Q#19 (Past Exam Paper – June 2018 Paper 12 Q9)
An elastic collision occurs between two bodies X and Y. The mass of body X is m and the mass of body Y is 4m. Body X travels at speed v before the collision and speed 3v / 5 in the opposite direction after the collision. Body Y is stationary before the collision.
What is the kinetic energy of body Y after the collision?
A 8/10 mv2 B 34/50 mv2 C 16/50 mv2 D 1/5 mv2
Solution:
Answer: C.
For an elastic collision, both the momentum and the kinetic energy in the system are conserved.
Taking into account these two quantities conserved, it can be shown that
Relative speed of approach (before collision) = Relative speed of separation (after collision)
Before the collision, X is moving towards Y, which is stationary.
Relative speed of approach = v + 0 = v
After the collision, X and Y are moving away from each other – they are separating.
Let the speed of Y be y.
Relative speed of separation = 3v / 5 + y
Relative speed of approach = Relative speed of separation
v = 3v / 5 + y
Speed y = v – (3v/5) = 2v / 5
Kinetic energy = ½ mv2
Kinetic energy of Y = ½ × 4m × (2v/5)2 = 16 mv2 / 50
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