3bImpulse and Collisions
Apply the impulse-momentum theorem and classify collisions as elastic, inelastic, or perfectly inelastic.
Impulse and collision analysis are essential for designing safety systems (airbags, helmets, crumple zones), understanding sports physics, and analyzing particle collisions in nuclear and particle physics.
Why do airbags save lives — and why does a baseball player "follow through" when hitting a ball?
Lesson Overview
Impulse (J = FΔt) is the product of force and the time over which it acts. The impulse-momentum theorem states that the impulse equals the change in momentum: J = Δp = mΔv. This explains why airbags (longer collision time → smaller force) and follow-through (longer contact time → greater impulse) are effective. Collisions are classified as elastic (both momentum and kinetic energy conserved), inelastic (momentum conserved, KE lost), or perfectly inelastic (objects stick together, maximum KE loss).
Key Equations
Worked Examples
A 0.5 kg ball moving at 10 m/s is brought to rest in 0.02 s by a wall. Find (a) the impulse and (b) the average force exerted by the wall.
A 60 kg person falls and hits the ground, stopping in 0.01 s (no airbag) vs 0.1 s (with padding). If they were moving at 5 m/s, find the average force in each case.
A 2 kg ball moving at 6 m/s east collides elastically with a 2 kg ball at rest. Find the velocities after the collision.
A 3 kg ball moving at 4 m/s east collides perfectly inelastically with a 1 kg ball at rest. Find the final velocity and the kinetic energy lost.
A 0.15 kg baseball moving at 40 m/s is hit by a bat and leaves at 50 m/s in the opposite direction. The contact time is 0.001 s. Find the impulse and average force.
Guided Problems
A 1000 N force acts on a 2 kg object for 0.5 s. Find the impulse and the change in velocity.
Hint: J = FΔt for impulse. Then J = mΔv → Δv = J/m.
A 0.2 kg ball bounces off a wall: it hits at 8 m/s east and leaves at 6 m/s west. Find the impulse on the ball.
Hint: J = Δp = m(v_f − v_i). Assign east as positive: v_i = +8, v_f = −6. Calculate Δp.
A 4 kg cart moving at 5 m/s east collides perfectly inelastically with a 6 kg cart at rest. Find the final velocity and the fraction of kinetic energy lost.
Hint: Use v_f = (m₁v₁ + m₂v₂)/(m₁+m₂). Then find KE_i and KE_f, and compute ΔKE/KE_i.
In a 1D elastic collision, a 3 kg ball at 6 m/s hits a stationary 1 kg ball. Find both final velocities. (Use: v₁f = (m₁−m₂)v₁ᵢ/(m₁+m₂) and v₂f = 2m₁v₁ᵢ/(m₁+m₂))
Hint: Substitute m₁ = 3, m₂ = 1, v₁ᵢ = 6 into both formulas. Check that momentum and KE are conserved.
A 70 kg person jumps from a 1.5 m height and lands on a hard floor (Δt = 0.01 s) vs a foam mat (Δt = 0.5 s). Find the landing speed and the average force in each case. (g = 9.8 m/s²)
Hint: Find landing speed: v = √(2gh). Then F = Δp/Δt = mv/Δt for each case.
Key Vocabulary
Impulse (J)
The product of the net force and the time interval over which it acts: J = FΔt. Impulse equals the change in momentum: J = Δp. Units: N·s = kg·m/s.
Example: A 500 N force acting for 0.1 s delivers an impulse of 50 N·s, changing the object's momentum by 50 kg·m/s.
Impulse-Momentum Theorem
The impulse on an object equals its change in momentum: J = FΔt = Δp = mΔv. This connects force, time, and the resulting change in motion.
Example: Airbags increase collision time Δt, reducing the force F needed to produce the same impulse (same Δp).
Elastic Collision
A collision in which both momentum AND kinetic energy are conserved. Objects bounce off each other without permanent deformation.
Example: Billiard ball collisions and atomic/molecular collisions are approximately elastic.
Inelastic Collision
A collision in which momentum is conserved but kinetic energy is NOT conserved. Some KE is converted to heat, sound, or deformation.
Example: A car crash is inelastic — the cars deform and KE is lost to heat and sound.
Perfectly Inelastic Collision
A collision in which the objects stick together after impact. Momentum is conserved but the maximum possible kinetic energy is lost.
Example: A football tackle where the tackler and ball-carrier move together after impact is perfectly inelastic.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A 2 kg ball moving at 5 m/s is stopped in 0.1 s. What is the average force?
In a perfectly inelastic collision, which quantity is conserved?
A 0.1 kg ball bounces off a wall: v_i = 10 m/s east, v_f = 10 m/s west. What is the impulse?
Two equal-mass balls collide elastically. Ball 1 is moving; ball 2 is at rest. After the collision:
Why do airbags reduce injury in a car crash?
Independent Practice
A 500 N force acts on a 5 kg object for 2 s. Find the impulse and the change in velocity.
A 0.3 kg ball moving at 12 m/s east bounces off a wall and moves at 8 m/s west. Find the impulse on the ball.
A 2 kg cart at 6 m/s east collides perfectly inelastically with a 4 kg cart at rest. Find the final velocity and KE lost.
A 1 kg ball moving at 8 m/s east collides elastically with a 3 kg ball at rest. Find both final velocities using the elastic collision formulas.
★ A 0.05 kg bullet moving at 400 m/s embeds in a 2 kg wooden block at rest. (a) Find the block's velocity after impact. (b) Find the KE before and after. (c) Find the fraction of KE lost. (d) Explain where the lost KE went.
ChallengeCommon Mistakes
Thinking kinetic energy is conserved in all collisions
KE is only conserved in elastic collisions. In inelastic collisions, KE is converted to heat, sound, and deformation. Momentum is always conserved (in isolated systems)
Forgetting that impulse and momentum change are vectors
J = Δp = m(v_f − v_i). When a ball bounces back, v_f is negative — the change in momentum is larger than if the ball just stopped
Confusing impulse (J = FΔt) with work (W = FΔx)
Impulse involves force × time and changes momentum. Work involves force × displacement and changes kinetic energy. They are different quantities
Applying elastic collision formulas when the collision is inelastic
Only use v₁f = (m₁−m₂)v₁ᵢ/(m₁+m₂) for elastic collisions. For perfectly inelastic, use v_f = (m₁v₁+m₂v₂)/(m₁+m₂)
Math Tips
Impulse-momentum theorem: J = FΔt = Δp = m(v_f − v_i). To reduce force, increase time (airbags, foam mats, follow-through).
Elastic collision check: verify both Σp and ΣKE are conserved. For equal masses: the balls exchange velocities.
Perfectly inelastic: v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁+m₂). This gives maximum KE loss while conserving momentum.
When a ball bounces off a wall: Δp = m(v_f − v_i). If it reverses direction, |Δp| = m(v_f + v_i) — larger than if it just stopped.