Pearson Edexcel International Advanced Level Mathematics: Momentum and impulse, equilibrium of a particle and moments (YMA01) – Revision Notes

These notes condense the study guide for the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019, M1 topics 4.3, 5.1, 5.2, 5.3 and 6.1. The specification lists M1 as an optional unit for IAS and IAL Mathematics and Further Mathematics. Use the practice questions to test yourself, tick off topics on the checklist, start from the course hub and take a free 10-minute diagnostic.

Key results

Topic	Result	Notes
4.3	Momentum = mv	Vector. Units N s or kg m s⁻¹.
4.3	Impulse = mv − mu	Change in momentum. Also F × t for a constant force.
4.3	m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂	Direct collision, one dimension. Impulses are equal and opposite.
5.1	Component along the angle = F cos θ, perpendicular = F sin θ	Check which angle is given.
5.2	Resultant force = 0	Resolve in two perpendicular directions.
5.3	F ≤ μR	F = μR only when limiting.
6.1	Moment = force × perpendicular distance to the pivot	Unit N m. Give the sense: clockwise or anticlockwise.
6.1	Clockwise moments = anticlockwise moments	About any point, plus forces balance.

4.3 Momentum and impulse in steps

1. Draw a diagram and mark a positive direction.
2. Give each velocity a sign. A particle that rebounds changes sign.
3. For impulse, write I = mv − mu with signs, then interpret the sign.
4. For a collision, write the conservation equation with all four terms, then substitute.
5. If particles join, use one common velocity.
6. Check by finding the impulse on each particle. The two values should be equal and opposite.

Worked reminder: a 0.2 kg ball moving at 6.5 m s⁻¹ is hit back so it moves at 2 m s⁻¹ in the opposite direction. Taking the first direction as positive, I = 0.2 × (−2) − 0.2 × 6.5 = −1.7. The impulse has magnitude 1.7 N s, opposite to the original motion.

The specification says Newton’s law of restitution is not required and problems are one-dimensional. A problem that gives you speeds before and after (or enough to find them) is solvable with conservation alone.

5.1 Resolving forces

Situation	Horizontal	Vertical
Force F at θ above horizontal	F cos θ	F sin θ
Force F at θ to the vertical	F sin θ	F cos θ
Force written as (ai + bj)	a	b

For a resultant (xi + yj), the magnitude is √(x² + y²). For the direction, find tan⁻¹(y/x) with a sketch to fix the quadrant, then state the angle relative to a named direction such as the positive x-axis.

Worked reminder: F₁ = (4i + 7j) N and F₂ = (−10i + 3j) N give a resultant of (−6i + 10j) N. Its magnitude is √(36 + 100) = 11.7 N (3 s.f.). The angle above the negative i direction is tan⁻¹(10/6) = 59.0°. For equilibrium you would add the force (6i − 10j) N.

5.2 Equilibrium in steps

1. Draw all forces on the particle. Include weight, any normal reaction, tension, thrust and friction.
2. Choose two perpendicular directions.
3. Resolve in each direction and set the sums to zero.
4. Solve. For two unknown tensions, resolve so one tension disappears where possible.
5. Round to 3 significant figures unless the question says otherwise, and state the g you used.

If the forces are written with i and j, total the i parts and the j parts and make each total zero.

Must-know distinctions:

• Tension pulls along a string. Thrust pushes along a rod. A string can only pull, a rod can do either.
• Normal reaction is at 90° to the surface. Friction lies along it.
• Weight is mg in newtons. Mass is in kg.

5.3 Friction at rest

Statement in the question	What to write
“in equilibrium” or “at rest”	F ≤ μR, and F balances the other forces
“on the point of slipping” or “limiting”	F = μR
“least value of μ”	Use limiting friction and solve for μ
“greatest force that can be applied”	Use limiting friction and solve for the force

Method: resolve at right angles to the surface for R, resolve along the surface for F, then compare F with μR. An upward component of an applied force lowers R. A downward component raises it.

Worked reminder: a block weighs 80 N and is pulled by a horizontal force of 25 N without moving. R = 80 and F = 25, so 25 ≤ 80μ and μ ≥ 0.3125. The friction is 25 N.

6.1 Moments in steps

1. Draw the body with every vertical force, including weights at the correct points.
2. State a pivot point. Choose the point where the most unknowns act.
3. Write clockwise moments equal to anticlockwise moments.
4. Resolve vertically to find the remaining force.
5. For “about to tilt”, make the reaction at the support that lifts off zero and take moments about the other support.

Must-know distinctions:

• Uniform: weight at the midpoint. Light: no weight. Non-uniform: the centre of mass is given.
• The support that is about to lose contact is the one with zero reaction.
• Moment uses perpendicular distance, not distance along the rod.

Worked reminder (tilting): a uniform rod of weight 60 N and length 4 m rests on supports 1 m from each end. A particle of weight 20 N is placed on the rod beyond the right support, a distance d from it. The rod is about to tilt about that support when the left reaction is zero. The rod’s weight is 1 m from the support on the left side, so 60 × 1 = 20 × d and d = 3 m. That is more than the 1 m of rod beyond the support, so the rod cannot tilt for any position on it.

Worked reminder (two supports): a light rod of length 3 m rests on supports at its two ends. A load of 45 N is 1 m from A. Moments about B: R_A × 3 = 45 × 2, so R_A = 30 N. Then R_B = 45 − 30 = 15 N.

Quick self-test

1. Find the momentum of a 2.5 kg particle moving at 6 m s⁻¹.
2. A 0.2 kg ball changes velocity from 6.5 m s⁻¹ to −2 m s⁻¹. State the impulse.
3. A (2 kg, 3 m s⁻¹) hits a resting B (1 kg) and the two stick together. Find their common speed.
4. Resolve a 30 N force acting at 60° to the vertical into horizontal and vertical components.
5. Forces (2i + 3j), (pi − 5j) and (−6i + qj) N keep a particle in equilibrium. Find p and q.
6. The normal reaction on a block is 60 N and μ = 0.35. Find the greatest friction force available.
7. Perpendicular forces of 6 N and 8 N act on a particle that is held in equilibrium by a third force. What is the size of that force?
8. Find the moment of a 15 N force about a point at perpendicular distance 0.8 m from its line of action.
9. A uniform rod of weight 30 N and length 2 m rests on supports at its ends. Find each reaction.
10. A light rod rests on supports at its ends, 3 m apart. A load of 45 N is placed 1 m from end A. Find both reactions.

Answers

1. 15 N s.
2. I = 0.2 × (−2) − 0.2 × 6.5 = −1.7, so 1.7 N s opposite to the original direction.
3. 2 × 3 = 3v, so v = 2 m s⁻¹.
4. Horizontal 30 sin 60° = 26.0 N (3 s.f.), vertical 30 cos 60° = 15 N.
5. i: 2 + p − 6 = 0, so p = 4. j: 3 − 5 + q = 0, so q = 2.
6. 0.35 × 60 = 21 N.
7. √(6² + 8²) = 10 N.
8. 15 × 0.8 = 12 N m.
9. 15 N each, by symmetry.
10. Moments about B: R_A × 3 = 45 × 2, so R_A = 30 N. Then R_B = 15 N. The support nearer the load takes more.

Where marks are usually lost

• Forgetting a sign on a velocity, so a rebound is treated as a speed reduction.
• Giving impulse as a bare number with no direction or unit.
• Not drawing a force diagram, which leads to a missing weight, thrust or friction term.
• Writing F = μR when the particle is merely at rest.
• Resolving with the wrong trigonometric function and not checking it against a sketch.
• Taking moments about a point but using a distance that is not perpendicular to the force.
• Leaving a reaction out of a vertical resolution after using moments.
• Quoting answers to the wrong accuracy or without units when the question asks for a numerical value.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019 (Pearson Education Limited), unit M1 topics 4.3, 5.1, 5.2, 5.3 and 6.1.