Particles A and B, of masses 2kg and 6kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley and the particles hang vertically below the pulley. Both particles are initially held at rest at a height of 3.2m above horizontal ground (see diagram). Particle A is projected vertically downwards with a speed of 1.2ms⁻¹. [Figure 5.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [1] Attempt at N2L for either A or B or the system. Correct number of dimensionally correct terms, allow sign errors. Condone using a=±10(or another incorrect value of a) in either the equation for A or B for this mark only.
2. [1] For any two correct equations in a and T (or implied by correct answers) – note that T must be positive in one equation and negative in the other. Note that the three equations: 2g-T = 2a,T-6g=6a and 2g-6g=8a score M1 A1 and lead to T =3g and a=-0.5g.
3. [1] Solving for T From two N2L equations (both containing a) with the correct number of dimensionally correct terms OE.
4. [1] a = 0.5g or 5 m s⁻² and T = 3g or 30 N a must be positive – allow those who say a=-0.5g⇒a=0.5g (or a=-5⇒a=5) without justification (but must be from correct working).
5. [1] Using s = ut+at² with s = ±3.2, u = ± 1.2 and their a from 5(a) (not ±g but allow ± 10 if come from wrong working in 5(a)) – dependent on either a system equation with the correct number of dimensionally correct terms from 5(a) or from equations for A and B that both contain the correct number of dimensionally correct terms or other complete method to find t.
6. [1] 1.4(0) s AWRT 1.4 or (6+2√209)/25 (For reference: 1.39654...).
7. [1] Correct method to find the speed or speed² of A/B after travelling 3.2 metres (oe) – using 1.2 and their a from 5(a) - same dependency as in part (b) for the value of a from 5(a). If using v=u+at then t must come from considering t₂ – t₁ (see reference on previous page).
8. [1] Attempt to find the distance travelled by A after B hits the ground, using their w, a=-g and v = 0.
9. [1] Difference in distances is s₁ + 3.2 + s₂ = 5.016 m AWRT 5.02 or 5.01 CWO.