A system is the group of objects you’re paying attention to. Forces between objects inside the system are internal. Forces from outside, like friction from the floor or a push from someone who isn’t part of the system, are external. A system with no net external force is called isolated.
The law of conservation of momentum says that in an isolated system, the total momentum doesn’t change. It doesn’t matter how violent the interaction is:
p_total, before = p_total, after
For two objects, written out with velocities (and signs):
m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′
💡The primes mean “after.” Individual objects can gain or lose momentum, but what one loses, the other gains. Only the total stays the same.
BMath
Why It’s True
Newton’s third law says that when object 1 pushes on object 2, object 2 pushes back on object 1 with an equal and opposite force. They touch for exactly the same amount of time. So by the impulse-momentum theorem from Activity 5-1-2:
F₁₂ = −F₂₁ (third law)
F₁₂ Δt = −F₂₁ Δt (same time Δt)
Δp₂ = −Δp₁ (impulse = change in momentum)
Δp₁ + Δp₂ = 0 (the total doesn’t change)
🔑You don’t need to know the force. It might be huge, it might change moment by moment, and it might crumple both objects. It still comes in equal and opposite pairs, so it can’t change the total.
💡What about friction? Collisions happen so fast that the huge internal forces swamp any outside forces, so momentum is conserved during the collision even on a rough road. Friction then slows the wreck afterward, which is how investigators use skid marks. You’ll use both ideas in Problem 5-3-1.
CMath
Objects That Stick Together
In a perfectly inelastic collision, the objects stick and move off as one. With one final velocity, the conservation equation becomes:
v_f = (m₁v₁ + m₂v₂) ÷ (m₁ + m₂)
Example 1: one cart at rest
Given
A 2.0 kg cart moves at +3.0 m/s and hits a 1.0 kg cart at rest
They stick together
Solve
p_before = 2.0 × 3.0 + 1.0 × 0 = 6.0 kg·m/s
v_f = 6.0 ÷ (2.0 + 1.0)
v_f = +2.0 m/s
✓ The pair moves off at 2.0 m/s in the original direction.
Example 2: a head-on collision
Given
A 1.0 kg cart moves at +4.0 m/s and a 3.0 kg cart at −2.0 m/s
✓ The heavier cart wins. The pair moves off at 0.5 m/s in the heavier cart’s direction.
Example 3: what happened to the kinetic energy?
Given
The same collision as Example 1
Solve
KE_before = ½ × 2.0 × 3.0² + 0 = 9.0 J
KE_after = ½ × (2.0 + 1.0) × 2.0² = 6.0 J
Lost = 9.0 − 6.0 = 3.0 J (one third)
✓ Momentum was conserved, but 3.0 J of kinetic energy became thermal energy, sound, and deformation.
🔑When two equal masses stick and one starts at rest, the final speed is exactly half the initial speed, and exactly half of the kinetic energy is lost.
DExample
Collision Explorer
Try the carts from the examples, then try something new: carts moving toward each other, a heavy cart hitting a light one, and so on. In every case, the total momentum bars before and after match. Then switch to a bounce.
Set up a head-on collision. Positive velocity is to the right. Compare the total momentum bars before and after, then check the kinetic energy. Switch between carts that stick together and carts that bounce apart.
✓ The 60 kg skater moves at 2.0 m/s the other way. The 300 J of kinetic energy came from the skaters’ muscles, so this time the kinetic energy increased.
Example 2: throwing a ball
Given
A 50 kg student (with skateboard) stands at rest
She throws a 2.0 kg ball forward at +6.0 m/s
Solve
0 = 50 × v + 2.0 × (+6.0)
v = −12 ÷ 50
v = −0.24 m/s
✓ The student rolls backward at 0.24 m/s. That backward push is recoil.
FConcept
The Semi and the Compact Car
Here is the unit’s big question. A 20,000 kg semi at +10 m/s hits a 1,000 kg car at rest, and they stick together.
Which one feels the bigger force?
Given
Semi: 20,000 kg at +10 m/s. Car: 1,000 kg at rest
They stick, and the collision lasts 0.10 s
Solve
v_f = (20,000 × 10) ÷ 21,000 = +9.52 m/s
Δp_car = 1,000 × (9.52 − 0) = +9,524 kg·m/s
Δp_semi = 20,000 × (9.52 − 10) = −9,524 kg·m/s
F = Δp ÷ Δt = ±9,524 ÷ 0.10 = ±95,238 N
a_car = 95,238 ÷ 1,000 = 95 m/s² (about 10 g)
a_semi = 95,238 ÷ 20,000 = 4.8 m/s² (about 0.5 g)
✓ The forces are equal and opposite. The car’s speed changes by 9.5 m/s and the semi’s by only 0.48 m/s.
🔑Neither vehicle feels the bigger force. Newton’s third law says the forces are equal. What differs is the effect. The same force on a small mass causes a huge change in velocity, so the people in the car feel about twenty times the acceleration of the people in the truck. That is why the car’s occupants are in far more danger.
GWatch Out
Three Common Mix-Ups
⚠️“Each object keeps its own momentum.” No. Individual momenta change in a collision. It’s the total that stays the same. What one object loses, another gains.
⚠️“The bigger object pushes harder in a collision.” It doesn’t. The two forces are always equal and opposite. The smaller object just changes velocity more.
⚠️“Momentum disappears when things stick together or stop.” It doesn’t. If a moving cart sticks to a stationary one, the momentum is shared between them. If something stops against the ground, the momentum goes into the Earth, which is too massive for the change in its velocity to ever be noticed.
HConcept
Quick Check
Four questions. They are not graded. Answer them to find out what still needs work.
Q1A 2.0 kg cart at +3.0 m/s collides with a 1.0 kg cart at rest, and they stick together. What is their final velocity?
Q2A 60 kg skater and a 40 kg skater, both at rest, push apart. The 40 kg skater moves at +3.0 m/s. What is the 60 kg skater’s velocity?
Q3A truck and a car collide head-on. Which statement is true?
Q4Two identical carts collide and stick together, and one was at rest. What fraction of the moving cart’s original kinetic energy is left?
Score: 0 of 4
IConcept
Looking Ahead
You now have the three tools for Unit 5’s first half: momentum, impulse, and conservation. Next you’ll use them to protect an egg.