perfectly inelastic: the objects stick together and move with a common velocity v = (m₁v₁ + m₂v₂) / (m₁ + m₂). Maximum kinetic energy is lost for a collision that conserves momentum.
  • Inelastic: the objects bounce, but some kinetic energy is lost to heat, sound and deformation.
  • Elastic: kinetic energy is also conserved. In a head-on elastic collision between equal masses, with one initially at rest, the moving mass stops and the other leaves with its speed.

Momentum is conserved in all three cases; kinetic energy is conserved only in the elastic one. Explosions and recoil work the same way in reverse: a system starting at rest has zero total momentum, so the pieces move apart with momenta that cancel.

Centre of mass

The centre of mass of point masses on a line is x = (m₁x₁ + m₂x₂ + …) / (m₁ + m₂ + …). It lies closer to the heavier mass.

Worked example

A 0.20 kg cart at 3.0 m/s hits a stationary 0.30 kg cart, and they couple.

  1. Momentum before: 0.20 × 3.0 = 0.60 kg·m/s.
  2. Common speed: v = 0.60 / (0.20 + 0.30) = 1.20 m/s.
  3. Kinetic energy before: ½ × 0.20 × 3.0² = 0.90 J. After: ½ × 0.50 × 1.2² = 0.36 J.
  4. Energy lost: 0.54 J, which is the energy turned into heat and deformation.

Common mistakes

  • Ignoring direction when velocities oppose.
  • Assuming kinetic energy is conserved in every collision.
  • Using the speed of the object after the collision in the momentum before.
  • Confusing impulse (force times time) with work (force times distance).

Practice questions

  1. A 0.145 kg ball travels at 40 m/s. What is the size of its momentum?

  2. A constant 30 N net force acts on a 2.0 kg cart, initially at rest, for 0.50 s. What is the cart's final speed?

  3. A 3.0 kg cart moving at 4.0 m/s sticks to a stationary 1.0 kg cart. Ignoring friction, what is their common speed afterwards?

  4. A 60 kg skater at rest on frictionless ice throws a 2.0 kg ball horizontally at 9.0 m/s (relative to the ice). Roughly what is the skater's recoil speed?

  5. A 2.0 kg puck moving at 5.0 m/s makes a perfectly elastic head-on collision with a stationary 2.0 kg puck on frictionless ice. What is the first puck's speed afterwards?

  6. A 3.0 kg cart moving at 4.0 m/s sticks to a stationary 1.0 kg cart. How much kinetic energy is lost in the collision?

Momentum and collisions explained: College Physics I study guide | CoLabEx