M.A by Tutoring the nation A - Level Mechanics Connected Particles — A Two - Phase Problem A complete step - by - step tutorial October 2021 A - Level Exam · Question 3 1. Spot the phase change before you start. Look for phrases like ‘Q does not rebound’ or ‘P hits the ground’. These signal the end of Phase 1. 2. State what changes between phases. Write clearly: ‘When Q hits the ground, the string goes slack and tension = 0. P decelerates under gravity alone.’ Examiners reward correct reasoning. 3. Carry v² forward, not v. At the end of Phase 1 you get v² = 6gh/7. Use this directly in Phase 2 — don’t find v (avoid the square root). 4. Check the answer is ‘in terms of h only’. If g or m appears in your final answer, something has gone wrong — they should always cancel in this type of question. 5. Draw a Phase 2 diagram. A quick sketch showing only P with gravity acting downward takes 30 seconds and prevents sign errors. 6. Remember: height is cumulative. P’s final height = (height at end of Phase 1) + (extra height in Phase 2). Forgetting the 3h from Phase 1 is the most common error. M.A by Tutoring the nation M.A by Tutoring the nation What Makes This Problem Different? This is a two - phase connected - particle problem. Most pulley questions involve one continuous motion — but here, Q hits the ground and stops. At that moment the string goes slack, tension drops to zero, and P continues upward under gravity alone. You must s plit your solution into two separate phases. Q (mass 5m) is much heavier than P (mass 2m), so Q falls and P rises when released. Q starts at height h; P starts at height 2h. Key Terminology — read this first! Equation of motion: Newton’s Second Law (F = ma) written out for one specific object with actual forces substituted in. Inextensible string: Cannot stretch — both balls move at the same speed and magnitude of acceleration while the string is taut. String goes slack: When Q lands and stops, the string is no longer taut. Tension instantly drops to zero. Instantaneous rest: The moment when velocity = 0. This is the highest point P reaches. SUVAT equations: Five equations for constant acceleration. Key ones here: v² = u² + 2as. Phase 1: While the string is taut and both balls are moving together. Phase 2: After Q hits the ground — string is slack, only gravity acts on P (deceleration = g). The Question (a)(i) Write down an equation of motion for P. [2 marks] (a)(ii) Write down an equation of motion for Q. [2 marks] (b) Find, in terms of h only, the height above the ground at which P first comes to instantaneous rest. [7 marks] (c) State one limitation of modelling the balls as particles that could affect your answer to part (b). [1 mark] (d) State how a non - inextensible (stretchy) string would affect the accelerations of the particles. [1 mark] M.A by Tutoring the nation Strategy Apply Newton’s Second Law (F = ma) to each ball separately. Q is heavier (5m > 2m), so Q falls and P rises. Take the direction of motion as positive for each ball separately. (i ) Equation of motion for P (moving upward) — take upward as positive: Forces on P: Tension T upward, Weight 2mg downward. T − 2mg = 2m × a i.e. T − 2mg = 2ma (ii) Equation of motion for Q (moving downward) — take downward as positive: Forces on Q: Weight 5mg downward, Tension T upward. 5mg − T = 5m × a i.e. 5mg − T = 5ma Why is acceleration 'a' the same for both? Because the string is inextensible. Whatever speed one ball moves at, the other must match. The directions are opposite (P up, Q down), but the magnitude of acceleration is the same. Part (a) — Equations of Motion [4 marks] M.A by Tutoring the nation This question has two separate phases. Work through them in order. PHASE 1 — String taut: both balls moving (Q has not yet hit the ground) Step 1 — Find the acceleration a Add the two equations of motion. This eliminates T (it appears with opposite signs in each equation): 1 Write both equations: (1) T − 2mg = 2ma (2) 5mg − T = 5ma 2 Add (1) and (2). T cancels: 5mg − 2mg = 2ma + 5ma 3mg = 7ma 3 Divide both sides by 7m (m also cancels, as it appears in every term): a = 3g/7 This is the acceleration of both balls throughout Phase 1. Acceleration a = 3g/7 (Phase 1) Step 2 — Find the speed of P when Q hits the ground Q starts at height h and falls to the ground, travelling a distance h. Because the string is inextensible, P also travels exactly h upward. Both start from rest (u = 0). Apply v² = u² + 2as: 4 Known values for Phase 1: u = 0 (system starts from rest) a = 3g/7 (found above) s = h (Q falls distance h; P rises the same distance h) 5 Apply v² = u² + 2as: v² = 0 + 2 × (3g/7) × h v² = 6gh/7 We leave this as v² — we don’t need v itself. 6 Position of P when Q hits the ground: P started at 2h and rose h, so P is now at height 3h. At end of Phase 1: v² = 6gh/7 and P is at height 3h PHASE 2 — String slack: only gravity acts on P (after Q hits the ground) Part (b) — Height Where P First Stops [7 marks] M.A by Tutoring the nation When Q hits the ground it stops and does not rebound. The string immediately goes slack — tension drops to zero. P still has upward velocity but now only gravity acts on it, decelerating P at rate g. What changes in Phase 2? Tension = 0 (string is slack, not pulling P up any more). Acceleration of P = −g (gravity decelerates P; negative because it opposes upward motion). P’s initial speed² for this phase = 6gh/7 (carried over from end of Phase 1). P comes to rest when v = 0 (this is the answer we want). Step 3 — Find how much further P rises after Q stops 7 Known values for Phase 2 (for P): u² = 6gh/7 (from end of Phase 1) v = 0 (P comes to instantaneous rest) a = −g (gravity decelerates P; deceleration opposes motion) 8 Apply v² = u² + 2as and solve for s (extra distance P travels upward): 0 = 6gh/7 + 2 × (−g) × s 0 = 6gh/7 − 2gs 9 Rearrange for s: 2gs = 6gh/7 s = (6gh/7) ÷ (2g) s = 6h/14 = 3h/7 Note: g cancels here. The answer is in h only, as the question requires. 10 Total height of P: Height at end of Phase 1: 3h Extra height in Phase 2: 3h/7 Total = 3h + 3h/7 = 21h/7 + 3h/7 = 24h/7 Height of P when it first comes to rest = 24h/7 (≈ 3.43h above the ground) Why does g cancel? Every term in the equation contained a factor of g, so we divided through by g. This is why the answer is ‘in terms of h only’ as required. Always check whether g or m will cancel before you start. Common error: forgetting to add the 3h from Phase 1, giving only 3h/7 instead of 24h/7. M.A by Tutoring the nation A particle is a mathematical point with mass but no size. Real balls have a physical diameter. Any one of the following earns the mark: Limitation How it affects part (b) Real balls have size — they are not point particles. Ball Q effectively lands slightly earlier (when its base, not its centre, reaches the ground), changing the speed P has at the start of Phase 2. P might collide with the pulley. A particle passes through a small smooth pulley. A real ball of finite size could strike the pulley during Phase 2, stopping P earlier. Part (d) — Effect of a Non - Inextensible String [1 mark] A non - inextensible (extensible) string can stretch, like an elastic band. The question asks how this changes the accelerations. In the model (inextensible) In reality (extensible string) Force is transmitted instantly. Both balls accelerate immediately at 3g/7. The string stretches when released. Some energy goes into stretching rather than accelerating the balls. The accelerations are smaller initially, and the string contracts later causing additional oscillations. The accelerations of the particles would be reduced / smaller than 3g/7 Summary of All Results Part Result Marks (a)(i) Equation for P T − 2mg = 2ma 2 (a)(ii) Equation for Q 5mg − T = 5ma 2 (b) Acceleration (Phase 1) a = 3g/7 — (b) Speed when Q lands v² = 6gh/7 (P at height 3h) — (b) Final height of P 24h/7 7 (c) Particle limitation Balls have size / P could hit pulley 1 (d) Elastic string effect Accelerations would be smaller 1 Total 13 Part (c) — Limitation of the Particle Model [1 mark] M.A by Tutoring the nation Glossary of Key Terms Acceleration How quickly velocity changes. Measured in m/s². Deceleration Acceleration in the opposite direction to motion (slowing down). In Phase 2, gravity decelerates P. Equation of motion Newton’s Second Law written for one object: (net force) = mass × acceleration. Inextensible string Cannot stretch, so both ends always move at the same speed and magnitude of acceleration. Instantaneous rest The single moment when velocity = 0. For P, this is its highest point. Particle model Treating an object as a point with mass but no size. Simplifies calculations but ignores rotation and collisions. Phase A distinct period of motion where the same forces apply. A new phase begins when conditions change. SUVAT equations Five equations for constant acceleration. Key one here: v² = u² + 2as. String goes slack When tension drops to zero because one end is no longer being pulled. Happens when Q lands. Tension The pulling force in a string. Acts upward on a hanging ball. Same throughout a light string over smooth pulley. Weight Gravitational force = mass × g. Always acts downward. Ball P weighs 2mg; ball Q weighs 5mg. M.A by Tutoring the nation 1. Spot the phase change before you start. Look for phrases like ‘Q does not rebound’ or ‘P hits the ground’. These signal the end of Phase 1. 2. State what changes between phases. Write clearly: ‘When Q hits the ground, the string goes slack and tension = 0. P decelerates under gravity alone.’ Examiners reward correct reasoning. 3. Carry v² forward, not v. At the end of Phase 1 you get v² = 6gh/7. Use this directly in Phase 2 — don’t find v (avoid the square root). 4. Check the answer is ‘in terms of h only’. If g or m appears in your final answer, something has gone wrong — they should always cancel in this type of question. 5. Draw a Phase 2 diagram. A quick sketch showing only P with gravity acting downward takes 30 seconds and prevents sign errors. 6. Remember: height is cumulative. P’s final height = (height at end of Phase 1) + (extra height in Phase 2). Forgetting the 3h from Phase 1 is the most common error. A - Level Mechanics Tutorial · Connected Particles (Two - Phase) · October 2021 · Question 3 Exam Tips for Two - Phase Problems