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.
Impulse and force–time area 6(a)
Question 1
A force–time graph is a triangle of base 0.12 s and height 500 N above the time axis. Find the impulse and state what physical change it equals.
Check the model response
Impulse is signed area under the graph: J = ½(0.12)(500) = +30 N s. It equals the body’s momentum change, Δp.
repair
2. Repair the six common breaks
Use only the repair matching an error, then repeat the corresponding graph or system model.
Impulse and force–time area 6(a)
Check this idea
Misconception: Impulse is the maximum force shown on a force–time graph.
Repair: Impulse is the signed area between the force curve and the time axis. Its unit is N s and it equals Δp; graph height alone is force.
Check this idea
Misconception: Area below the force–time axis contributes a positive impulse.
Repair: Choose a positive force direction first. Area below the time axis is negative and must be combined algebraically with positive area.
worked example
3. Follow three worked models
Follow how each solution declares graph sign, system boundary and collision type before calculating.
Impulse and force–time area 6(a)
Model 1
A force rises linearly from 0 to 400 N in 0.020 s, stays at 400 N until 0.060 s, then falls linearly to zero at 0.080 s. It acts on a 0.40 kg ball initially moving at −30 m s⁻¹. Find its final velocity.
Check the model response
Impulse is the two triangles plus rectangle: ½(0.020)(400) + (0.040)(400) + ½(0.020)(400) = 24 N s. Initial momentum is 0.40(−30) = −12 kg m s⁻¹, so final momentum is −12 + 24 = +12 kg m s⁻¹ and v = +30 m s⁻¹.
guided practice
4. Guided practice
Use each hint only to choose the signed graph area, closed system or elastic condition.
Impulse and force–time area 6(a)
Question 1
A triangular force pulse below the time axis has base 0.050 s and magnitude 240 N. A 0.30 kg object initially moves at +12 m s⁻¹. Find its final velocity.
Hint: The graph area is negative for the declared positive direction.
Check the model response
J = −½(0.050)(240) = −6.0 N s. Initial momentum = 0.30(12) = 3.6 kg m s⁻¹, so final momentum = −2.4 kg m s⁻¹ and v = −8.0 m s⁻¹.
independent practice
5. Independent practice
Solve without repair notes and show the sign convention, system condition and energy comparison.
Impulse and force–time area 6(a)
Question 1
A force–time graph has +12 N s of area above the axis followed by 5.0 N s below it. Find the net impulse and the momentum change.
Check the model response
Net signed area is +12 − 5.0 = +7.0 N s. Therefore the body’s momentum change is +7.0 kg m s⁻¹ in the declared positive direction.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Impulse and force–time area 6(a)
Question 1
A force–time graph is a triangle above the axis with base 0.016 s and peak 1200 N. It acts on a 0.30 kg ball initially moving at −8.0 m s⁻¹. Find the impulse, average force over the interval and final velocity.
Check the model response
Impulse = ½(0.016)(1200) = +9.6 N s. Average force = J/Δt = 9.6/0.016 = 600 N. Initial momentum is −2.4 kg m s⁻¹, so final momentum is +7.2 kg m s⁻¹ and v = +24 m s⁻¹.
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.
Impulse and force–time area 6(a)
Question 1
A constant −300 N force acts for 0.040 s on a 0.60 kg object initially moving at +15 m s⁻¹. Find the impulse and final velocity.
Check the model response
Impulse = FΔt = −300(0.040) = −12 N s. Initial momentum = 0.60(15) = +9.0 kg m s⁻¹, so final momentum = −3.0 kg m s⁻¹ and v = −5.0 m s⁻¹.