AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/4/4-3)
Unit 4 · Topic 4.3
4.3 Conservation of Linear Momentum
When the net external force on a system is zero, its total momentum stays constant, so any momentum one part gains, another part loses. This lets you find velocities right before and right after collisions and explosions, and it means the system's center of mass keeps moving at a constant velocity.
Key terms
- conservation of momentum
- total momentum
- isolated system
- center-of-mass velocity
- net external force
The conservation law
Momentum is never created or destroyed; it only moves from one object to another. So whether a system's momentum stays the same comes down to the system you draw. With zero net external force, nothing outside can add or remove momentum, and Σp_before = Σp_after. With a net external force, momentum flows in or out, and the system's change in momentum equals the external impulse: Δp_system = F_net,ext Δt.
During a collision, outside forces like friction or gravity are tiny compared with the forces between the colliding objects, and they act only briefly. So momentum is very nearly conserved from just before to just after the collision, even if friction acts afterward.
Why it works: the third law
When two objects interact, they exert equal and opposite forces on each other for the same time (2.3). So they receive equal and opposite impulses, and their momentum changes are equal and opposite. Whatever momentum one loses, the other gains, and the total doesn't change.
Choosing the system matters. A ball falling toward Earth gains downward momentum, so the ball alone doesn't conserve momentum: gravity is external. The ball–Earth system does, because Earth gains an equal upward momentum, far too small to notice.
Center-of-mass velocity
A system of moving objects has one center-of-mass velocity: v_cm = Σmv/Σm = p_total/m_total. With no net external force, v_cm is constant, so the center of mass glides along steadily before, during and after any collision or explosion inside the system. For two objects that stick together, their shared final velocity is exactly v_cm.
Solving one-dimensional problems
- Choose a system that has no net external force (or only a negligible one during the interaction).
- Choose a positive direction and give every velocity its sign.
- Write m₁v₁ + m₂v₂ (before) = m₁v₁′ + m₂v₂′ (after), where the primes mean after.
- Solve for the one unknown in symbols, then substitute. A negative answer means that object moves in the negative direction.
- For explosions from rest, the total before is zero, so the pieces' momenta after are equal and opposite.
Two dimensions: set up, don't solve
Momentum is conserved separately in the x and y directions. If nothing moved in the y-direction before a collision, the y-momenta of the pieces afterward must add to zero: if one piece goes up and to the right, another must have a downward component.
AP Physics 1 treats two-dimensional momentum only semiquantitatively. You may need to write the x and y equations correctly and reason about how changing a mass, speed or angle affects another quantity. You won't need to solve simultaneous equations for unknown velocities.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Recoil on ice
A 60 kg skater stands at rest on frictionless ice holding a 2.0 kg ball. She throws the ball forward at 9.0 m/s. What is her velocity afterward?
Show the solutionHide the solution
- Step 1: System: skater + ball. Ice is frictionless, and gravity and the normal force cancel vertically, so horizontal momentum is conserved.
- Step 2: Before: everything at rest, p_total = 0.
- Step 3: After: 0 = (2.0)(+9.0) + (60)v, so v = −18 ÷ 60 = −0.30 m/s.
- Step 4: She slides backward at 0.30 m/s. The center of mass stays at rest.
Answer: 0.30 m/s, opposite to the throw
- Example 2Calculator allowed
Carts that stick together
A 2.0 kg cart moving at +4.0 m/s collides with a 3.0 kg cart at rest, and they stick together. Find their velocity afterward and the system's center-of-mass velocity before the collision.
Show the solutionHide the solution
- Step 1: Momentum before: (2.0)(4.0) + (3.0)(0) = 8.0 kg·m/s.
- Step 2: After, they move together: (2.0 + 3.0)v′ = 8.0, so v′ = 1.6 m/s.
- Step 3: Center of mass before: v_cm = 8.0 ÷ 5.0 = 1.6 m/s. It's the same before and after, as it must be when no net external force acts.
Answer: 1.6 m/s in the original direction; v_cm = 1.6 m/s throughout
- Example 3Calculator allowed
Setting up a two-dimensional collision
A puck of mass m moving at v₀ along the +x axis strikes an identical puck at rest. Afterward, the first puck moves at speed v₁ at 60° above the +x axis. Write the momentum equations for each direction, and explain which way the second puck must move.
Show the solutionHide the solution
- Step 1: x-direction: mv₀ = mv₁ cos 60° + mv₂ₓ.
- Step 2: y-direction: 0 = mv₁ sin 60° + mv₂ᵧ.
- Step 3: From the y equation, v₂ᵧ = −v₁ sin 60°, which is negative: the second puck must move below the x-axis, so the two y-momenta cancel.
- Step 4: From the x equation, v₂ₓ = v₀ − v₁ cos 60°. A collision can't add kinetic energy, so v₁ ≤ v₀ and v₁ cos 60° = v₁/2 is less than v₀. That makes v₂ₓ positive: the second puck moves forward, so it heads down and to the right.
- Step 5: That's the depth AP Physics 1 expects: correct equations and reasoning about signs and sizes.
Answer: x: mv₀ = mv₁ cos 60° + mv₂ₓ; y: 0 = mv₁ sin 60° + mv₂ᵧ; the second puck moves below the x-axis, forward and down
Common mistakes
- Choosing a system with a large net external force and still assuming its momentum is conserved.
- Dropping signs for objects moving in the negative direction.
- Assuming kinetic energy is conserved along with momentum. Check it separately (4.4).
- Thinking momentum can't be conserved if friction acts after the collision. Apply conservation only from just before to just after the impact.
On the exam
- Free-response questions often ask you to derive a final velocity "starting with conservation of momentum." Write the general law first, then substitute the masses and velocities as symbols.
- Expect questions about the center of mass of an exploding object, such as a firework: its center of mass keeps following the same parabola it was on, because the explosion's forces are internal and only gravity acts from outside. That holds until a piece hits something.
- For Qualitative/Quantitative Translation questions, use your derived equation to explain a limiting case, such as what happens if one mass is much larger than the other.
Connected topics
Videos
Check yourself
5 questions on 4.3 Conservation of Linear Momentum. Pick an answer to see if you got it, and why.
Two carts are at rest on a level, frictionless track with a compressed spring between them. When it is released, the 2.0 kg cart moves left at 1.5 m/s. How fast does the 3.0 kg cart move to the right?
A 1200 kg car moving at 20 m/s rear-ends an 800 kg car at rest at a stoplight, and the cars lock together. What is their speed just after the collision?
A 2.0 kg cart moving right at 3.0 m/s collides head-on with a 1.0 kg cart moving left at 4.0 m/s. They stick together. What is their velocity just after the collision?
On a level, frictionless air track, a 3.0 kg glider moves right at 4.0 m/s toward a 1.0 kg glider moving left at 2.0 m/s.
Described situation
What is the velocity of the center of mass of the two-glider system before they collide?
After the collision, the gliders move apart. What happens to the velocity of the center of mass?
0 of 5 answered