Skaters of masses 80 kg and 40 kg move directly towards each other and collide.
Before the collision, the heavier skater is moving to the right at a speed of 2.0 m s⁻¹ and the lighter skater is moving to the left at a speed of 1.0 m s⁻¹.
After the collision, the heavier skater moves to the right at a speed of 0.80 m s⁻¹.
What is the relative speed of separation of the two skaters?
A 0.6 m s⁻¹
B 1.4 m s⁻¹
C 2.2 m s⁻¹
D 2.6 m s⁻¹
ANSWER: A
The Physics behind the answer:
- Momentum is conserved in this situation. The total momentum before collision and the total momentum after are equal.
- This conservation gives: (80)(2.0) + (40)(-1.0) = (80)(0.80) + 40x where x is the velocity of 40-kg skater after collision.
- After doing the mathematics, x = 1.4 m s⁻¹.
- But the answer is not B as what is being asked is the relative speed of separation e.g. the speed of one skater as seen by the other.
- Relative speed is calculated by taking the difference between speeds when objects are moving in the same direction or taking the sum of the speed when moving in the opposite directions.
- We therefore should do 1.4 - 0.80.... the answer is 0.6 m s⁻¹ [A].
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