The law hasn’t changed since Activity 5-1-3: in an isolated system, the total momentum before equals the total after. Hard problems are the same rule with more going on. A reliable routine keeps it manageable:
1Draw before and after.Sketch both objects, with arrows for their velocities.
2Choose a positive direction.Every velocity pointing the other way gets a minus sign.
3Write the momentum of each object, before and after.Use p = mv, with signs. Leave unknowns as letters.
4Set the total before equal to the total after.That is one equation. An elastic collision gives a second one.
5Solve, then check.Does the answer have the right sign and a sensible size? Does KE ever grow? It shouldn’t.
💡Step 5 matters. A kinetic energy that comes out bigger after an ordinary collision, or a momentum that doesn’t add up, means a sign slipped somewhere.
BMath
Both Moving, and the Unknown
When the objects stick, there is one final velocity, and the equation is a single line:
v_f = (m₁v₁ + m₂v₂) ÷ (m₁ + m₂)
Example 1: head-on
Given
A 2.0 kg cart at +3.0 m/s and a 1.0 kg cart at −2.0 m/s collide head-on and stick
✓ The heavier cart wins and the pair moves off in its direction.
Example 2: one chasing the other
Given
The same carts, but the 1.0 kg cart moves at +1.0 m/s, so the 2.0 kg cart catches up and sticks
Solve
p_before = 2.0(+3.0) + 1.0(+1.0) = +7.0 kg·m/s
v_f = 7.0 ÷ 3.0
v_f = +2.33 m/s
✓ Both momenta have the same sign, so they add instead of partly cancelling.
Example 3: finding an unknown mass
Given
A 1.5 kg cart at +4.0 m/s hits a stationary cart of unknown mass
They stick and move off at +1.5 m/s
Solve
m_A v_A = (m_A + m_B) v_f
m_B = m_A (v_A − v_f) ÷ v_f
m_B = 1.5 × (4.0 − 1.5) ÷ 1.5 = 2.5 kg
✓ The hidden cart has a mass of 2.5 kg. This is the idea behind the Mystery Mass lab.
⚠️An unknown mass found this way is only as good as the speeds you measured. When v_f is small, a small timing error is a large fraction of it, and the answer gets less reliable.
CMath
Elastic Collisions with Both Objects Moving
In Activity 5-2-1 you solved elastic collisions where one object started at rest. When both are moving, the same two ideas still work: momentum is conserved, and the relative velocity reverses. Written as a pair of equations:
The second equation gives v₂′ = v₁′ + (v₁ − v₂). Substitute that into the momentum equation, solve for v₁′, then find v₂′. If you prefer, the results can be written directly:
v₁′ = [(m₁ − m₂)v₁ + 2m₂v₂] ÷ (m₁ + m₂)
v₂′ = [(m₂ − m₁)v₂ + 2m₁v₁] ÷ (m₁ + m₂)
An elastic head-on collision
Given
A 3.0 kg cart at +2.0 m/s and a 1.0 kg cart at −2.0 m/s collide head-on, elastic
Check KE: before ½(3.0)(2.0²) + ½(1.0)(2.0²) = 8.0 J, after ½(1.0)(4.0²) = 8.0 J ✓
✓ The heavy cart stops and the light cart rebounds at 4.0 m/s. Momentum and kinetic energy are both conserved.
DMath
Systems That Split or Throw Things
The same equation handles an object that starts out moving and then breaks apart, or throws something. The only difference from a standing start is that the total momentum before is not zero.
Example 1: a cart that splits
Given
A 2.0 kg cart rolls at +1.0 m/s and releases a spring, splitting into two 1.0 kg halves
One half moves at +3.0 m/s
Solve
p_before = 2.0(+1.0) = +2.0 kg·m/s
2.0 = 1.0(+3.0) + 1.0(v) → v = −1.0 m/s
KE before = ½(2.0)(1.0²) = 1.0 J
KE after = ½(1.0)(3.0²) + ½(1.0)(1.0²) = 5.0 J
✓ The other half moves backward at 1.0 m/s. The extra 4.0 J of kinetic energy came from the compressed spring.
Example 2: a skater throws a ball forward
Given
A 60 kg skater, including a 2.0 kg ball, glides at +2.0 m/s
She throws the ball forward at +10 m/s (relative to the ground)
Solve
p_before = 60 × (+2.0) = +120 kg·m/s
120 = 58 × v + 2.0 × (+10)
v = (120 − 20) ÷ 58
v = +1.72 m/s
✓ Throwing the ball forward slows the skater from 2.0 to 1.72 m/s. That is recoil.
EMath
Collide, Then Skid
A crash investigator often has two facts: the masses and a skid mark. The skid is stage 2, and the collision is stage 1, and you handle them with different tools, working backward:
Stage 2, the skid. After the cars lock together, friction brings them to a stop. The road’s friction force is F_f = μMg and acts over the skid length d, so the work-energy theorem says all of the kinetic energy is removed:
μ M g d = ½ M v² → v_after = √(2μgd)
The mass cancels, so the speed just after the collision depends only on the road and the skid length.
Stage 1, the collision. The collision is so brief that outside forces don’t change the momentum noticeably, so momentum is conserved across it. For car A hitting parked car B:
m_A v_A = (m_A + m_B) v_after
Reconstructing a crash
Given
A 1,200 kg car hits a parked 1,000 kg car, and they lock together
Stage 1: p = (1,200 + 1,000)(12.8) = 28,230 kg·m/s
v_A = 28,230 ÷ 1,200 = 23.5 m/s
23.5 m/s × 3.6 = 85 km/h (about 53 mph)
✓ Car A was going about 23.5 m/s, or roughly 85 km/h, before the collision.
Typical friction ratios for locked, skidding tires. They depend on the tires and the surface:
Road surface
Typical μ
Dry asphalt
about 0.7
Wet asphalt
about 0.4
Packed snow
about 0.2
Ice
about 0.1
⚠️Don’t mix up the tools. Momentum is conserved across the collision, but not across the skid, because the road’s friction is an external force acting the whole time. And the work-energy theorem is the right tool for the skid, but not for the collision, where kinetic energy is lost to crumpling.
FExample
Skid Reconstructor
Try different roads and skid lengths. Notice how much the answer depends on the road surface, and how a small error in μ changes the answer.
Car A hits stationary car B, and they lock together and skid to a stop. Set the masses, the skid length, and the road. The tool works backward to car A’s speed before the crash.
If the road’s μ is off by ±0.10, the answer ranges from 21.8 to 25.1 m/s (49 to 56 mph).
KE kept by the pair: 181,104 JKE lost in the crash: 150,920 J (45%)
GConcept
Reporting a Range, Not a Number
Nobody knows the road’s friction ratio exactly, and a skid mark can be measured only so well. So a responsible reconstruction gives a range, not a single speed. For the crash above, suppose the investigator can only say that μ was between 0.60 and 0.80:
A range of speeds
Given
Same crash: 12 m skid, masses 1,200 kg and 1,000 kg
✓ Car A was going between about 21.8 and 25.2 m/s (78 to 91 km/h).
🔑A reconstruction isn’t weaker for giving a range. It’s more honest. If the speed limit was 22 m/s, a range of 21.8 to 25.2 m/s says the driver was probably speeding, but not by how much. In Problem 5-3-1, reporting your conclusion this way is a good habit.
HWatch Out
Three Common Mix-Ups
⚠️“Use one tool for the whole crash.” A crash with a skid has two stages that need two tools. Momentum handles the collision, and work-energy handles the skid. Using momentum on the skid, or energy on the collision, gives wrong answers.
⚠️“Add the speeds.” Momentum is a signed quantity. If one object moves the opposite way, its momentum is negative and has to be subtracted. Dropping a sign is the most common mistake in these problems.
⚠️“You can’t find a mass you can’t weigh.” You can, if you can measure speeds before and after a collision. Conservation of momentum lets you find any one missing quantity.
IConcept
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 and a 1.0 kg cart at −2.0 m/s collide head-on and stick. What is their final velocity?
Q2A 1.5 kg cart at +4.0 m/s hits a stationary cart, and they stick and move off at +1.5 m/s. What is the second cart’s mass?
Q3Two cars lock together and skid 12 m to a stop on a dry road. Which tool gives their speed just after the collision?
Q4Why can’t you use conservation of momentum across the skid itself?
Score: 0 of 4
JConcept
Looking Ahead
You now have the full toolkit for Unit 5. Next you’ll test conservation of momentum yourself, and then put it all to work on a real crash.