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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow 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 g = 9.8 m s⁻² and give decimal answers to 3 significant figures unless a question says otherwise.

Start with the study guide and revision notes if a topic is new. See also the course hub, printable checklist and free 10-minute diagnostics.

Questions

1. A particle of mass 0.6 kg moves at 9 m s⁻¹ along a smooth horizontal surface. It receives an impulse of magnitude 7.2 N s in the direction opposite to its motion. Find the speed and direction of the particle afterwards. [3]

2. A ball of mass 0.45 kg hits a fixed vertical wall at 14 m s⁻¹, perpendicular to the wall, and rebounds at 9 m s⁻¹.

(a) Find the magnitude of the impulse the wall exerts on the ball. [2] (b) The contact lasts 0.015 s. Find the average force on the ball. [2]

3. Particle A of mass 2.5 kg moves at 7 m s⁻¹ on a smooth horizontal line. It collides directly with particle B of mass 1.5 kg moving towards it at 3 m s⁻¹. After impact A keeps moving the same way, at 1 m s⁻¹.

(a) Determine B’s velocity just after impact. [3] (b) Find the magnitude and direction of the impulse on B. [2]

4. A particle of mass 3 kg moving at 8 m s⁻¹ collides directly with a stationary particle B of mass m kg. The particles join and move at 2 m s⁻¹.

(a) Find m. [2] (b) Find the magnitude of the impulse on B. [2]

5. A block of weight 90 N lies on rough horizontal ground, with μ = 0.3 between the block and the ground. A horizontal force P acts on it and it does not move.

(a) When P = 20 N, find the friction force and the normal reaction. [2] (b) Find the largest P that keeps the block at rest. [2]

6. Three forces (3i − 2j) N, (pi + 5j) N and (−7i + qj) N act on a particle in equilibrium.

(a) Find p and q. [3] (b) Calculate the magnitude of the resultant of the first and third forces. [2]

7. A particle of mass 7 kg hangs in equilibrium from two light strings. One string makes 25° with the horizontal and the other 55°, on opposite sides of the particle. Find the tension in each string. [6]

8. A 15 kg crate stands on a rough floor where the coefficient of friction is 0.25. A force P pushes it, directed 35° below the horizontal.

(a) Find the value of P at which the crate is just about to slip. [5] (b) Show that if P = 40 N the crate stays at rest, and state the friction force. [3]

9. A uniform rod AB of length 4.8 m and mass 36 kg rests horizontally on supports at C and D. C is 0.6 m from A and D is 1.2 m from B.

(a) Calculate the reaction at each support. [4] (b) State where the weight acts, and why. [1]

10. A uniform plank AB of length 6 m and mass 24 kg rests horizontally on supports at C and D. C is 1 m from A and D is 4 m from A. A child of mass 18 kg stands on the plank 4.5 m from A.

(a) Calculate the reaction at each support. [5] (b) The child walks towards B. Find the greatest distance from A at which the plank stays in equilibrium on both supports. [3]

11. A uniform plank AB of length 4 m and mass 30 kg hangs horizontally from two vertical ropes attached at C and D, each 0.5 m from the nearer end. A load of mass 20 kg is placed at E, x m from A. Rope C can bear a tension of at most 330 N and rope D at most 350 N.

(a) For x = 3.5, find the tension in each rope. [4] (b) Find every x for which neither rope is overloaded. [6]

Answers

1. Take the original direction as positive. −7.2 = 0.6v − 0.6 × 9 [1]. 0.6v = 5.4 − 7.2 = −1.8 [1]. v = −3, so the particle moves at 3 m s⁻¹ in the opposite direction [1]. [3] Examiner insight: State a positive direction and put the sign of the impulse into the equation. A bare “3” omits the direction.

2. (a) Take away from the wall as positive. I = 0.45 × 9 − 0.45 × (−14) [1], so I = 10.35 N s (10.4 N s to 3 s.f.) [1]. (b) Force = I / t = 10.35 / 0.015 [1] = 690 N [1]. [4] Examiner insight: Writing the initial velocity as negative is the step people skip. Showing the substitution line makes a sign error easy to spot.

3. (a) Take A’s direction as positive. 2.5 × 7 + 1.5 × (−3) [1] = 2.5 × 1 + 1.5v [1]. 13 = 2.5 + 1.5v, so v = 7. B moves at 7 m s⁻¹ in A’s original direction [1]. (b) I = 1.5 × 7 − 1.5 × (−3) [1] = 15 N s in A’s original direction [1]. [5] Examiner insight: Give the direction of B’s velocity in words. The impulse on A is −15 N s, so quote one of the two and check it against the other.

4. (a) 3 × 8 = (3 + m) × 2 [1], so 3 + m = 12 and m = 9 [1]. (b) I = 9 × 2 − 9 × 0 [1] = 18 N s [1]. [4] Examiner insight: Treat joined particles as one common velocity. Check with the 3 kg particle: 3 × 2 − 3 × 8 = −18 N s.

5. (a) Vertical: R = 90 N [1]. Horizontal: friction balances P, so F = 20 N [1]. (b) F ≤ 0.3 × 90 = 27 [1], so the greatest P is 27 N [1]. [4] Examiner insight: In part (a) friction is 20 N, not μR. Writing F = μR when the block is only at rest gives 27 N.

6. (a) i: 3 + p − 7 = 0 [1], so p = 4 [1]. j: −2 + 5 + q = 0, so q = −3 [1]. (b) (3i − 2j) + (−7i − 3j) = (−4i − 5j) [1]. Magnitude = √(16 + 25) = √41 = 6.40 N [1]. [5] Examiner insight: Resolve each component separately and show both equations. In part (b) the answer equals the magnitude of the second force, which is a quick consistency check.

7. Weight = 7 × 9.8 = 68.6 N [1]. Horizontal: T₁ cos 25° = T₂ cos 55° [1]. Vertical: T₁ sin 25° + T₂ sin 55° = 68.6 [1]. Substituting T₁ = T₂ cos 55° / cos 25°: T₂(cos 55° tan 25° + sin 55°) = 68.6 [1]. T₂ = 63.1 N [1], T₁ = 40.0 N [1]. [6] Examiner insight: Mark the angles on a force diagram. Keep full values until the last line.

8. (a) Vertical: R = 15 × 9.8 + P sin 35° = 147 + P sin 35° [1]. Horizontal: F = P cos 35° [1]. Limiting: F = 0.25R [1]. P cos 35° = 0.25(147 + P sin 35°), so P(cos 35° − 0.25 sin 35°) = 36.75 [1]. P = 54.4 N [1]. (b) R = 147 + 40 sin 35° = 169.9 [1]. Greatest friction = 0.25 × 169.9 = 42.5 N [1]. Friction needed = 40 cos 35° = 32.8 N, which is less than 42.5 N, so the crate stays at rest and the friction force is 32.8 N [1]. [8] Examiner insight: A downward push raises R. Say in words that the friction needed is less than the greatest available, since “show that” wants the reasoning.

9. (a) CD = 4.8 − 0.6 − 1.2 = 3 m and the weight acts 2.4 m from A, which is 1.8 m from C. Weight = 36 × 9.8 = 352.8 N. Moments about C: R_D × 3 = 352.8 × 1.8 [1], with both lever arms correct [1]. R_D = 212 N (211.68 N) [1]. R_C = 352.8 − 211.68 = 141 N [1]. (b) At the midpoint, 2.4 m from each end, because the rod is uniform [1]. [5] Examiner insight: Taking moments about C removes the unknown R_C. “Uniform” is the modelling assumption that justifies the weight’s position, so quote it.

10. (a) Weights: plank 24 × 9.8 = 235.2 N at 3 m from A, child 18 × 9.8 = 176.4 N at 4.5 m from A. Distances from C: 2 m and 3.5 m. Moments about C: R_D × 3 = 235.2 × 2 + 176.4 × 3.5 [1], with correct distances [1]. R_D × 3 = 1087.8, so R_D = 362.6 N [1]. Vertical: R_C + R_D = 411.6 [1], so R_C = 49 N [1]. (b) About to tilt about D, so R_C = 0. Let the child be y m beyond D. Moments about D: 235.2 × 1 = 176.4y [1]. y = 4/3 [1]. Distance from A = 4 + 4/3 = 5.33 m (3 s.f.) [1]. [8] Examiner insight: “Greatest distance” signals a limiting condition, so set the reaction at the other support to zero. Check that your answer lies on the plank, here 5.33 m is less than 6 m.

11. (a) Weights: plank 294 N, load 196 N, total 490 N. E is above D. Moments about C: T_D × 3 = 294 × 1.5 + 196 × 3 [1], with correct distances [1]. T_D = 343 N [1]. T_C = 490 − 343 = 147 N [1]. (b) Moments about C with signed distance x − 0.5: 3T_D = 441 + 196(x − 0.5) [1]. T_D ≤ 350 gives 441 + 196(x − 0.5) ≤ 1050, so x ≤ 3.607 [1]. T_C = 490 − T_D, and T_C ≤ 330 needs T_D ≥ 160 [1]. Then 441 + 196(x − 0.5) ≥ 480, so x ≥ 0.699 [1]. Both limits needed: 0.699 ≤ x ≤ 3.61 (3 s.f.) [1]. Check that neither rope goes slack: both tensions stay positive in this range [1]. [10] Examiner insight: Convert each limit to a condition on x before solving, and give a two-sided range. An upper limit alone answers half the question.

Where marks are usually lost

• Using a speed where a signed velocity is needed in an impulse or collision equation.
• Leaving out the direction of an impulse or a final velocity.
• Writing F = μR for a block that is only at rest.
• Treating R as the weight when the applied force has a vertical part.
• Taking a moment with the distance along a rod rather than the perpendicular distance.
• Applying “about to tilt” without setting the correct reaction to zero.
• Leaving out the final comparison in a “show that” question.

Next steps

• Review the revision notes for the key results and methods.
• Reread the study guide for the worked examples behind each topic.
• Return to the Edexcel A Level Mathematics hub and tick topics on the checklist.
• Take one of the free 10-minute diagnostics to find the next gap.
• Book a free trial class if you want help from a teacher.

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.