Question

A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear, as seen in part (a) of the figure below. ((a) before collision, (b) after collision) One has a mass of m1 = 462 kg and the other m2 = 546 kg, owing to differences in passenger mass. If the lighter one approaches at v1 = 4.48 m/s and the other is moving at v2 = 3.63 m/s, calculate the velocity of the lighter car after the collision.

Calculate the velocity of the heavier car after the collision.

Calculate the change in momentum of the lighter car.

Calculate the change in momentum of the heavier car.

Answer 1

(a) As we now for an elastic, head-on collision,

Relative velocity of approach = Relative velocity of separation,

=> 4.48 m/s - 3.63 m/s = v - u [Where u and v are the post-collision velocities of the lighter and heavier cars, respectively]

=> v - u = 0.85 m/s

=> v = 0.85 m/s + u -------------------------------------------------(i)

Now, conserve momentum:

462 kg * 4.48 m/s + 546 kg * 3.63 m/s = 462 kg * u + 546 kg * v

plugging in for v in the above equation from equation (i) -

2069.76 + 1981.98 = 462 * u + 546 * (0.85 + u)

=> 4051.74 = 464.1 + 1008 * u

=> 1008 * u = 4051.74 - 464.1 = 3587.64

=> u = 3587.64 / 1008 = 3.56 m/s

Hence, velocity of the lighter car after the collision, u = 3.56 m/s (Answer)

(b) v = u + 0.85 m/s

= 3.56 + 0.85 = 4.41 m/s

Hence, velocity of heavier car after the collision, v = 4.41 m/s.

(c) Change in momentum of the ligher car,

Δp = mΔv = 462 kg * (3.56 - 4.48)m/s = -425.04 kg·m/s (Answer)

(d) Change in momentum of the heavier car,

Δp = 546 kg * (4.41 - 3.63) m/s = 425.88 kg. m/s (Answer)