A particle P of mass 0.1 kg is projected vertically upwards with speed 30 m s⁻¹ from horizontal ground. At the same instant a particle Q of mass 0.4 kg is projected vertically upwards with speed 10 m s⁻¹ from a height of 15 m above the ground. P and Q move in the same vertical line. When P and Q collide, they coalesce.

Mark scheme

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

1. [*M1] [Height gained by P (s_P) =]30t - ½gt² [Height lost by Q (s_Q) =]10t - ½gt²; For use of s = ut + ½at² at least once with a=±g and u=±30 or u=±10. Allow this M1 only if using a found value of t.
2. [DM1] Meet when 30t - ½gt² = 10t - ½gt² + 15; For use of s_P = s_Q ±15 with s_P and s_Q of the correct form which would lead to a linear equation in t (so in the expressions for s_P and s_Q the signs of the ½gt² terms must be the same).
3. [A1] t = ¾ or 0.75; OE. CWO.
4. [A1] Height = 19.6875 m or 315/16 m; Allow 19.7 or better. CWO. DO NOT ISW.
5. [*B1FT] [v_P =]30 - ¾g [=22.5] [v_Q =][10 - ¾g][= ±2.5]; For either expression for the speed of P or Q before impact FT their t = ¾ and/or FT their height. 315/16 = 19.6875. The 2 M marks in 4(a) must have been awarded.
6. [DM1] 0.1×22.5+0.4×2.5=(0.1+0.4)×V_PQ; Use of conservation of momentum; correct number of non-zero terms, allow sign errors. 22.5 and 2.5 coming from a correct method. Do not allow with v_P = ±30 or with v_Q = ±10. If using mg rather than m, then do not allow subsequent A marks.
7. [A1] V_PQ = 6.5; SOI, allow to 2sf.
8. [DM1] v² = their 6.5² + 2g×their19.6875 OR 0=their 6.5² +2(-g)s ⇒s=2.1125, height above ground =2.1125+19.6875=21.8 hence v² = 0² + 2g×21.8; Dependent on previous M1 and B1. Complete method to get an equation in speed or (speed)². May see t=2.738061302.
9. [A1] Speed = 20.9 ms⁻¹ or 2√109 ms⁻¹; 20.88061303... Use of g in momentum equation can be awarded B1M1A0M1A0, 3 marks max.