Source: ENGR 216 Lectures, Labs 3–4
Tags: momentum, conservation of momentum, impulse, collisions, elastic collision, inelastic collision, perfectly inelastic, impulse-momentum theorem, ENGR 216, PHYS 206
Momentum (p = mv) is conserved in any system where no net external force acts. Collisions are classified by what happens to kinetic energy: elastic collisions conserve it, inelastic collisions do not, and perfectly inelastic collisions (objects stick together) lose the maximum amount. The impulse-momentum theorem connects force applied over time to change in momentum.
Linear momentum (p)
The product of mass and velocity: p = mv. A vector quantity, measured in kg·m/s. Direction matters.
Conservation of momentum
In a closed system with no net external force, total momentum before an event equals total momentum after: Σp_before = Σp_after. This holds for all collision types.
Impulse (J)
The change in an object's momentum: J = Δp = FΔt (for constant force). Impulse is a vector, measured in N·s (equivalent to kg·m/s).
Impulse-momentum theorem
The net impulse on an object equals the change in its momentum: ΣF · Δt = m · Δv. This is Newton's second law in integrated form.
Elastic collision
A collision in which both momentum and kinetic energy are conserved. Objects bounce off each other with no energy lost to deformation, heat, or sound. Rare in everyday life but a good approximation for very hard objects (billiard balls, air-table pucks).
Inelastic collision
A collision in which momentum is conserved but kinetic energy is not. Some kinetic energy is converted to heat, sound, or deformation.
Perfectly inelastic collision
The extreme case where the objects stick together after colliding. Maximum kinetic energy is lost. The combined object moves at: v_f = (m₁v₁ + m₂v₂) / (m₁ + m₂).
Coefficient of restitution (e)
The ratio of relative speed after collision to relative speed before: e = |v₂f − v₁f| / |v₁i − v₂i|. For elastic collisions e = 1; for perfectly inelastic e = 0.
Centre of mass (COM)
The mass-weighted average position of a system. The velocity of the centre of mass is constant when external forces are zero: v_COM = Σ(m_i · v_i) / Σm_i.
Newton's third law guarantees it. During a collision, object A exerts a force on object B and B exerts an equal-and-opposite force on A. These internal forces change the momentum of each object but cancel when you look at the system as a whole. The total momentum of the system remains unchanged.
External forces (gravity, friction with the surface) can break conservation over longer time intervals. In collision problems, the collision happens so quickly that external impulses are negligible during the event itself.
The approach is the same regardless of collision type:
Define the system (all objects involved in the collision).
Write momentum conservation for each relevant direction: m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f.
If elastic, add kinetic energy conservation: ½m₁v₁i² + ½m₂v₂i² = ½m₁v₁f² + ½m₂v₂f².
If perfectly inelastic, set v₁f = v₂f = v_f.
Solve for the unknowns.
For elastic collisions in 1D with one object initially at rest (v₂i = 0):
v₁f = v₁i(m₁ − m₂) / (m₁ + m₂)
v₂f = 2m₁v₁i / (m₁ + m₂)
Special cases worth knowing: if m₁ = m₂, the first object stops and the second moves off with the first's original velocity. If m₁ ≫ m₂, the light object bounces back at nearly twice the heavy object's speed.
Impulse equals the area under a force-time graph during the collision. A longer collision time for the same momentum change means a smaller average force. This is the physics behind crumple zones, airbags, and bending your knees when landing: extending Δt reduces the peak force.
For constant force: J = F · Δt = Δp = m · Δv.
In two dimensions, momentum is conserved independently in x and y:
Σp_x before = Σp_x after
Σp_y before = Σp_y after
This gives two equations. For elastic 2D collisions, kinetic energy conservation provides a third. Glancing collisions (where objects deflect at angles) require careful component work.
The fraction of kinetic energy lost in a perfectly inelastic collision:
ΔKE / KE_initial = m₂ / (m₁ + m₂) (when object 2 is initially at rest)
This is relevant to Lab 4 (Collisions), where you measure pre- and post-collision velocities on the air table and compare measured energy loss to theoretical predictions.
Momentum: p = mv
Impulse: J = FΔt = Δp
Momentum conservation: m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f
Perfectly inelastic (stick together): v_f = (m₁v₁i + m₂v₂i) / (m₁ + m₂)
Elastic 1D (v₂i = 0): v₁f = v₁i(m₁ − m₂)/(m₁ + m₂) v₂f = 2m₁v₁i/(m₁ + m₂)
Kinetic energy: KE = ½mv²
Coefficient of restitution: e = |v₂f − v₁f| / |v₁i − v₂i|
⚠️ Momentum is always conserved in collisions (no net external force during the brief collision). Kinetic energy is only conserved in elastic collisions.
⚠️ Momentum is a vector. In 2D problems, conserve components separately. Do not add magnitudes of velocities at different angles.
⚠️ In perfectly inelastic collisions, the objects stick together. This means one equation (momentum) and one unknown (v_f). If they bounce, you need more information (energy or coefficient of restitution).
⚠️ The exam frequently tests whether students can identify which type of collision is described and apply the correct constraints.
⚠️ Lab 4 (Collisions) on the air table is a direct experimental test of these principles. Expect questions connecting lab observations to theory.
Q: A 2 kg ball moving at 4 m/s collides head-on with a 3 kg ball at rest. They stick together. What is the final velocity?
A: v_f = (2 × 4 + 3 × 0) / (2 + 3) = 8/5 = 1.6 m/s in the original direction of the 2 kg ball.
Q: Is kinetic energy conserved in the collision above?
A: No. Initial KE = ½(2)(4²) = 16 J. Final KE = ½(5)(1.6²) = 6.4 J. Energy lost = 9.6 J, converted to heat and deformation.
Q: A 0.15 kg baseball is pitched at 40 m/s and hit back at 50 m/s. What impulse did the bat deliver?
A: J = m(v_f − v_i) = 0.15(50 − (−40)) = 0.15 × 90 = 13.5 N·s (taking the hit direction as positive).
Q: Why do airbags reduce injury in car crashes?
A: They increase the time over which momentum changes (larger Δt), which reduces the average force on the occupant for the same impulse.
Q: In an elastic collision between equal masses where one is initially at rest, what happens?
A: The moving mass stops completely and the stationary mass moves off with the original velocity. All momentum and kinetic energy transfer to the second mass.
momentum, linear momentum, conservation of momentum, impulse, impulse-momentum theorem, collision, elastic collision, inelastic collision, perfectly inelastic, coefficient of restitution, kinetic energy, energy loss, centre of mass, Newton's third law, air table, ENGR 216, PHYS 206, Lab 4, mechanics, Texas A&M