One end of a light inextensible string is attached to a particle A of mass 3 kg. The other end of the string is attached to a particle B of mass 4kg. Particle A is in contact with a rough plane inclined at 30° to the horizontal, and particle B is in contact with a smooth horizontal plane. A second light inextensible string is attached to B. The other end of this second string is attached to a particle C of mass 5 kg which hangs vertically. Both strings are taut and pass over small smooth pulleys that are fixed at the ends of the horizontal plane. The part of the string from A to the pulley is parallel to a line of greatest slope of the inclined plane, and A, B and C are in the same vertical plane (see diagram). The system is released from rest. In the subsequent motion, C moves vertically downwards with acceleration 2ms⁻², and neither A nor B reach a pulley. [Figure 5.1]

Mark scheme

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

1. [NOTE] The mark scheme for this question was not used by examiners. All candidates were awarded full marks for this question due to a series-specific issue.
2. [1] (a) 5g-TBC = 5×2 Attempt at N2L for C - correct number of terms but allow sign errors (but must be using correct mass).
3. [1] (a) TBC-TAB =4×2 Attempt at N2L for B – correct number of terms but allow sign errors (but must be using correct mass). Allow with their TBC .
4. [1] (a) TBC = 40 N and TAB =32 N Both correct.
5. [1] (b) TAB-F-3g sin30=3×2 Attempt at N2L on A correct number of terms but allow sign errors; allow sin/cos mix (but must be using correct mass). For reference: F=11.
6. [1] (b) R=3gcos 30 Correct expression for normal contact force at A.
7. [1] (b) their 32-μ×3gcos30-3g sin30=3×2 and attempt to solve for μ Use of F = µR (where R is a component of weight) and their TAB to obtain an equation in μ only and solve for μ.
8. [1] (b) μ = 0.423 11√3 / 45
9. [1] (c) Distance travelled by A in first 1.5 seconds is 1/2×2×1.5² [= 2.25]
10. [1] (c) When string breaks A is moving at a speed of 3 (m s⁻¹)
11. [1] (c) -F-3g sin 30 = 3a a = -26/3 Attempt at N2L for A – correct number of terms but allow sign errors, and cos/sin mix.
12. [1] (c) 0=3²+2×(their a)×s s=27/52 Attempt at finding the distance travelled by A up the plane after the string breaks using v² =u²+2as (or other complete method) with v=0,u=3 and their negative acceleration.
13. [1] (c) Total distance travelled by A up the plane is [2.25+27/52=] 2.77 m 36/13 , 2.769230789....