Two particles A and B, of masses 3.2kg and 2.4kg respectively, lie on a smooth horizontal table. A moves towards B with a speed of vms¯¹ and collides with B, which is moving towards A with a speed of 6 ms¯¹. In the collision the two particles come to rest.
- (a)Find the value of v.[2]
- (b)Find the loss of kinetic energy of the system due to the collision.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- 2(a)
- [3.2v+2.4×(-6)]=0
- [M1] Attempt at conservation of momentum; 2 non-zero terms; allow sign errors.
- v = 4.5
- [A1] M1A0 for use of mgv. v = -4.5 is A0.
- 2(b)
- KE =±½×3.2×(their 4.5)² OR ±½×2.4×6²
- [M1] Attempt at either KE term, using their v. Do not allow ½×3.2×(their 4.5±6)², or ½×2.4×(their 4.5±6)², or ½×(3.2+2.4)×(their 4.5±6)², or ½×3.2×(their 4.5-0)², or ½×2.4×(6-0)².
- KEloss = 75.6 J
- [A1] Allow -75.6. Note ½×(3.2+2.4)×6² or ½×(3.2+2.4)×(their 4.5)² is M1A0.
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