Two particles A and B, of masses 2.4kg and 1.2kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. A is held at a distance of 2.1 m above a horizontal plane and B is 1.5 m above the plane. The particles hang vertically and are released from rest. In the subsequent motion A reaches the plane and does not rebound and B does not reach the pulley.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [M1] Attempt at Newton's second law on either particle or the system with correct number of terms; allow sign errors.
2. [A1] 2.4g-T = 2.4a T-1.2g =1.2a 2.4g-1.2g = (2.4+1.2) a Any 2 consistent and correct May have an a in opposite direction to our a
3. [M1] From equation(s) with correct number of relevant terms. If g missing then M0A0M1A0, maximum1/4. Must get a = or T =
4. [A1] T=16 N and a = 10/3 ms⁻² Both correct; allow a = 3.33. AG for T =16. Assuming T = 16 and only one equation is M1A0M0A0 maximum 1/4. Withhold A mark if T =15.9... ≈ 16, but condone T =1.2×3.33+12=16 or T = 24-2.4×3.33=16
5. [M1] Use of suvat or use energy to find v or v², using their a ≠ ±g (unless 10 comes from their attempt at a) from (a), s = 2.1; v² = 2 × (10/3) × 2.1 [=14] [v=√14 = 3.741.....] OR ½×2.4×v² = 2.4g×2.1-16×2.1 OR ½×2.4×v² + ½×1.2×v² = 2.4g×2.1-1.2g×2.1
6. [M1] Attempt to use v² = u² + 2as (or other complete method), using a = -g, to find additional height after string slack, using their v or v². 0=14-2g×s → s = ... or ½(1.2)(√14)² = (1.2)g×h → h=...
7. [A1] s = [1.5+2.1+0.7 =] 4.3 m AWRT 4.3(0); Allow use of a = 3.33 to give s = 4.2993 ≈ 4.3(0) Allow use of v = 3.74 to give s = 4.29938 ≈ 4.3 (0)