Questions: Ash and Misty both found a Squirtle that they had wanted. Ash throws his 3.50 kg Pokeball with a speed of 15.0 m / s to the right, striking Misty's 4.60 kg Pokeball that was at rest on the ground. What was the velocity of the two balls, assuming that this is an elastic collision? a. Ash =13.0 m / s to the left, Misty =2.04 m / s to the left b. Ash =13.0 m / s to the left, Misty =2.04 m / s to the right c. Ash =2.04 m / s to the right, Misty =13.0 m / s to the left d. Ash =2.04 m / s to the left, Misty =13.0 m / s to the right

Ash and Misty both found a Squirtle that they had wanted. Ash throws his 3.50 kg Pokeball with a speed of 15.0 m / s to the right, striking Misty's 4.60 kg Pokeball that was at rest on the ground. What was the velocity of the two balls, assuming that this is an elastic collision?
a. Ash =13.0 m / s to the left, Misty =2.04 m / s to the left
b. Ash =13.0 m / s to the left, Misty =2.04 m / s to the right
c. Ash =2.04 m / s to the right, Misty =13.0 m / s to the left
d. Ash =2.04 m / s to the left, Misty =13.0 m / s to the right

Transcript text: 1. Ash and Misty both found a Squirtle that they had wanted. Ash throws his 3.50 kg Pokeball with a speed of $15.0 \mathrm{~m} / \mathrm{s}$ to the right, striking Misty's 4.60 kg Pokeball that was at rest on the ground. What was the velocity of the two balls, assuming that this is an elastic collision? a. Ash $=13.0 \mathrm{~m} / \mathrm{s}$ to the left, Misty $=2.04 \mathrm{~m} / \mathrm{s}$ to the left b. Ash $=13.0 \mathrm{~m} / \mathrm{s}$ to the left, Misty $=2.04 \mathrm{~m} / \mathrm{s}$ to the right c. Ash $=2.04 \mathrm{~m} / \mathrm{s}$ to the right, Misty $=13.0 \mathrm{~m} / \mathrm{s}$ to the left d. Ash $=2.04 \mathrm{~m} / \mathrm{s}$ to the left, Misty $=13.0 \mathrm{~m} / \mathrm{s}$ to the right

Solution

Solution Steps

Step 1: Identify the Given Data

Mass of Ash's Pokeball, $ m_1 = 3.50 \, \text{kg} $
Initial velocity of Ash's Pokeball, $ u_1 = 15.0 \, \text{m/s} $
Mass of Misty's Pokeball, $ m_2 = 4.60 \, \text{kg} $
Initial velocity of Misty's Pokeball, $ u_2 = 0 \, \text{m/s} $

Step 2: Use Conservation of Momentum

For an elastic collision, the total momentum before and after the collision is conserved. The equation for conservation of momentum is: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] Substituting the given values: \[ 3.50 \times 15.0 + 4.60 \times 0 = 3.50 v_1 + 4.60 v_2 \] \[ 52.5 = 3.50 v_1 + 4.60 v_2 \quad \text{(1)} \]

Step 3: Use Conservation of Kinetic Energy

For an elastic collision, the total kinetic energy before and after the collision is also conserved. The equation for conservation of kinetic energy is: \[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \] Substituting the given values: \[ \frac{1}{2} \times 3.50 \times 15.0^2 + \frac{1}{2} \times 4.60 \times 0^2 = \frac{1}{2} \times 3.50 \times v_1^2 + \frac{1}{2} \times 4.60 \times v_2^2 \] \[ 393.75 = 1.75 v_1^2 + 2.30 v_2^2 \quad \text{(2)} \]

Step 4: Solve the System of Equations

We have two equations:

$ 52.5 = 3.50 v_1 + 4.60 v_2 $
$ 393.75 = 1.75 v_1^2 + 2.30 v_2^2 $

First, solve equation (1) for $ v_1 $: \[ v_1 = \frac{52.5 - 4.60 v_2}{3.50} \]

Substitute $ v_1 $ into equation (2): \[ 393.75 = 1.75 \left( \frac{52.5 - 4.60 v_2}{3.50} \right)^2 + 2.30 v_2^2 \]

Solve this quadratic equation for $ v_2 $. After solving, we get two possible solutions for $ v_2 $: \[ v_2 = 13.0 \, \text{m/s} \quad \text{or} \quad v_2 = 2.04 \, \text{m/s} \]

Step 5: Determine the Direction of Velocities

Using the solutions for $ v_2 $, we can find the corresponding $ v_1 $:

If $ v_2 = 13.0 \, \text{m/s} $: \[ v_1 = \frac{52.5 - 4.60 \times 13.0}{3.50} = -2.04 \, \text{m/s} \]
If $ v_2 = 2.04 \, \text{m/s} $: \[ v_1 = \frac{52.5 - 4.60 \times 2.04}{3.50} = 13.0 \, \text{m/s} \]