Question

CARA CARB 5.0 m/s 3.2 m/s

The bumper cars at an amusement park collide as one approaches the other directly from the rear. Car A has a mass of 426 kg and a velocity of 5.0 m/s. Car B has a mass of 470 kg and a velocity of 3.2 m/s. The coefficient of restitution is 0.72.

A) Determine the velocity of car A immediately after the collision.
(include units with answer)

B) Determine the velocity of car B immediately after the collision.
(include units with answer)

C) Determine the change in momentum of car A.
(include units with answer)

D) Determine the change in momentum of car B.
(include units with answer)

Answer 1

Let the speed of Car A after collision be Va

and that of Car B be Vb

A) By the conservation of momentum we have

m1u1 + m2u2 = m1v1 + m2v2

426*5 + 470*3.2 = 426Va + 470Vb

426Va + 470Vb = 3634

213Va + 235Vb = 1817 ---------(1)

Velocity of approach = Velocity of recess

Vb - Va = 5 - 3.2

Vb - Va = 1.8   

213Vb - 213Va = 383.4 -------(2)

Adding (1) and (2) we have

448Vb = 2200.4

Vb = 4.91 m/s

Va = Vb - 1.8

= 4.91- 1.8 = 3.11

A) 3.11 m/s

B) 4.91 m/s

C) change in momentum of car A= ma*(Va - Ua) = 426*(5 - 3.11) = 804.45 kg m/s

D)change in momentum of Car B = mb*(Vb - Ub) = 470*(3.2 - 4.91) = -803.7 kg m/s