Two bumper cars in an amusement park ride collide elastically as one approaches the other directly fromthe rear. Car A has a mass of 435 kg and car B a mass of 495 kg, owing to differences in passenger mass. Ifcar A approaches at 4.50 m/s and car B is moving at 3.70 m/s calculatea) their velocities after the collisionb) the change in momentum of each.

a) The velocity of car B after the collision is 4.45 m/s.

The velocity of car A after the collision is 3.65 m/s.

b) The change of momentum of car A is - 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

Explanation:

Hi there!

Since the cars collide elastically, the momentum and kinetic energy of the system do not change after the collision.

The momentum of the system is calculated adding the momenta of each car:

initial momentum = final momentum

mA · vA + mB · vB = mA · vA' + mB · vB'

Where:

mA = mass of car A

vA = initial velocity of car A

mB = mass of car B

vB = initial velocity of car B

vA' = final velocity of car A

vB' = final velocity of car B

Let's replace with the data we have and solve the equation for vA':

mA · vA + mB · vB = mA · vA' + mB · vB'

435 kg · 4.50 m/s + 495 kg · 3.70 m/s = 435 kg · vA' + 495 kg · vB'

3789 kg · m/s = 435 kg · vA' + 495 kg · vB'

3789 kg · m/s - 495 kg · vB' = 435 kg · vA'

(3789 kg · m/s - 495 kg · vB') / 435 kg = vA'

Let's write this expression without units for a bit more clarity:

vA' = (3789 - 495 vB') / 435

The kinetic energy of the system is also conserved, then, the initial kinetic energy is equal to the final kinetic energy:

initial kinetic energy of the system = final kinetic energy of the system

1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · (vA') ² + 1/2 · mB · (vB') ²

Replacing with the dа ta:

initial kinetic energy = 1/2 · 435 kg · (4.50 m/s) ² + 1/2 · 495 kg · (3.70) ²

initial kinetic energy = 7792.65 kg · m²/s²

7792.65 kg · m²/s² = 1/2 · 435 kg · (vA') ² + 1/2 · 495 kg · (vB') ²

multiply by 2 both sides of the equation:

15585.3 kg · m²/s² = 435 kg · (vA') ² + 495 kg · (vB') ²

Let's replace vA' = (3789 - 495 vB') / 435

I will omit units for clarity in the calculation:

15585.3 = 435 · (vA') ² + 495 · (vB') ²

15585.3 = 435 · (3789 - 495 vB') ² / 435² + 495 (vB') ²

15585.3 = (3789² - 3751110 vB' + 245025 vB²) / 435 + 495 (vB') ²

multiply both sides of the equation by 435:

6779605.5 = 3789² - 3751110 vB' + 245025 vB² + 215325 vB'²

0 = - 6779605.5 + 3789² - 3751110 vB' + 460350 vB'²

0 = 7576915.5 - 3751110 vB' + 460350 vB'²

Solving the quadratic equation:

vB' = 4.45 m/s

vB' = 3.70 m/s (the initial velocity)

a) The velocity of car B after the collision is 4.45 m/s

The velocity of car A will be teh following:

vA' = (3789 - 495 vB') / 435

vA' = (3789 - 495 (4.45 m/s)) / 435

vA' = 3.65 m/s

The velocity of car A after the collision is 3.65 m/s

b) The change of momentum of each car is calculated as the difference between its final momentum and its initial momentum:

ΔpA = final momentum of car A - initial momentum of car A

ΔpA = mA · vA' - mA · vA

ΔpA = mA (vA' - vA)

ΔpA = 435 kg (3.648387097 m/s - 4.50 m/s) (I have used the value of vA' without rounding).

ΔpA = - 370.45 kg · m/s

The change of momentum of car A is - 370.45 kg · m/s

ΔpB = mB (vB' - vB)

ΔpB = 495 kg (4.448387097 m/s - 3.70 m/s) (I have used the value of vB' without rounding).

ΔpB = 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

I have used the values of the final velocities without rounding so we can notice that the change of momentum of both cars is equal but of opposite sign.