A particle A is projected vertically upwards from level ground with an initial speed of 30 m s⁻¹. At the same instant a particle B is released from rest 15 m vertically above A. The mass of one of the particles is twice the mass of the other particle. During the subsequent motion A and B collide and coalesce to form particle C.

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

1. [B1] Use of constant acceleration equations to find expressions for displacements of A or B.
2. SA = ±(30t-5t²) or SB = ±5t²
3. [B1] Use SA+SB = 15 to find time at which particles collide.
4. SA+SB = 15 leading to 15=30t leading to t=0.5
5. [B1] Find speed of particles at t = 0.5 before collision.
6. t=0.5 leading to vA = ±25 and vB = ±5
7. [B1] Find position of A or B at which collision occurs at t = 0.5 Alternatively allow hB = ±1.25 as displacement of B
8. t=0.5 leading to hA = ±(30×0.5-1/2 g×0.5²) = ±13.75
9. [M1] Use of conservation of momentum, either case, using their vA and vB ≠ 0 or 30, with 3 terms.
10. 25×(2m)-5(m)=(3m)v -> v1 = 15
11. 25(m)-5×(2m)=(3m)v -> v2 = 5
12. [A1] Both values of v correct
13. [M1] Use of s = ut + 1/2 at^2 OE to find t, using either their numerical v1 or numerical v2 from a relevant conservation of momentum equation.
14. Particle C1 -13.75=15t-5t²
15. Particle C2 -13.75=5t-5t²
16. [A1] tc1, tc2 = 3.74,2.23 leading to T =1+√5−√3=1.50