Question

A 120-g rubber ball is thrown with a speed of 10 m/s at an oncoming 900-kg car which is approaching at 30 m/s, and undergoes a one-dimensional elastic collision with the car. What is the speed of the ball after the collision?

A) 30 m/s B) 70 m/s C) 40 m/s D)50 m/s E) 10 m/s

Answer 1

The velocity of the ball after the collision is 50 m/s

Step-by-step explanation:

Hi there!

Since the collision is elastic, the kinetic energy and momentum of the system ball-car is the same before and after the collision. The momentum of the system is the sum of the momentums of each object:

initial momentum = final momentum

mb · vb + mc · vc = mb · vb´ + mc · vc´

Where:

mb = mass of the ball (0.120 kg)

vb = velocity of the ball before the collision (10 m/s).

mc = mass of the car (900 kg).

vc = velocity of the car before the collision (30 m/s).

vb´ = velocity of the ball after the collision.

vc´ = velocity of the car after the collision.

Let´s calculate the initial momentum of the system:

initial momentum = 0.120 kg · 10 m/s + 900 kg · 30 m/s

initial momentum = 27001.2 kg · m/s

Since intial momentum = final momentum

27001.2 kg · m/s = 0.120 kg · vb´ + 900 kg · vc´

The kinetic energy of the system is also conserved:

initial kinetic energy = final kinetic energy

1/2 mb · vb² + 1/2 mc · vc² = 1/2 mb · vb´² + 1/2 mc · vc´²

Let´s calculate the initial kinetic energy:

initial kinetic energy = 1/2 · 0.120 kg · (10 m/s)² + 1/2 · 900 kg · (30 m/s)²

initial kinetic energy = 405006 kg · m²/s². Then:

initial kinetic energy = final kinetic energy

405006 kg · m²/s² = 1/2 · 0.120 kg · vb´² + 1/2 · 900 kg · vc´²

405006 kg · m²/s² = 0.06 kg · vb´² + 450 kg · vc´²

So, we have a system of two equations with two unknowns:

405006 kg · m²/s² = 0.06 kg · vb´² + 450 kg · vc´²

27001.2 kg · m/s = 0.120 kg · vb´ + 900 kg · vc´

Let´s solve the second equation for vc´ (I will omit units for clarity):

27001.2 = 0.120 vb´ + 900 vc´

(27001.2 - 0.120 vb´)/ 900 = vc´

Now let´s replace vc´ in the first equation and solve it for vb´.

405006 = 0.06 vb´² + 450 vc´²

405006 = 0.06 vb´² + 450 [(27001.2 - 0.120 vb´)/ 900]²

405006 = 0.06 vb´² + 450(27001.2² - 6480.288 vb´ + 0.0144 vb´²)/900²

405006 = 0.06 vb´² + 27001.2²/1800 - 3.60016 vb´ + 8 × 10⁻⁶ vb´²

0 = 0.060008 vb´² + 27001.2²/1800 - 405006 - 3.60016 vb´

0 = 0.060008 vb´² - 3.60016 vb´ + 30.0008

Solving the quadratic equation with the quadratic formula:

vb´ = 50 m/s

and

vb´ = 10 m/s

Since 10 m/s is the velocity of the ball before the collision, the solution is vb´ = 50 m/s