A bead, A, of mass 0.1kg is threaded on a long straight rigid wire which is inclined at sin¯¹(7/25) to the horizontal. A is released from rest and moves down the wire. The coefficient of friction between A and the wire is µ. When A has travelled 0.45m down the wire, its speed is 0.6ms¯¹.

Mark scheme

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

1. [M1] 0.6² = 0 + 2a × 0.45 (Use of constant acceleration equations to find a.)
2. [A1] a = 0.4
3. [B1] R = 0.1g × cos α = 0.1g × 24/25 [= 0.96] (Must use a value for cos α.)
4. [M1] 0.1g × 7/25 - F = 0.1 × 0.4 [0.28 – F = 0.04 → F = 0.24] (Newton's second law, 3 terms.)
5. [M1] F = µ × 0.1g × 24/25 [F = µR, where R is a component of 0.1g]
6. [A1 AG] µ = 0.25 (Must be from exact working µ = 0.25 only)
7. [M1] 0.1×0.6 = 0.5v (Use of conservation of momentum, 2 terms.)
8. [A1] v = 0.12
9. [B1] For B 0.5g × 7/25 - 0.275 × 0.5g × 24/25 = 0.5a [leading to a = 0.16] (Apply Newton's second law for particle B, 3 terms. Allow correct unsimplified expression in a only.)
10. [*M1] Attempt an expression for either sA or sB. (Must see uA = 0 and uB ≠ 0 but uB must have been found from a momentum equation.)
11. [DM1] For both sA and sB and attempt to solve sA = sB to find t (Must be from 3 terms leading to a 2-term quadratic. If energy used in 7(a) then must find a = 0.4 for A. Their working must be leading to a positive t value.)
12. [A1] Required time is t = 1 s