Collisions: impulse, momentum and kinetic energy
Key idea: Use a declared system and sign convention to connect force–time area to momentum change, then distinguish momentum conservation from kinetic-energy conservation.
Before you start: Momentum & ImpulseElastic & Inelastic Collisions
By the end, you can
- Find impulse as signed area under a force–time graph and use it as momentum change.
- State and apply conservation of momentum to a closed system in one-dimensional elastic and inelastic interactions.
- Use relative speed of approach equal to relative speed of separation for a perfectly elastic two-body collision.
- Explain why momentum remains conserved in a closed system while kinetic energy usually changes.
Starting-point self-check
1. Check your starting point
Attempt all three groups without notes and mark the first graph area, system boundary or collision condition you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Question 1
State the principle of conservation of momentum, including the condition on the chosen system.
Check the model response
The total momentum of a closed system remains constant when the resultant external force is zero, or when external impulse is negligible during the interaction.
Question 2
A 2.0 kg trolley moving at +5.0 m s⁻¹ collides with a 3.0 kg trolley moving at −1.0 m s⁻¹. They stick together. Find their common velocity.
Check the model response
For the two-trolley system, initial momentum = 2.0(5.0) + 3.0(−1.0) = 7.0 kg m s⁻¹. Thus 7.0 = (2.0 + 3.0)v and v = +1.4 m s⁻¹.
repair
2. Repair the six common breaks
Use only the repair matching an error, then repeat the corresponding graph or system model.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Check this idea
Misconception: Momentum is conserved separately for each body in a collision.
Repair: Each body receives an impulse and changes momentum. Total momentum is conserved only for the chosen closed system when external impulse is negligible.
Check this idea
Misconception: Speeds can be inserted as positive values because momentum conservation is a scalar equation in one dimension.
Repair: Momentum is a vector. Declare a positive direction and use signed velocities throughout the one-dimensional equation.
worked example
3. Follow three worked models
Follow how each solution declares graph sign, system boundary and collision type before calculating.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Model 1
A 2.0 kg body at +5.0 m s⁻¹ and a 3.0 kg body at −1.0 m s⁻¹ stick together. Find their final velocity and the kinetic-energy decrease.
Check the model response
Momentum conservation gives v = [2.0(5.0) + 3.0(−1.0)]/5.0 = +1.4 m s⁻¹. Initial kinetic energy = ½(2.0)(5.0²) + ½(3.0)(1.0²) = 26.5 J. Final kinetic energy = ½(5.0)(1.4²) = 4.9 J, so the decrease is 21.6 J, transferred to other stores.
guided practice
4. Guided practice
Use each hint only to choose the signed graph area, closed system or elastic condition.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Question 1
A stationary object explodes into 0.80 kg and 1.20 kg fragments. The 0.80 kg fragment moves at +6.0 m s⁻¹. Find the other velocity.
Hint: The closed system starts with zero total momentum.
Check the model response
0 = 0.80(6.0) + 1.20v, so v = −4.0 m s⁻¹. The negative sign means the second fragment moves in the opposite direction.
independent practice
5. Independent practice
Solve without repair notes and show the sign convention, system condition and energy comparison.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Question 1
A 0.30 kg body at +10 m s⁻¹ collides with a 0.20 kg body at −5.0 m s⁻¹. Afterwards the first moves at −2.0 m s⁻¹. Find the second body’s velocity, assuming negligible external impulse.
Check the model response
Initial momentum = 0.30(10) + 0.20(−5.0) = 2.0 kg m s⁻¹. Thus 2.0 = 0.30(−2.0) + 0.20v, giving v = +13 m s⁻¹.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Question 1
A 1200 kg car at +18 m s⁻¹ collides with an 800 kg car at −6.0 m s⁻¹ and they lock together. Find their velocity and state the condition needed to conserve momentum.
Check the model response
For the two-car system, momentum = 1200(18) + 800(−6.0) = 16 800 kg m s⁻¹. Common velocity = 16 800/2000 = +8.4 m s⁻¹. External impulse must be negligible during the collision.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Momentum conservation and one-dimensional interactions 6(b)–(c)
Question 1
An 8.0 kg object initially at rest separates into 3.0 kg and 5.0 kg parts. The 3.0 kg part moves at −10 m s⁻¹. Find the other velocity and justify the conservation equation.
Check the model response
With negligible external impulse, total momentum remains zero: 0 = 3.0(−10) + 5.0v. Hence v = +6.0 m s⁻¹.