Pearson Edexcel International Advanced Level Mathematics: Work, energy and power, and collisions with the coefficient of restitution (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 Pearson Edexcel International Advanced Level Mathematics (YMA01), Specification Issue 3, April 2019, unit M2 sections 3.1, 4.1, 4.2 and 4.3. The specification states that M2 is an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. Teaching is in the study guide and revision notes. See also the hub, the checklist and the free 10-minute diagnostics. Use g = 9.8 m s⁻². State answers to three significant figures unless told otherwise.

Questions

1. A sledge of mass 25 kg starts from rest on horizontal ground. It is pulled 12 m by a rope that makes an angle of 30° with the horizontal, with constant tension 80 N. The total resistance to motion is a constant 30 N. Find the speed of the sledge after 12 m. [4]

2. A cyclist and bicycle have total mass 80 kg. The cyclist works at a constant rate of 250 W. The resistance to motion is a constant 20 N.

(a) Find the steady speed on level ground. [2] (b) Find the steady speed up a road inclined at α to the horizontal, where sin α = 1/25. [3]

3. A particle of mass 0.6 kg is launched at 7 m s⁻¹ up a plane that is inclined at 30° to the horizontal, moving along the steepest direction.

(a) The plane is smooth. How far up the plane does it travel before it is instantaneously at rest? [3] (b) The plane is now rough, with a constant frictional resistance of 2 N. Find the new distance it travels up the plane. [2] (c) Find the speed of the particle when it returns to the point of projection in the situation in part (b). [2]

4. Particle A of mass 3 kg moves at 7 m s⁻¹ along a straight line on a smooth horizontal surface. Particle B of mass 2 kg moves along the same line at 2 m s⁻¹ in the opposite direction. They collide directly. The coefficient of restitution is 0.5.

(a) Find the velocities of A and B after the collision, stating their directions. [4] (b) Find the magnitude of the impulse exerted on B. [2] (c) Find the kinetic energy lost. [2]

5. Particle P of mass m, moving at speed u on a smooth horizontal surface, collides directly with particle Q of mass 3m at rest. After the collision P moves in the opposite direction with speed u/4.

(a) Show that the coefficient of restitution is 2/3. [4] (b) Find the fraction of the initial kinetic energy lost in the collision. [3]

6. A ball is dropped from rest at a height of 3.6 m above a horizontal floor. The coefficient of restitution between ball and floor is 0.7.

(a) Find the speed of the ball just before it first hits the floor. [2] (b) Find the greatest height reached after the first bounce. [2] (c) Find the greatest height reached after the second bounce. [2]

7. Particles A, B and C, of masses 3 kg, 1 kg and 2 kg, lie in that order on a smooth horizontal line. B and C are at rest. A moves towards B at 4 m s⁻¹. The coefficient of restitution is 0.5 for every collision.

(a) Find the speeds of A and B after the first collision. [3] (b) B then collides with C. Find the speeds of B and C after this collision. [3] (c) Show that A and B collide again, and find their speeds afterwards. [4] (d) Explain whether B and C collide again. [1]

8. Particle A of mass m moves at 10 m s⁻¹ on a smooth horizontal surface and hits particle B of mass 4m, at rest between A and a fixed vertical wall. The coefficient of restitution between A and B is 0.5. The coefficient of restitution between B and the wall is e.

(a) Find the velocities of A and B after their first collision. [3] (b) Find the range of values of e for which A and B collide a second time. [4]

9. A van of mass 1500 kg climbs a hill whose steepest direction makes an angle α with the horizontal, where sin α = 1/12. It moves at constant speed 18 m s⁻¹ with the engine working at 36 kW.

(a) Find the constant resistance to the motion of the van. [4] (b) The engine is switched off. Assuming the resistance is unchanged, find the further distance the van travels up the road before stopping. [3] (c) Find the work done by the engine in the 2 minutes before it was switched off. [1]

10. Particle A of mass 0.3 kg has velocity (2i − 4j) m s⁻¹ and particle B of mass 0.5 kg has velocity (−3i + j) m s⁻¹. They collide, and immediately afterwards A has velocity (−i + 2j) m s⁻¹.

(a) Find the velocity of B after the collision. [3] (b) Find the magnitude of the impulse exerted by B on A. [3]

Answers

1. Work done by the tension = 80 cos 30° × 12 = 831.4 J [1]. Work done against resistance = 30 × 12 = 360 J [1]. Work-energy: 831.4 − 360 = ½ × 25 × v² [1], so v² = 2 × 471.4 / 25 and v = 6.14 m s⁻¹ [1]. Examiner insight: A force at an angle does work through its component along the motion, so a bare 80 × 12 is likely to lose credit.

2. (a) Driving force = resistance = 20 N [1]. v = 250 / 20 = 12.5 m s⁻¹ [1]. (b) Driving force = 20 + 80 × 9.8 × (1/25) [1] = 51.36 N [1]. v = 250 / 51.36 = 4.87 m s⁻¹ [1]. Examiner insight: On a hill the driving force must also overcome the weight component along the slope; finding it from power alone and then forgetting that term gives the level-ground speed.

3. (a) KE at start = ½ × 0.6 × 49 = 14.7 J [1]. Gain in PE = 0.6 × 9.8 × d sin 30° = 2.94d [1]. 14.7 = 2.94d, so d = 5 m [1]. (b) 14.7 = 2d + 2.94d [1], so d = 2.98 m [1]. (c) Down the slope: ½ × 0.6 × v² = 2.94d − 2d = 0.94 × 2.976 [1], so v = 3.05 m s⁻¹ [1]. Examiner insight: Carry the unrounded distance from part (b) into part (c), since a rounded value can change the third significant figure.

4. (a) A’s direction positive. Momentum: 3vA + 2vB = 21 − 4 = 17 [1]. Restitution: vB − vA = 0.5 × 9 = 4.5 [1]. So 5vA + 9 = 17 [1]. vA = 1.6 m s⁻¹ and vB = 6.1 m s⁻¹, both in A’s original direction [1]. (b) Impulse on B = 2[6.1 − (−2)] [1] = 16.2 N s [1]. (c) KE before = ½(3)(49) + ½(2)(4) = 77.5 J, KE after = ½(3)(2.56) + ½(2)(37.21) = 41.05 J [1]. Loss = 36.45 J [1]. Examiner insight: Opposite directions mean the speed of approach is the sum of the speeds; a sign diagram makes this clear.

5. (a) Take P’s original direction as positive. Momentum: mu = −m(u/4) + 3mvQ [1]. So vQ = 5u/12 [1]. Restitution: vQ − (−u/4) = eu [1]. So e = 5/12 + 1/4 = 8/12 = 2/3 [1]. (b) KE after = ½m(u/4)² + ½(3m)(5u/12)² [1] = mu²/32 + 25mu²/96 = 7mu²/24 [1]. Initial KE = mu²/2 = 12mu²/24, so the fraction lost is 5/12 [1]. Examiner insight: A “show that” needs both equations written in full, with exact fractions rather than decimals.

6. (a) v² = 2 × 9.8 × 3.6 = 70.56 [1], so v = 8.4 m s⁻¹ [1]. (b) Rebound speed = 0.7 × 8.4 = 5.88 m s⁻¹ [1]. Height = 5.88² / (2 × 9.8) = 1.76 m (1.764 m) [1]. (c) Speed after the second impact = 0.7 × 5.88 × 0.7 = 0.49 × 8.4 = 4.116 m s⁻¹ [1]. Height = 4.116² / 19.6 = 0.864 m [1]. Examiner insight: Noting that each bounce multiplies the height by e² saves time.

7. (a) Momentum: 3(4) = 3vA + vB [1]. Restitution: vB − vA = 0.5 × 4 = 2 [1]. So 4vA + 2 = 12: vA = 2.5 m s⁻¹ and vB = 4.5 m s⁻¹, both towards C [1]. (b) Momentum: 1(4.5) = vB′ + 2vC [1]. Restitution: vC − vB′ = 0.5 × 4.5 = 2.25 [1]. So 3vC = 6.75: vC = 2.25 m s⁻¹ and B is brought to rest [1]. (c) B is at rest and A approaches at 2.5 m s⁻¹, so they collide again [1]. Momentum: 3(2.5) = 3a + b [1]. Restitution: b − a = 0.5 × 2.5 = 1.25 [1]. So a = 1.5625 ≈ 1.56 m s⁻¹ and b = 2.8125 ≈ 2.81 m s⁻¹, both towards C [1]. (d) B (2.81 m s⁻¹) is behind and faster than C (2.25 m s⁻¹), so they collide again [1]. Examiner insight: List all three velocities after each collision, and give a reason in words for any claim about a further collision.

8. (a) Take towards the wall as positive. Momentum: 10m = ma + 4mb [1]. Restitution: b − a = 0.5 × 10 = 5 [1]. Then 5a + 20 = 10: a = −2 m s⁻¹ (A rebounds) and b = 3 m s⁻¹ [1]. (b) B returns from the wall at 3e m s⁻¹ [1], A moves away at 2 m s⁻¹ [1]. B must be faster: 3e > 2 [1]. So 2/3 < e ≤ 1 [1]. Examiner insight: Equal speeds mean no further meeting, so the inequality is strict; the upper bound comes from e ≤ 1.

9. (a) Driving force F = P/v = 36 000 / 18 = 2000 N [1]. Weight component down the slope = 1500 × 9.8 / 12 = 1225 N [1]. Constant speed: 2000 = R + 1225 [1]. R = 775 N [1]. (b) KE = ½ × 1500 × 18² = 243 000 J [1]. Work against resistance and weight component: (775 + 1225)d = 243 000 [1], so d = 121.5 m [1]. (c) Work = Pt = 36 000 × 120 = 4.32 × 10⁶ J [1]. Examiner insight: Convert kilowatts and minutes to watts and seconds before substituting, and give units with final answers.

10. (a) Momentum before = 0.3(2i − 4j) + 0.5(−3i + j) = −0.9i − 0.7j [1]. After: 0.3(−i + 2j) + 0.5vB = −0.9i − 0.7j [1]. So vB = (−1.2i − 2.6j) m s⁻¹ [1]. (b) I = 0.3[(−i + 2j) − (2i − 4j)] [1] = −0.9i + 1.8j [1]. |I| = √(0.81 + 3.24) = √4.05 = 2.01 N s [1]. Examiner insight: Keep the i and j equations separate and write each vector in full, so that a slip in one component is easy to see.

Where marks are usually lost

• Putting the length of the slope into mgh in place of the vertical rise.
• Leaving the weight component out of the driving-force equation on an inclined road.
• Writing the restitution equation with approach and separation the wrong way round.
• Using the momentum equation for a collision with a wall.
• Not checking each particle’s direction after a collision before deciding on another.
• Rounding too early or omitting units.

Next steps

• Re-read the revision notes and study guide.
• Open the course hub and printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

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 M2: Mechanics 2, sections 3.1, 4.1, 4.2 and 4.3.