Pearson Edexcel International Advanced Level Mathematics: Newton's laws, pulleys, slopes and friction (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 cover topics 4.1, 4.2 and 4.4 of Unit M1 (Mechanics 1) in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification states that M1 is an optional unit for IAS Mathematics and Further Mathematics and for IAL Mathematics and Further Mathematics. Marks shown are Marlbridge’s own allocations. Take g = 9.8 m s⁻² and give answers to 3 significant figures unless told otherwise. Strings are light and inextensible, and pulleys are smooth and fixed.

Questions

1. A particle P of mass 2 kg moves under two forces only, (5i + 2j) N and (−11i + 14j) N.

(a) Find the acceleration of P in the form pi + qj. [2] (b) Find the magnitude of the acceleration and the angle between its direction and the vector j. [3]

2. A crate of mass 120 kg is raised by a vertical cable. It starts from rest and reaches a speed of 3 m s⁻¹ after 5 s, with constant acceleration.

Find the tension in the cable during these 5 s. [3]

3. A lorry of mass 3000 kg pulls a trailer of mass 1000 kg along a straight horizontal road, using a horizontal coupling. The driving force on the lorry is 9200 N. The resistance to motion is 1800 N on the lorry and 400 N on the trailer.

(a) Find the acceleration of the lorry. [2] (b) Find the tension in the coupling. [2] (c) The driving force is removed and the resistances stay the same. Find the magnitude of the force in the coupling and state whether it is a tension or a thrust. [3]

4. Particles of mass 9 kg and 5 kg are attached to the ends of a string that passes over a smooth fixed pulley. The particles hang vertically and are released from rest.

(a) Find the acceleration of the particles and the tension in the string. [4] (b) The 9 kg particle is 1.4 m above the ground at release. Find the time it takes to reach the ground. [2]

5. Particles A and B, of masses 6 kg and 2 kg, are attached to the ends of a string that passes over a smooth fixed pulley. A is held 1.8 m above a horizontal floor with the string taut. The system is released from rest.

(a) Find the acceleration of A and the tension before A lands. [3] (b) Find the speed with which A hits the floor. [2] (c) A does not bounce. Find the greatest height of B above its starting position. [3]

6. A particle is projected up a line of greatest slope of a smooth plane inclined at 20° to the horizontal, with speed 7 m s⁻¹.

(a) Find the deceleration of the particle. [2] (b) Find the distance the particle travels up the plane before it first comes to rest. [2]

7. A block of mass 8 kg is released from rest on a rough plane inclined at 30° to the horizontal. It slides 6 m down a line of greatest slope in 2.5 s.

(a) Find the acceleration of the block. [2] (b) Find the coefficient of friction between the block and the plane. [3]

8. A crate of mass 10 kg is pulled up a line of greatest slope of a rough plane by a constant force of 80 N acting parallel to the plane. The plane is inclined at an angle α to the horizontal, where tan α = 5/12, and the coefficient of friction is 0.3. The crate starts from rest.

(a) Find the acceleration of the crate. [4] (b) After 4 s the pulling force is removed. Assuming the crate is still moving up the plane, find the further distance it travels before it comes to rest. [4]

9. A particle A of mass 5 kg lies on a rough horizontal table, with coefficient of friction 0.4. It is attached by a string passing over a smooth pulley at the edge of the table to a particle B of mass 3 kg hanging freely 0.8 m above the floor. The system is released from rest with the string taut.

(a) Find the acceleration of the system and the tension in the string. [5] (b) Find the speed of B as it reaches the floor. [2] (c) B does not bounce and A does not reach the pulley. Find the total distance moved by A from release until it comes to rest. [4]

10. A particle A of mass 3 kg lies on a rough plane inclined at 30° to the horizontal, with coefficient of friction 0.2. It is attached by a string over a smooth pulley at the top of the plane to a particle B of mass 7 kg hanging freely. The system is released from rest with the string taut and parallel to a line of greatest slope.

(a) Find the acceleration of the system. [4] (b) Find the tension in the string. [2] (c) After 1.5 s the string breaks. Find the further distance A moves up the plane before it first comes to rest. [3]

11. A box of mass 8 kg is at rest on rough horizontal ground, with coefficient of friction 0.2. A horizontal force of 40 N acts on it for 5 s and is then removed.

(a) Find the acceleration of the box while the force acts. [3] (b) Find the speed of the box and the distance it has travelled when the force is removed. [2] (c) Find the further distance the box travels before it comes to rest. [3]

Answers

1. (a) Resultant = (5 − 11)i + (2 + 14)j = (−6i + 16j) N [1]; a = (−6i + 16j)/2 = (−3i + 8j) m s⁻² [1] [2] (b) |a| = √(9 + 64) = √73 = 8.54 m s⁻² [1]; tan θ = 3/8 where θ is measured from j [1]; θ = 20.6° [1] [3] Examiner insight: Write the resultant on its own line, and say which axis your angle is measured from.

2. 3 = 0 + 5a gives a = 0.6 m s⁻² [1]; T − 120g = 120 × 0.6 [1]; T = 120(9.8 + 0.6) = 1248 N [1] [3] Examiner insight: Fix the positive direction on your diagram first, so weight and tension take opposite signs.

3. (a) Whole system: 9200 − 1800 − 400 = 4000a [1]; a = 1.75 m s⁻² [1] [2] (b) Trailer: T − 400 = 1000 × 1.75 [1]; T = 2150 N [1] [2] (c) Without the driving force: −2200 = 4000a, so a = −0.55 [1]; trailer: T − 400 = 1000(−0.55) [1]; T = −150, so the force is 150 N and it is a thrust [1] [3] Examiner insight: A negative tension needs a sentence of interpretation: say thrust and give the positive size.

4. (a) 9g − T = 9a [1]; T − 5g = 5a [1]; adding: 4g = 14a, so a = 2.8 m s⁻² [1]; T = 9(9.8 − 2.8) = 63 N [1] [4] (b) 1.4 = ½ × 2.8 × t² [1]; t = 1 s [1] [2] Examiner insight: Write one equation per particle, so your working is easy to follow and to check.

5. (a) 6g − T = 6a [1]; T − 2g = 2a [1]; 4g = 8a, so a = 4.9 m s⁻² and T = 2(9.8 + 4.9) = 29.4 N [1] [3] (b) v² = 0 + 2 × 4.9 × 1.8 = 17.64 [1]; v = 4.2 m s⁻¹ [1] [2] (c) After A lands the string is slack, and B moves freely with deceleration g [1]; further rise = 17.64/(2 × 9.8) = 0.9 m [1]; greatest height above start = 1.8 + 0.9 = 2.7 m [1] [3] Examiner insight: “Above its starting position” means add the first 1.8 m.

6. (a) Along the slope: ma = −mg sin 20° [1]; deceleration = 3.35 m s⁻² [1] [2] (b) 0 = 7² − 2 × 3.352 × s [1]; s = 7.31 m [1] [2] Examiner insight: Show the equation of motion along the slope with the sign of the deceleration clear.

7. (a) 6 = ½ × a × 2.5² [1]; a = 1.92 m s⁻² [1] [2] (b) R = 8g cos 30° [1]; 8a = 8g sin 30° − μ × 8g cos 30° [1]; 1.92 = 4.9 − 8.487μ, so μ = 0.351 [1] [3] Examiner insight: The mass cancels, so write the equation first and cancel visibly before substituting.

8. (a) sin α = 5/13, cos α = 12/13 [1]; R = 10g × 12/13 = 90.46 N, so F = 0.3R = 27.14 N [1]; 10a = 80 − 10g × 5/13 − 27.14 [1]; a = 1.52 m s⁻² [1] [4] (b) v = 4 × 1.517 = 6.07 m s⁻¹ [1]; deceleration = g sin α + μg cos α = 6.483 m s⁻² [1]; 0 = 6.068² − 2 × 6.483 × s [1]; s = 2.84 m [1] [4] Examiner insight: After the pull is removed, friction still acts down the slope, so both terms keep the same sign.

9. (a) R = 5g = 49 N [1]; F = 0.4 × 49 = 19.6 N [1]; A: T − 19.6 = 5a [1]; B: 3g − T = 3a [1]; adding: 9.8 = 8a, so a = 1.23 m s⁻² (1.225) and T = 25.7 N [1] [5] (b) v² = 2 × 1.225 × 0.8 = 1.96 [1]; v = 1.4 m s⁻¹ [1] [2] (c) A has moved 0.8 m by the time B lands [1]; the string is then slack and the deceleration is μg = 3.92 m s⁻² [1]; 0 = 1.96 − 2 × 3.92 × s, so s = 0.25 m [1]; total distance = 0.8 + 0.25 = 1.05 m [1] [4] Examiner insight: “Total distance” signals two stages, each with its own acceleration.

10. (a) R = 3g cos 30° = 25.46 N, so F = 0.2R = 5.09 N [1]; A: T − 3g sin 30° − F = 3a [1]; B: 7g − T = 7a [1]; adding: 68.6 − 14.7 − 5.09 = 10a, so a = 4.88 m s⁻² [1] [4] (b) T = 7(9.8 − 4.881) [1]; T = 34.4 N [1] [2] (c) v = 1.5 × 4.881 = 7.32 m s⁻¹ [1]; deceleration = g sin 30° + 0.2g cos 30° = 6.597 m s⁻² [1]; s = 7.321²/(2 × 6.597) = 4.06 m [1] [3] Examiner insight: After the string breaks, B is no longer part of the problem, so its weight should not appear.

11. (a) R = 8g = 78.4 N [1]; F = 0.2 × 78.4 = 15.68 N [1]; 8a = 40 − 15.68, so a = 3.04 m s⁻² [1] [3] (b) v = 3.04 × 5 = 15.2 m s⁻¹ [1]; s = ½ × 3.04 × 25 = 38 m [1] [2] (c) The deceleration is μg = 1.96 m s⁻² [1]; 0 = 15.2² − 2 × 1.96 × s [1]; s = 58.9 m [1] [3] Examiner insight: Carry the unrounded speed into the second stage to protect the third significant figure.

Where marks are usually lost

• Using R = mg on an inclined plane instead of R = mg cos α.
• Showing friction in the wrong direction, or including it once the particle has stopped.
• Writing one equation for two particles whose forces differ.
• Keeping the tension after a particle has hit the ground.
• Giving one stage’s distance when the question asks for the total.
• Mixing signs in T − mg = ma and mg − T = ma.
• Rounding intermediate values and losing the third significant figure.

Next steps

• Revisit the revision notes and the study guide.
• Check the Edexcel mathematics hub and the printable checklist.
• Take one of the all free 10-minute diagnostics.
• Book a free trial class if you want help with mechanics.

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: Mechanics 1, topics 4.1, 4.2 and 4.4.