Unit 4 · Topic 4.3

4.3 Conservation of Linear Momentum

If the net external force on a system is zero, its total momentum stays constant: whatever momentum one part gains, another part loses. The system's center of mass then moves at a constant velocity. Conservation of momentum lets you find velocities just before and after collisions and explosions in one or two dimensions.

Key terms

conservation of momentum
isolated system
center-of-mass velocity
net external force
two-dimensional collision

Why momentum is conserved

When two objects in a system push on each other, Newton's third law makes the forces equal and opposite, and they act for the same time. So the impulses are equal and opposite, and the momentum one object gains is exactly the momentum the other loses. The total doesn't change.

Only an external force can change a system's total momentum: $\vec{F}_{net,ext} = \frac{d\vec{p}_{sys}}{dt}$ . If the net external force is zero, the system is isolated (in the momentum sense) and:

$\sum m_i\vec{v}_{i,\text{before}} = \sum m_i\vec{v}_{i,\text{after}}$

Momentum is always conserved for the universe as a whole. The skill is choosing a system where it's constant. In a car crash, the car alone is not a good choice (the other car pushes it), but both cars together is.

Conservation one component at a time

Momentum is a vector, so the rule applies to each component separately. If external forces act only vertically (gravity and a normal force), horizontal momentum is still conserved. For instance, when a person jumps horizontally off a skateboard that was at rest, the board's and person's horizontal momenta stay balanced even though gravity acts.

In two dimensions, write one conservation equation for x and one for y. That gives two equations, enough to solve for two unknowns, like a final speed and angle.

The center of mass keeps moving

The velocity of a system's center of mass is the total momentum divided by the total mass:

$\vec{v}_{cm} = \frac{\sum m_i\vec{v}_i}{\sum m_i} = \frac{\vec{p}_{sys}}{m_{sys}}$

So if total momentum is constant, $\vec{v}_{cm}$ is constant too. In any collision or explosion inside an isolated system, the center of mass moves on exactly as before, even as the pieces fly apart. A firework shell bursting in midair has its center of mass keep following the parabola the whole shell was on (until pieces hit the ground).

Explosions and recoil

In an explosion, a system that starts at rest has zero momentum, so afterward the pieces' momenta must add to zero. Two pieces fly apart in opposite directions with equal-size momenta, and the lighter piece moves faster. A skater who throws a ball forward slides backward; a gun recoils when it fires.

Kinetic energy is not conserved in an explosion: it increases, supplied by stored energy such as chemical energy or a compressed spring. Momentum still is.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Recoil on ice
A 50 kg skater standing still on frictionless ice throws a 2.0 kg ball forward at 10 m/s. Find the skater's velocity afterward.
Show the solutionHide the solution
1. Step 1: System: skater + ball. The horizontal net external force is zero, and the total momentum starts at zero.
2. Step 2: $0 = (2.0)(10) + (50)v$ , so $v = -0.40$ m/s.
3. Step 3: The skater moves backward at 0.40 m/s. The center of mass of the skater–ball system stays put.
Answer: 0.40 m/s backward.
Example 2Calculator allowed
A two-dimensional crash
A 1500 kg car moving east at 20 m/s collides with a 2000 kg truck moving north at 15 m/s. They lock together. Find the velocity of the wreck just after the collision.
Show the solutionHide the solution
1. Step 1: System: car + truck. During the brief crash, the external forces (friction from the road) are tiny compared with the collision forces, so momentum is conserved in each direction.
2. Step 2: East (x): $(1500)(20) = 30{,}000$ kg·m/s. North (y): $(2000)(15) = 30{,}000$ kg·m/s.
3. Step 3: After: $v_x = \frac{30{,}000}{3500} \approx 8.57$ m/s and $v_y = \frac{30{,}000}{3500} \approx 8.57$ m/s.
4. Step 4: Speed: $\sqrt{8.57^2 + 8.57^2} \approx 12.1$ m/s. Direction: equal components, so 45° north of east.
Answer: About 12.1 m/s at 45° north of east.
Example 3
Center-of-mass velocity stays the same
A 2.0 kg cart moving at 3.0 m/s hits a 1.0 kg cart at rest on a frictionless track. Find the center-of-mass velocity before the collision. If the 2.0 kg cart moves at 1.0 m/s just afterward, find the 1.0 kg cart's velocity and check the center-of-mass velocity after.
Show the solutionHide the solution
1. Step 1: Before: $v_{cm} = \frac{(2.0)(3.0) + (1.0)(0)}{3.0} = 2.0$ m/s.
2. Step 2: Momentum: $6.0 = (2.0)(1.0) + (1.0)v_2$ , so $v_2 = 4.0$ m/s.
3. Step 3: After: $v_{cm} = \frac{(2.0)(1.0) + (1.0)(4.0)}{3.0} = 2.0$ m/s, unchanged, as it must be with no external force.
Answer: $v_{cm} = 2.0$ m/s before and after; the 1.0 kg cart moves at 4.0 m/s.

Common mistakes

Choosing a system that has a significant external force acting on it, like one car in a two-car crash.
Adding momenta in two dimensions without components. Conserve x-momentum and y-momentum separately.
Thinking kinetic energy is conserved whenever momentum is. In explosions, kinetic energy increases; in most collisions, it decreases.
Dropping signs for objects moving in opposite directions.

On the exam

Free-response questions often ask you to justify why momentum is conserved. State the system and that the net external force on it is zero (or negligible during the collision).
Lab questions may give before-and-after velocities from carts on a track. Check conservation by comparing total momentum before and after, allowing for measurement uncertainty.

Connected topics

Videos

Conservation of linear momentum | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Topic 4.3 - Conservation of Momentum
Lessons With LondotWatch on YouTube (opens in a new tab)
Introduction to Conservation of Momentum with Demonstrations
Flipping PhysicsWatch on YouTube (opens in a new tab)
High School Physics - Conservation of Momentum
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
16.1 Cases of Constant Momentum
MIT OpenCourseWareWatch on YouTube (opens in a new tab)
Conservation of Linear Momentum
Bozeman ScienceWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 4.3 Conservation of Linear Momentum. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A 4.0 kg rifle, at rest and free to move, fires a 0.010 kg bullet at 400 m/s. What is the rifle's recoil speed?

Question 2 of 4Calculator allowed

A 3.0 kg object at rest on a frictionless surface breaks into two pieces when a spring inside it is released. A 1.0 kg piece moves east at 6.0 m/s. What is the velocity of the other piece?

Question 3 of 4Calculator allowed

A 2.0 kg cart moves at 5.0 m/s to the right toward a 3.0 kg cart moving at 5.0 m/s to the left. What is the velocity of the two-cart system's center of mass?

Question 4 of 4Calculator allowed

A 60 kg person stands at one end of a 120 kg boat at rest in still water. She walks 4.0 m toward the other end of the boat. Ignoring the water's resistance, how far does the boat move relative to the water?

0 of 4 answered