Dynamics & Momentum
Momentum describes how hard it is to stop a moving object. Impulse links force and time to a change in momentum. Conservation of momentum explains what happens when two objects collide — essential for crash analysis, machine guarding and conveyor design.
2Topics
8Examples
4Calculators
12Practice Qs
📶Momentum & Impulsep = mv, FΔt = mΔv
🔁ConservationBefore = After
⚙️Applications4 workshop calculators
🎮Game12 randomised questions
📝Practice12 graded questions
Why this matters in engineeringReal world
Vehicle crash testing relies on conservation of momentum to predict post-collision velocities. Machine guard design uses impulse calculations to ensure stopping forces are within safe limits. Conveyor engineers use momentum to calculate the jolt when a load drops onto a moving belt.
📶
Momentum & Impulse
Momentum is mass times velocity. Impulse is the force multiplied by the time it acts — equal to the change in momentum.
📶Momentum & Impulse CalculatorChoose what to find, enter values, press Calculate
Choose what to find, enter the other values, then press Calculate.
📚 Key vocabulary
- Momentum (p)
- the product of mass and velocity — a measure of how hard it is to stop a moving object. Unit: kg m/s.
- Impulse
- force multiplied by the time for which it acts. Equal to the change in momentum. Unit: N s = kg m/s.
- Change in velocity (Δv)
- final velocity minus initial velocity. Can be negative if the object slows or reverses.
- Vector quantity
- a quantity with both magnitude and direction. Momentum and velocity are vectors; mass is not.
- Newton's second law
- F = ma, which can also be written as F = Δp / t — force equals rate of change of momentum.
- Newton s (N s)
- the unit of impulse — equivalent to kg m/s, since 1 N = 1 kg m/s².
👁️ Visual aid — momentum: a 1200 kg car moves at 15 m/s to the right; p = m × v = 18 000 kg m/s
✍️Worked example 1 — momentum of a vehicle
QuestionA car of mass 1200 kg travels at 15 m/s. Calculate its momentum.
Formula
p = m × vSubstitute
p = 1200 × 15Multiply
1200 × 15 = 18 000Answerp = 18 000 kg m/s
✍️Worked example 2 — impulse and force
QuestionA 900 kg vehicle accelerates from rest to 20 m/s in 6 s. Find the average force.
Formula
F = m × Δv ÷ tStep 1
Δv = 20 − 0 = 20 m/sSubstitute
F = 900 × 20 ÷ 6Step 2
900 × 20 = 18 000Divide
18 000 ÷ 6 = 3000AnswerF = 3000 N
✍️Worked example 3 — stopping time
QuestionA 600 kg machine component moving at 4 m/s is brought to rest by a braking force of 800 N. How long does it take to stop?
Formula
t = m × Δv ÷ FStep 1
Δv = 4 − 0 = 4 m/sSubstitute
t = 600 × 4 ÷ 800Step 2
600 × 4 = 2400Divide
2400 ÷ 800 = 3Answert = 3 s
✍️Worked example 4 — change in velocity from impulse
QuestionA force of 500 N acts on a 250 kg trolley for 4 s. Find the change in velocity.
Formula
Δv = F × t ÷ mSubstitute
Δv = 500 × 4 ÷ 250Step 2
500 × 4 = 2000Divide
2000 ÷ 250 = 8AnswerΔv = 8 m/s
🔁
Conservation of Momentum
In a closed system with no external forces, total momentum before a collision equals total momentum after.
🔁Collision CalculatorChoose collision type, enter values, press Calculate
Choose collision type, enter masses and initial velocities, then press Calculate.
📚 Key vocabulary
- Conservation of momentum
- the total momentum of a closed system does not change unless an external force acts on it.
- Perfectly inelastic collision
- a collision in which the objects stick together and move as one after impact.
- Perfectly elastic collision
- an ideal collision in which both momentum and kinetic energy are conserved.
- Closed system
- a system on which no external forces act, so total momentum is constant.
- Initial velocity (u)
- velocity before a collision or event.
- Final velocity (v)
- velocity after a collision or event.
👁️ Visual aid — perfectly inelastic collision: 2000 kg at 10 m/s hits stationary 1000 kg; they join and move at 6.67 m/s
✍️Worked example 5 — perfectly inelastic collision
QuestionA 2000 kg car travelling at 10 m/s collides with a stationary 1000 kg car. They join together. Find their common velocity after impact.
Before
p = m₁u₁ + m₂u₂ = 2000×10 + 1000×0 = 20 000 kg m/sAfter
(m₁ + m₂)v = 20 000Total mass
2000 + 1000 = 3000 kgDivide
v = 20 000 ÷ 3000 = 6.67Answerv = 6.67 m/s
✍️Worked example 6 — finding velocity after collision (object initially moving)
QuestionA 500 kg trolley moving at 6 m/s collides with a 300 kg trolley moving at 2 m/s in the same direction. They join. Find the final velocity.
Before
p = 500×6 + 300×2 = 3000 + 600 = 3600 kg m/sAfter
v = 3600 ÷ (500 + 300) = 3600 ÷ 800Divide
3600 ÷ 800 = 4.5Answerv = 4.5 m/s
✍️Worked example 7 — finding unknown mass
QuestionA 400 kg object at 5 m/s collides with a stationary object. After the perfectly inelastic collision the combined velocity is 2 m/s. Find the mass of the stationary object.
Before
p = 400 × 5 = 2000 kg m/sAfter
(400 + m₂) × 2 = 2000Rearrange
400 + m₂ = 2000 ÷ 2 = 1000Solve
m₂ = 1000 − 400 = 600Answerm₂ = 600 kg
✍️Worked example 8 — head-on collision (opposite directions)
QuestionA 1000 kg car moves at +8 m/s (positive direction). A 600 kg car moves at −5 m/s (opposite). They collide and stick together. Find the final velocity.
Before
p = 1000×8 + 600×(−5) = 8000 − 3000 = 5000 kg m/sAfter
v = 5000 ÷ (1000 + 600) = 5000 ÷ 1600Divide
5000 ÷ 1600 = 3.125Answerv = 3.125 m/s in the positive direction
⚙️
Engineering Applications
Four momentum and impulse calculators for vehicle, machine and conveyor scenarios.
🚗 Vehicle braking — stopping force
Calculate the average braking force needed to bring a vehicle to rest within a given time.
F = m × Δv ÷ t
🧮 Calculator
Enter mass, initial velocity and stopping time, then press Calculate.
🔁 Conveyor collision — joining velocity
A loaded pallet drops onto a moving conveyor belt. Find the combined velocity after impact.
(m₁ + m₂)v = m₁u₁ + m₂u₂
🧮 Calculator
Enter pallet and belt values, then press Calculate.
📶 Machine guard — safe stopping time
Given a maximum safe braking force, find the minimum time to stop a moving component safely.
t = m × Δv ÷ F
🧮 Calculator
Enter mass, velocity and max force, then press Calculate.
🔨 Post-collision velocity
Two objects collide and join. Find the combined velocity using conservation of momentum.
v = (m₁u₁ + m₂u₂) ÷ (m₁ + m₂)
🧮 Calculator
Enter masses and velocities (negative = opposite direction), then press Calculate.
🎮
Dynamics Challenge
12 randomised questions per round. New numbers every time.
📶 Quick-fire round
Question 1 / 12
Score: 0
Streak: 0 🔥
Press Start to begin!
📝
Practice Questions
12 graded questions. Attempt on paper first, then reveal the full worked solution.
Show the formula, the substitution and every step. State the positive direction for collision questions. Units: kg m/s for momentum, N s for impulse.
Question 1 · Momentum1 mark
A 750 kg motorbike travels at 20 m/s. Calculate its momentum.
p = m × v = 750 × 20 = 15 000 kg m/s
Question 2 · Momentum2 marks
A forklift truck of mass 3500 kg has a momentum of 14 000 kg m/s. Find its velocity.
v = p ÷ m = 14 000 ÷ 3500 = 4 m/s
Question 3 · Impulse2 marks
A 1200 kg car accelerates from 5 m/s to 25 m/s in 8 s. Calculate the average driving force.
F = m × Δv ÷ t
Δv = 25 − 5 = 20 m/s
F = 1200 × 20 ÷ 8 = 24 000 ÷ 8 = 3000 N
Δv = 25 − 5 = 20 m/s
F = 1200 × 20 ÷ 8 = 24 000 ÷ 8 = 3000 N
Question 4 · Impulse2 marks
A 2000 N braking force acts on a 500 kg trolley for 3 s. Find the change in velocity.
Δv = F × t ÷ m
Δv = 2000 × 3 ÷ 500 = 6000 ÷ 500 = 12 m/s
Δv = 2000 × 3 ÷ 500 = 6000 ÷ 500 = 12 m/s
Question 5 · Impulse2 marks
A component of mass 120 kg moving at 6 m/s must be stopped in 1.5 s. Find the required braking force.
F = m × Δv ÷ t
F = 120 × 6 ÷ 1.5 = 720 ÷ 1.5 = 480 N
F = 120 × 6 ÷ 1.5 = 720 ÷ 1.5 = 480 N
Question 6 · Impulse2 marks
A 300 N force acts for 5 s on a 150 kg object initially at rest. Calculate the final velocity.
Δv = F × t ÷ m = 300 × 5 ÷ 150 = 1500 ÷ 150 = 10 m/s
v = u + Δv = 0 + 10 = 10 m/s
v = u + Δv = 0 + 10 = 10 m/s
Question 7 · Conservation2 marks
A 1500 kg car at 12 m/s rear-ends a stationary 1000 kg car. They join. Find the combined velocity.
p before = 1500 × 12 + 1000 × 0 = 18 000 kg m/s
v = 18 000 ÷ (1500 + 1000) = 18 000 ÷ 2500 = 7.2 m/s
v = 18 000 ÷ (1500 + 1000) = 18 000 ÷ 2500 = 7.2 m/s
Question 8 · Conservation2 marks
A 600 kg trolley at 4 m/s collides with a 200 kg trolley at 1 m/s (same direction). They join. Find the final velocity.
p = 600 × 4 + 200 × 1 = 2400 + 200 = 2600 kg m/s
v = 2600 ÷ 800 = 3.25 m/s
v = 2600 ÷ 800 = 3.25 m/s
Question 9 · Conservation2 marks
A 500 kg object at 8 m/s collides with a stationary object. After the perfectly inelastic collision the combined velocity is 2 m/s. Find the mass of the stationary object.
p before = 500 × 8 = 4000 kg m/s
(500 + m₂) × 2 = 4000
500 + m₂ = 2000
m₂ = 1500 kg
(500 + m₂) × 2 = 4000
500 + m₂ = 2000
m₂ = 1500 kg
Question 10 · Conservation2 marks
A 2000 kg lorry at +6 m/s collides with a 500 kg car at −4 m/s (head-on). They stick together. Find the velocity and direction.
p = 2000 × 6 + 500 × (−4) = 12 000 − 2000 = 10 000 kg m/s
v = 10 000 ÷ 2500 = 4 m/s in the positive direction
v = 10 000 ÷ 2500 = 4 m/s in the positive direction
Question 11 · Combined3 marks
A 900 kg vehicle decelerates from 18 m/s to 3 m/s. (a) Find the change in momentum. (b) If this takes 5 s, find the braking force.
Δp = m × Δv = 900 × (18 − 3) = 900 × 15 = 13 500 kg m/s
F = Δp ÷ t = 13 500 ÷ 5 = 2700 N
F = Δp ÷ t = 13 500 ÷ 5 = 2700 N
Question 12 · Combined3 marks
A 1200 kg car at 10 m/s collides with a 800 kg stationary car. They join. (a) Find their combined velocity. (b) Find the impulse on the stationary car during impact.
v = (1200 × 10) ÷ (1200 + 800) = 12 000 ÷ 2000 = 6 m/s
Impulse on car 2 = m₂ × Δv = 800 × (6 − 0) = 4800 N s
Impulse on car 2 = m₂ × Δv = 800 × (6 − 0) = 4800 N s