03Momentum and Collisions
Define momentum and impulse, apply the impulse-momentum theorem, and use conservation of momentum to analyze elastic and inelastic collisions.
Conservation of momentum is one of the most powerful tools in physics. It explains car crashes, rocket propulsion, and particle interactions.
How do the concepts of momentum and impulse explain what happens during a collision — and why do some collisions feel more violent than others?
Momentum is the product of an object's mass and velocity (p = mv). It is a vector quantity — direction matters. When a net external force acts on an object over a time interval, it delivers an impulse (J = FΔt) that equals the change in momentum (Δp). This relationship, the impulse-momentum theorem, explains why airbags save lives: by increasing the collision time, they reduce the peak force on a passenger.
The law of conservation of momentum states that, in a closed system with no net external force, the total momentum before an event equals the total momentum after. This law governs all collisions and explosions. Collisions are classified by whether kinetic energy is also conserved (elastic), partially lost (inelastic), or maximally lost while objects stick together (perfectly inelastic). In two dimensions, momentum is conserved independently along each axis. The center of mass of a system moves as if all the mass were concentrated there, and in rocket propulsion, momentum conservation explains how a spacecraft accelerates by expelling mass.
Collision Types
Elastic
Both momentum and kinetic energy are conserved. Objects bounce off each other. Use both the momentum equation and the KE equation simultaneously.
Inelastic
Momentum is conserved; kinetic energy is lost to heat, sound, or deformation. Objects do not stick together. Most real-world collisions fall here.
Perfectly Inelastic
Momentum is conserved; maximum KE is lost. Objects stick together and move with a common final velocity. Use v_f = (m₁v₁ + m₂v₂)/(m₁ + m₂).
Key Equations
p = mvLinear MomentumJ = FΔt = Δp = m·ΔvImpulseΣp_i = Σp_fConservation of Momentumv_f = (m₁v₁ + m₂v₂) / (m₁ + m₂)Perfectly Inelastic v_fx_cm = Σmᵢxᵢ / ΣmᵢCenter of Masse = |v₂f − v₁f| / |v₁i − v₂i|Coeff. of RestitutionWorked Examples
A 0.15 kg baseball traveling at 40 m/s is struck by a bat and leaves in the opposite direction at 55 m/s. (a) Find the impulse delivered to the ball. (b) If the contact time is 0.002 s, find the average force exerted on the ball.
Define positive direction as the initial direction of the ball (toward the bat). Initial velocity: v_i = +40 m/s. Final velocity: v_f = −55 m/s (opposite direction).
Apply the impulse-momentum theorem: J = Δp = m·(v_f − v_i).
J = (0.15 kg)(−55 − 40) m/s = (0.15)(−95) = −14.25 kg·m/s.
The magnitude of the impulse is |J| = 14.25 kg·m/s. The negative sign indicates the impulse is directed opposite to the ball's initial motion (i.e., the bat pushes the ball back).
For part (b), use J = F_avg · Δt → F_avg = J / Δt.
F_avg = −14.25 kg·m/s ÷ 0.002 s = −7125 N.
The magnitude of the average force is 7125 N, directed opposite to the ball's initial motion.
A 2 kg block moving at 6 m/s collides with a stationary 3 kg block on a frictionless surface. The blocks stick together after the collision (perfectly inelastic). Find their common final velocity.
Identify: m₁ = 2 kg, v₁ᵢ = 6 m/s; m₂ = 3 kg, v₂ᵢ = 0 m/s. Objects stick together → perfectly inelastic collision.
Apply conservation of momentum: m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)v_f.
(2)(6) + (3)(0) = (2 + 3)v_f.
12 = 5 v_f.
v_f = 12 / 5 = 2.4 m/s.
Check: Initial KE = ½(2)(6²) = 36 J. Final KE = ½(5)(2.4²) = 14.4 J. KE lost = 21.6 J — consistent with an inelastic collision.
A 4 kg ball moving at 8 m/s collides elastically with a stationary 2 kg ball on a frictionless surface. Find the final velocity of each ball.
Identify: m₁ = 4 kg, v₁ᵢ = 8 m/s; m₂ = 2 kg, v₂ᵢ = 0. Elastic collision → both momentum and KE are conserved.
Use the elastic collision formulas derived from simultaneous momentum and KE conservation:
v₁f = (m₁ − m₂)/(m₁ + m₂) · v₁ᵢ = (4 − 2)/(4 + 2) · 8 = (2/6)(8) = 16/6 ≈ 2.67 m/s.
v₂f = 2m₁/(m₁ + m₂) · v₁ᵢ = 2(4)/(4 + 2) · 8 = (8/6)(8) = 64/6 ≈ 10.67 m/s.
Wait — recheck v₂f: v₂f = [2(4)/(4+2)] × 8 = (8/6) × 8 = 64/6 ≈ 10.67 m/s.
Verify momentum: p_i = 4×8 = 32 kg·m/s. p_f = 4×(8/3) + 2×(32/3) = 32/3 + 64/3 = 96/3 = 32 kg·m/s. ✓
Verify KE: KE_i = ½(4)(64) = 128 J. KE_f = ½(4)(8/3)² + ½(2)(32/3)² = ½(4)(64/9) + ½(2)(1024/9) = 128/9 + 1024/9 = 1152/9 = 128 J. ✓
Three masses are placed along the x-axis: 3 kg at x = 0 m, 5 kg at x = 4 m, and 2 kg at x = 7 m. Find the center of mass of the system.
Use the center-of-mass formula: x_cm = Σmᵢxᵢ / Σmᵢ.
Numerator: Σmᵢxᵢ = (3)(0) + (5)(4) + (2)(7) = 0 + 20 + 14 = 34 kg·m.
Denominator: Σmᵢ = 3 + 5 + 2 = 10 kg.
x_cm = 34 / 10 = 3.4 m.
Sanity check: 3.4 m lies between x = 0 and x = 7, closer to the 5 kg mass at x = 4 m — makes sense since 5 kg is the largest mass.
A 70 kg astronaut at rest pushes off a 200 kg spacecraft. The astronaut moves away at 3 m/s. Find the recoil velocity of the spacecraft.
The system (astronaut + spacecraft) is initially at rest, so total initial momentum = 0.
Apply conservation of momentum: 0 = m_A · v_A + m_S · v_S.
0 = (70)(3) + (200)(v_S).
0 = 210 + 200 v_S.
v_S = −210 / 200 = −1.05 m/s.
The negative sign means the spacecraft moves in the direction opposite to the astronaut. This is the rocket-propulsion principle: expelling mass in one direction propels the remaining mass in the opposite direction.
Guided Practice
A 0.5 kg soccer ball is kicked from rest and reaches a speed of 20 m/s. The kick lasts 0.05 s. Find (a) the impulse delivered to the ball and (b) the average force of the kick.
Hint: Use J = Δp = m·Δv for part (a). Then rearrange J = F·Δt to find F for part (b). Remember the ball starts from rest, so v_i = 0.
A 1200 kg car traveling at 15 m/s rear-ends a stationary 900 kg car. The cars lock bumpers (perfectly inelastic). Find the speed immediately after the collision and the kinetic energy lost.
Hint: Use v_f = (m₁v₁ + m₂v₂)/(m₁ + m₂) with v₂ = 0. Then compute KE_i = ½m₁v₁² and KE_f = ½(m₁+m₂)v_f² to find ΔKE.
A 3 kg ball moving at 10 m/s collides elastically with a 3 kg stationary ball. What are the final velocities of both balls?
Hint: For equal masses in an elastic collision, the moving object stops and the stationary object moves off with the original velocity. Verify using the elastic collision formulas: v₁f = (m₁−m₂)/(m₁+m₂)·v₁ᵢ and v₂f = 2m₁/(m₁+m₂)·v₁ᵢ.
Two masses are on the x-axis: 6 kg at x = 2 m and 4 kg at x = 8 m. Find the center of mass. Then a third mass of 2 kg is placed at x = 5 m — find the new center of mass.
Hint: Apply x_cm = Σmᵢxᵢ / Σmᵢ twice. For the first part use only the two masses. For the second part include all three masses in both the numerator and denominator.
A 5 kg object moving at 4 m/s east collides with a 3 kg object moving at 2 m/s west. They stick together. Find the final velocity (magnitude and direction).
Hint: Choose east as positive. Then v₁ = +4 m/s and v₂ = −2 m/s. Apply conservation of momentum: (5)(4) + (3)(−2) = (5+3)v_f. Solve for v_f; the sign tells you the direction.
Key Vocabulary
Momentum
The product of an object's mass and velocity (p = mv). A vector quantity measured in kg·m/s. A heavier or faster object has greater momentum.
Example: A 2 kg ball at 5 m/s has momentum p = 10 kg·m/s.
Impulse
The product of the average net force and the time interval over which it acts (J = FΔt). Equal to the change in momentum (Δp). Measured in N·s = kg·m/s.
Example: A 500 N force acting for 0.1 s delivers an impulse of 50 N·s.
Conservation of Momentum
In a closed system with no net external force, the total momentum before an interaction equals the total momentum after: Σp_before = Σp_after.
Example: Two ice skaters push off each other — their combined momentum remains zero.
Elastic Collision
A collision in which both momentum and kinetic energy are conserved. Objects bounce off each other with no permanent deformation or heat generated.
Example: Billiard balls colliding approximate an elastic collision.
Inelastic Collision
A collision in which momentum is conserved but kinetic energy is not — some KE is converted to heat, sound, or deformation. Objects do not stick together.
Example: A car crash where the cars crumple but separate is inelastic.
Perfectly Inelastic Collision
A special case of inelastic collision where the objects stick together after impact, moving with a common final velocity. Maximum kinetic energy is lost.
Example: A football tackle where the tackler and ball-carrier move together.
Center of Mass
The weighted average position of all the mass in a system (x_cm = Σmᵢxᵢ / Σmᵢ). The system behaves as if all its mass is concentrated at this point.
Example: The center of mass of a uniform rod is at its midpoint.
Impulse-Momentum Theorem
States that the net impulse acting on an object equals its change in momentum: J = FΔt = Δp = mΔv. Connects force, time, and motion.
Example: Airbags increase collision time, reducing the force on a passenger.
Workbook Check
Interactive Practice — 5 Questions
A 3 kg object moving at 4 m/s has a momentum of:
Which statement about a perfectly inelastic collision is correct?
A 0.2 kg ball changes velocity from +10 m/s to −6 m/s. What is the magnitude of the impulse?
In an elastic collision between a moving object and an identical stationary object, what happens?
Masses of 2 kg at x = 1 m and 6 kg at x = 5 m are placed on the x-axis. Where is the center of mass?
Independent Practice
A 0.15 kg baseball moving at 40 m/s is hit by a bat and returns at 50 m/s in the opposite direction. Calculate the impulse delivered to the ball.
A 1,200 kg car moving at 15 m/s collides with a stationary 800 kg car. They stick together after the collision. Find their common final velocity.
A 3.0 kg ball moving at 6.0 m/s east collides elastically with a 3.0 kg ball at rest. What are the final velocities of both balls?
A 60 kg person jumps off a 200 kg boat at 3.0 m/s relative to the water. What is the recoil velocity of the boat? (Both start at rest.)
★ A 2.0 kg block moving at 8.0 m/s east collides with a 4.0 kg block moving at 2.0 m/s west. (a) If the collision is perfectly inelastic, find the final velocity. (b) Calculate the kinetic energy lost. (c) If instead the collision is perfectly elastic, find the final velocity of each block.
ChallengeCommon Mistakes
Applying conservation of momentum when an external force acts on the system
Momentum is conserved only when the NET external force is zero. Always define your system and check for external forces first
Assuming kinetic energy is conserved in all collisions
Only perfectly elastic collisions conserve KE. Inelastic collisions lose KE to heat/sound/deformation; perfectly inelastic collisions lose the most
Forgetting that momentum is a vector — ignoring direction signs in 1D problems
Assign positive/negative directions. If two objects move in opposite directions, their momenta have opposite signs
Confusing impulse (J = FΔt = Δp) with work (W = FΔx = ΔKE)
Impulse changes momentum; work changes kinetic energy. They use different formulas and different quantities (time vs displacement)
Math Tips
Conservation of momentum setup: p_before = p_after → m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f. For perfectly inelastic: both objects share the same final velocity v_f
Elastic collision final velocities: v₁f = (m₁−m₂)v₁ᵢ/(m₁+m₂), v₂f = 2m₁v₁ᵢ/(m₁+m₂). Special case: equal masses → objects exchange velocities
Impulse–momentum theorem: J = Δp = FΔt. A large force over a short time gives the same impulse as a small force over a long time — this is why airbags work
Center of mass velocity v_cm = (m₁v₁ + m₂v₂)/(m₁ + m₂). In the CM frame, total momentum = 0. This frame simplifies elastic collision analysis