A ball of mass 1.6 kg is released from rest at a point 5 m above horizontal ground. When the ball hits the ground it instantaneously loses 8 J of kinetic energy and starts to move upwards.

Mark scheme

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

1. [B1] Either finds PE loss or uses v² = u² + 2as to find the velocity and hence the kinetic energy on reaching the ground (PE = 1.6 × 10 × 5 [= 80J] or v↓ = √(2×10×5) [= 10], KE = ½ × 1.6 × 10² [= 80 J])
2. [M1] Using Initial PE = Final PE + Loss in KE or using KE = ½ mv² to find initial velocity upwards and either v² = u² + 2as or KE loss = PE gain to form equation in h. (1.6 × 10 × 5 = 1.6 × 10 ×h + 8 or ½ × 1.6 × v² = 80 – 8, v↑ = √90, 0 = 90 + 2×(-10) × h leading to h = ... or ½ m × 90 = m × 10 × h leading to h = ...)
3. [A1] h = 4.5 m
4. [M1] Use of s = ut + ½ gt² for downward motion or use of s = ½ (u + v)t for downward motion or use of v = u + gt for downward motion. (5 = 0 + ½ × 10 × t² leading to t = 1 or 5 = ½ × (0 + 10) × t leading to t = 1 or 10 = 10t leading to t = 1)
5. [M1] Use of s = vt – ½ (-g)t² for upward motion or use of s = ½ (u + v)t for upward motion or use of v = u – gt for upward motion. (4.5 = 0 - ½ × (-10) × t² leading to t = √0.9 or 4.5 = ½ × (√90 + 0) × t leading to t = √0.9 or 0 = √90 – 10t leading to t = √0.9)
6. [A1] t = 1.95 s