I guess we assume they stick together?
initial momentum = 350*2 + 0 = 700
final momentum = (350+250) v
final = initial
so
600 v = 700
v = 7/6 m/s = 1.17 m/s
If that is wrong, then perhaps they mean the collision is elastic and kinetic energy before = kinetic energy after but the velocities are different
in that case
initial momentum still 700
final momentum = 350 V1 + 250 V2
so
350 V1 + 250 V2 = 700 or V2 =2.8-1.4 V1
and now energy
.5*350 V1^2 + .5*250 V2^2 = .5*350 *4
or
v1^2 + .714 V2^2 = 4
V1^2 +.714 (2.8-1.4V1)^2 = 4
V1^2 +.714(7.84 -7.84 V1+1.96V1^2) = 4
V1^2 + 5.6 -5.6V1+1.4V1^2 = 4
2.4 V1^2 -5.6 V1 + 1.6 = 0
solve quadratic
V1 = 2 or V1 =.333333 (speed unchanged second car stationary)
2 is impossible so V1 = .3333 (slowed down)
2.8 - 1.4(.33333) = 2.33 = V2 of empty car
Your bumper car is travelling at 2 m/s. The total mass of you and your car is 350 kg. An empty motionless bumper car with a mass of 250 kg is in your path. How fast in m/s will the empty bumper car move after you collide with it head on?
I tried using (m1v1 + m2v2)/m1 + m2 = vf, to get 1.17 m/s, which was wrong. What should I be doing?
Thank you!
1 answer
1 answer