Maths › Further Mechanics 1 › Momentum and impulse
Momentum and impulse
Force applied for a time changes momentum by exactly that much. When two bodies push on each other the pushes are equal and opposite, so with no external impulse the total momentum is unchanged.
Builds on Forces and Newton's laws and Kinematics with constant acceleration.
IN THIS TOPIC
- Use impulse equals change in momentum, including for a rebound.
- Find an average force from an impulse and a contact time.
- Apply conservation of momentum to a direct collision between two spheres.
- Handle coalescence and explosion as special cases of the same principle.
COMMON MISCONCEPTION
A ball that bounces back off a wall at the same speed it arrived receives no impulse, since its speed is unchanged.
Force for a time
Momentum is mass times velocity. It is a vector, and its unit is the newton second, which is the same thing as the kilogram metre per second. Newton's second law rearranges into the impulse-momentum principle.
The booklet carries nothing on momentum or impulse, so learn it. Here F is the resultant force, taken as constant; when the force varies, Ft still gives the impulse provided F is read as the mean resultant force over the contact. Because momentum is a vector, a rebound is a change of sign, and the change is the sum of the two speeds, not their difference. A ball arriving at 20 and leaving at 20 the other way has had its momentum reversed, and that takes a large impulse.
When the force varies, the impulse is the area under the force-time graph. Divide it by the contact time and you have the average force. Longer contact therefore means a gentler force for the same change of momentum, which is the whole idea behind crumple zones and follow-through in a stroke.
WORKED EXAMPLE
A struck ball
A ball of mass 0.15 kg arrives at 20 m/s and is struck straight back at 25 m/s. The contact lasts 0.02 s. Find the impulse and the average force.
Taking the outgoing direction as positive: u = −20, v = 25.
I = 0.15(25) − 0.15(−20) = 3.75 + 3 = 6.75 N s.
Average force = 6.75/0.02 = 337.5 N, over two hundred times the ball's weight of 1.47 N.
Why the total cannot change
During a collision each body exerts a force on the other, and by Newton's third law those forces are equal and opposite for the same length of time. The impulses are therefore equal and opposite. Whatever momentum one body gains, the other loses, so, provided no external impulse acts on the pair, or the collision is so brief that any external impulse is negligible, the total is conserved whatever happens to the energy.
Conservation of linear momentum is not printed either. Learn it. Coalescence, where the two move off together, and explosion, where one body separates into two, are the same equation with the masses grouped differently. Signs matter throughout. Choose a positive direction, write every velocity with its sign, and let the algebra report which way anything is moving afterwards.
GUIDED PRACTICE
A direct collision
A sphere of mass 4 kg moving at 6 m/s meets a sphere of mass 3 kg moving at 2 m/s towards it. After the collision the 4 kg sphere continues in its original direction at 1 m/s. Find the velocity of the other sphere.
Show the working
Take the 4 kg sphere's direction as positive, so u₁ = 6 and u₂ = −2.
Total momentum before = 4(6) + 3(−2) = 24 − 6 = 18 N s.
After: 4(1) + 3v = 18, so 3v = 14 and v = 14/3 ≈ 4.67 m/s.
The sign is positive, so the 3 kg sphere has reversed and now moves in the original direction of the heavier one.
ASSESSMENT FOCUS
- Choose a positive direction, mark it on your diagram, and write every velocity with its sign.
- Treat impulse as a vector. A rebound adds the two speeds; it does not subtract them.
- Quote the units. N s for impulse and momentum, which is the same as kg m/s.
- For coalescence, use the combined mass on the right-hand side and one velocity.
- An explosion starts from zero momentum if the body was at rest, so the two fragments must have opposite signs.
CHECK YOURSELF
A particle of mass 3 kg moving at 5 m/s collides with a stationary particle of mass 2 kg and they move off together. Find their common speed.
Show a hint
Total momentum before equals combined mass times common velocity.
Show the answer
Momentum before = 3(5) = 15 N s. After: 5v = 15, so v = 3 m/s in the original direction.
Impulse is force times time and equals the change in momentum, so a rebound needs an impulse equal to the sum of the two speeds.
In a collision the internal forces are equal and opposite, so when the external impulse is negligible the total momentum before equals the total momentum after.
Divide an impulse by the contact time to get the average force.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the momentum and impulse questions page.
CHECK YOUR PROGRESS
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- Use impulse equals change in momentum, including for a rebound.
- Find an average force from an impulse and a contact time.
- Apply conservation of momentum to a direct collision between two spheres.
- Handle coalescence and explosion as special cases of the same principle.
Open the full revision checklist to see every objective in the course in one place.