Maths › Further Mechanics 1 › Impulse and momentum as vectors
Impulse and momentum as vectors
The same principle in two dimensions, where the i parts and the j parts each look after themselves. Nothing new is needed beyond the discipline of keeping components apart.
Builds on Momentum and impulse and Vectors in two dimensions.
IN THIS TOPIC
- Apply the impulse-momentum principle in component form, and give a magnitude and direction at the end.
- Conserve momentum in two dimensions by treating the components separately.
COMMON MISCONCEPTION
The magnitude of the impulse is the mass times the change in the speed of the particle.
Components look after themselves
Written as vectors, nothing changes.
Still not in the booklet, and still yours to remember. The i component of the impulse produces the change in the i component of the momentum, and likewise for j. Only at the end, once the vector is known, is a magnitude taken. Take magnitudes first and you throw away direction. A particle whose speed is unchanged but whose direction has turned has certainly received an impulse.
Drawn as a triangle, the initial momentum and the impulse placed nose to tail give the final momentum. That picture settles most sign questions faster than the algebra does.
WORKED EXAMPLE
An impulse in two dimensions
A particle of mass 0.5 kg moving with velocity (4i − 2j) m/s receives an impulse of (i + 3j) N s. Find its new velocity and speed.
Momentum before = 0.5(4i − 2j) = 2i − j. Adding the impulse: 3i + 2j.
Dividing by the mass: v = 6i + 4j m/s.
Speed = √(36 + 16) = √52 = 7.21 m/s. The speed before was √20 ≈ 4.47, so both size and direction have changed.
Conservation, component by component
In a two-dimensional collision with no external impulse, the total momentum vector is conserved. The i components balance and the j components balance, which gives two equations instead of one. Solve them together like any simultaneous pair.
Set the work out in a table, one row per particle and one column per component. That keeps the signs straight and makes the arithmetic checkable. A speed comes at the end from Pythagoras, and a direction from an inverse tangent with the quadrant checked against the signs.
GUIDED PRACTICE
Coalescing in two dimensions
A particle of mass 2 kg with velocity (3i + j) m/s collides with one of mass 3 kg with velocity (−i + 2j) m/s and they coalesce. Find the common velocity and its speed.
Show the working
Momenta: 2(3i + j) = 6i + 2j and 3(−i + 2j) = −3i + 6j.
Total = 3i + 8j N s, so 5v = 3i + 8j.
v = 0.6i + 1.6j m/s.
Speed = √(0.36 + 2.56) = √2.92 = 1.71 m/s, well below both original speeds of 3.16 and 2.24. Coalescence always loses kinetic energy.
ASSESSMENT FOCUS
- Work with momentum vectors throughout and divide by the mass only at the last step.
- Set the components out in columns. A single sign error in one component is the usual loss.
- Check a direction against the signs of the components before trusting the calculator's angle.
CHECK YOURSELF
A particle of mass 2 kg has velocity (5i − j) m/s. An impulse changes it to (i + 3j) m/s. Find the impulse.
Show a hint
Final momentum minus initial momentum.
Show the answer
I = 2(i + 3j) − 2(5i − j) = (2i + 6j) − (10i − 2j) = −8i + 8j N s, of magnitude 8√2 ≈ 11.3 N s.
In vector form I = mv − mu, so the i components and the j components each obey the principle on their own.
Take magnitudes only after the vector is known. A change of direction at constant speed still needs an impulse.
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 impulse and momentum as vectors questions page.
CHECK YOUR PROGRESS
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- Apply the impulse-momentum principle in component form, and give a magnitude and direction at the end.
- Conserve momentum in two dimensions by treating the components separately.
Open the full revision checklist to see every objective in the course in one place.