Conservation of Momentum
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
If no net external force acts on a system, its total momentum stays the same. Objects inside the system can trade momentum back and forth, but the total doesn’t change.
1. Pushing apart
2. Why it works
By Newton’s third law, the skaters push on each other with equal and opposite forces for the same amount of time. So they get equal and opposite impulses: whatever momentum one gains, the other loses.
Choose your system so the outside forces are zero or cancel. For carts on a level, frictionless track, gravity and the normal force cancel vertically, so horizontal momentum is conserved.
Momentum is conserved, not speed. In the figure, the speeds change from 0 to 2 and 3 m/s.
3. Example, then try it
A 4.0 kg rifle fires a 0.010 kg bullet forward at 400 m/s. What is the rifle’s recoil velocity?
Before: everything at rest, so ptotal = 0.
After: (0.010)(400) + (4.0)v = 0 → v = −4.0 ÷ 4.0 = −1.0 m/s (1.0 m/s backward).
On a frictionless track, a 2.0 kg cart moving right at 4.0 m/s hits a 1.0 kg cart at rest. Afterward, the 2.0 kg cart moves right at 2.0 m/s. Find the 1.0 kg cart’s velocity.
Choose + : right. System: both carts (no net horizontal external force).
Before: (2.0)(4.0) + (1.0)(0) = 8.0 kg·m/s.
After: (2.0)(2.0) + (1.0)v = 4.0 + v.
Set equal: 4.0 + v = 8.0 → v = +4.0 m/s.
Answer: 4.0 m/s to the right.
4. Common mistakes
A cannon on frictionless wheels fires a ball forward. Before firing, everything is at rest. Right after firing, what is the total momentum of the cannon and ball?
The explosion forces are internal, so the total momentum stays what it was before: zero. The ball moves forward and the cannon recoils backward with equal and opposite momentum.
Ready to practice?
5 questions on Conservation of Momentum, each with a hint and an explanation for every wrong answer.