Asked by Jane
Running at 2.0 m/s a 45.0 kg person collides with a 90.0 kg person who is traveling at 7.0 m/s in the other direction. Upon collision, the 90 kg person continues to travel forward at 1.0 m/s. How fast is the 45 kg person knocked backwards?
Answers
Answered by
Henry2
Given: m/s.
M 1 = 45 kg, V1 = 2 m/s.
M2 = 90 kg, V2 = -7 m/s.
V3 = Velocity of M1 after collision.
V4 = -1 m/s = Velocity of M2 after collision.
Momentum before = Momentum after
M1*V1 + M2*V2 = M1*V3 + M2*V4.
45*2 + 90*(-7) = 45V3 + 90*(-1),
90 - 630 = 45V3 - 90,
-450 = 45V3,
V3 = -10 m/s.
M 1 = 45 kg, V1 = 2 m/s.
M2 = 90 kg, V2 = -7 m/s.
V3 = Velocity of M1 after collision.
V4 = -1 m/s = Velocity of M2 after collision.
Momentum before = Momentum after
M1*V1 + M2*V2 = M1*V3 + M2*V4.
45*2 + 90*(-7) = 45V3 + 90*(-1),
90 - 630 = 45V3 - 90,
-450 = 45V3,
V3 = -10 m/s.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.