Two particles, A and B, of masses 3 kg and 5 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. The particles are held with the string taut and its straight parts vertical. Particle A is 1 m above a horizontal plane, and particle B is 2m above the plane (see diagram). The particles are released from rest. In the subsequent motion, A does not reach the pulley, and after B reaches the plane it remains in contact with the plane. [Figure 7]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

1. [*B1] For one correct equation.
2. [*B1] For any two correct consistent equations. If tensions T_A and T_B both stated must at some point state or imply that they are equal to score the second B1.
3. [DM1] May find a first [a = 2.5] Must get to 'T='. Dep on both B marks.
4. [A1] Allow without working.
5. [DM1] E.g. 2 = 0.5×2.5×t^2 Must get to 't='. Dep on both B marks or the B1 for the equation 5g-3g=±(3+5)a if this used to find a, but not on first M1 mark. Could find v = √10 then use 2 = (0+their√10)/2 * t.
6. [A1] t=1.2649... If candidates do not try to find T but do attempt to find the time, they can score B1B1M0A0M1A1. Do not allow if also give negative answer and do not discard. Note: t = √1.6 only, scores A0 .
7. [*M1] v² = 0 + 2×2.5×2. Must use s=2 and u=0, OR v=0+2.5×√1.6 Must use u=0, OR 2 = 1/2(0+v)√1.6 Must use s=2 and u=0. Must be complete method to find v or v² [v = √10]. Could have found v = √10 in part (a) and give M1 if used in part (b).
8. [*DM1] 3.25-3=(their√10)t+0.5×(-10)×t².
9. [DM1] If correct should get t = 0.093, 0.540 or (√10-√5)/10, (√10+√5)/10. 0.09262... or 0.53978... Could use formula and realise that the time for at least 3.25 m, = (√b²-4ac)/(2a) = (√10-4×5×0.25)/(2×5) which gets DM1.
10. [A1] Time=0.53978...-0.09262... = 0.44721... 2(√10-(√5/10)) = 0.44721...
11. [*M1] PE loss = KE gain: 5g×2-3g×2 = 1/2×5v²+1/2×3v², Or PE loss - WD by tension = KE gain: 5g×2-their37.5×2 = 1/2×5v², Or WD by tension – PE gain = KE gain: their37.5×2-3g×2=1/2×3v². Must be complete method to find v or v² [v = √10 or3.162...]. Dependent on both B marks only if candidate uses tension, but otherwise not dependent on either B mark.
12. [*DM1] w² = their√10²+2×(-g)×0.25. [w² = 5] or [w= √5].
13. [DM1] 0=√5+(-g)×t [t=√5/10].
14. [A1] For doubling.