A pair of bumper cars in an amusement park ride collide elastically as one appro

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 m1 = 460kg and the other m2 = 560kg, owing to differences in passenger mass. If the lighter one approaches at v1 = 4.22m/s and the other is moving at v2 = 3.50m/s,

A. calculate the velocity of the lighter car after the collision.

B. Calculate the velocity of the heavier car after the collision.

C. Calculate the change in momentum of the lighter car.

D. Calculate the change in momentum of the heavier car.

Explanation / Answer

according to conservation of linear momentum

m1*u1 + m2*u2 = m1*v1+m2*v2

m1*(u1-v1) = m2*(v2-u2)........(1)

according to conservation of energy

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

m1*(u1^2 - v1^2) = m2*(v2^2-u2^2).....(2)

from 1 &2

u1 + v1 = u2+v2

u1 - u2 = v2 - v1

v2 = u1 - u2 + v1........(3)

3 in 1

m1*(u1-v1) = m2*(u1 - u2 + v1 - u2)

v1 = u1*(m1-m2)/(m1+m2) + 2*m2*u2/(m1+m2)

v2 = u2*(m2-m1)/(m1+m2) + 2*m1*u1/(m1+m2)

m1 = 460 kg.....m2 = 560 kg

u1 = 4.22...........u2 = 3.5

A ) v1 = ((460*(4.22-3.5)) + (2*560*3.5))/(460+560) = 4.17 m/s

B) v2 = ((560*(3.5-4.22)) + (2*460*4.22))/(460+560) = 3.41 m/s

C) dP1 = m1*(v1-u1) = 460*(4.17-4.22) = -23 kg m /s

D) dP2 = m2*(v2-u2) = 560*(3.41-3.5) = -50.4 kg m/s

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