- #1
Drakon25th
I need help with this question:
A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear. One has a mass of 450kg and the other 550kg, owing to differences in passenger mass. if the lighter one approaches at 4.5m/s and the other is moving at 3.7m/s, calculate a)their velocities after the collision and b) the change in momentum of each
ok this is what i know
m1 = 450 kg, v1 = 4.5 m/s
m2 = 550 kg, v2 = 3.7 m/s
i know I'm suppose to use these formulas:
m1v1+m2v2 = m1v1' + m2v2'
and
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1^2' + (1/2)m2v2^2'
my teacher gave us this homework without really explaining anything and the example my book gives to me is very confusing
so i know m1v1+m2v2 = total momentum in this system, which is:
450kg(4.5m/s)+550kg(3.7m/s)=4060 Ns
and the total KE after the collision is:
.5(450kg)(4.5m/s)^2+.5(550kg)(3.7m/s)^2 = 8321J
then my book becomes confusing and i get lost
what do i do?
A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear. One has a mass of 450kg and the other 550kg, owing to differences in passenger mass. if the lighter one approaches at 4.5m/s and the other is moving at 3.7m/s, calculate a)their velocities after the collision and b) the change in momentum of each
ok this is what i know
m1 = 450 kg, v1 = 4.5 m/s
m2 = 550 kg, v2 = 3.7 m/s
i know I'm suppose to use these formulas:
m1v1+m2v2 = m1v1' + m2v2'
and
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1^2' + (1/2)m2v2^2'
my teacher gave us this homework without really explaining anything and the example my book gives to me is very confusing
so i know m1v1+m2v2 = total momentum in this system, which is:
450kg(4.5m/s)+550kg(3.7m/s)=4060 Ns
and the total KE after the collision is:
.5(450kg)(4.5m/s)^2+.5(550kg)(3.7m/s)^2 = 8321J
then my book becomes confusing and i get lost
what do i do?