A pair of bumper cars in an amusement park ride collide in a perfectly elastic collision as one approaches the other directly from the rear. One has a mass of m1 = 439 kg and the other m2 = 546 kg, owing to differences in passenger mass. The lighter one approaches at v1 = 4.36 m/s and the other is moving at v2 = 3.62 m/s.
What is the velocity of the lighter car after collision, the heavier car after collision, and the change in momentum of both the heavier and lighter car?
How would you determine the solution?
2006-12-01 10:31:18 · 2 answers · asked by gunder53534 2 in Science & Mathematics ➔ Physics
I will set it up but will not give you the answer.
V1f= [(m1-m2)/M]x v1i+ [2(m2)/M]x v2i
V2f= [2(m1)/M]x v1i+ [(m2-m1)/M]x v2i
V1f is the final velocity of m1
V2f is the final velocity of m2
In elastic collisions Momentum is conserved
2006-12-01 10:51:24 · answer #1 · answered by cascsiany 2 · 0⤊ 0⤋
Momentum is always conserved so the momemtum before and after the collision is equal. In an elastic collision, kinetic energy is conserved also. So that the kinetic energy before and after the collision is also equal.
This will give you 2 equations with 2 unknows. Solve them.
2006-12-01 10:35:29 · answer #2 · answered by rscanner 6 · 0⤊ 0⤋