Question

Assignment 2 Momentum and Energy
Your work must be clear to read. Clearly label all cases (a, b). Show all steps in GRASS format.
GRASS means given/required/analysis/solution/statement.

Problem:
A 4.0 kg ball moving to the right at 5 m/s collides head-on with a 2.0 kg ball moving to the left at 4.0 m/s. If the collision is elastic, determine the velocity of each ball after the collision.

Answer 1

To solve this problem using the GRASS format, we will break it down into clear sections.

### Given:
1. Mass of ball 1 (m1) = 4.0 kg
2. Initial velocity of ball 1 (u1) = 5 m/s (to the right, take this as positive)
3. Mass of ball 2 (m2) = 2.0 kg
4. Initial velocity of ball 2 (u2) = -4 m/s (to the left, take this as negative since it is in the opposite direction)

### Required:
1. Final velocity of ball 1 (v1) after the collision
2. Final velocity of ball 2 (v2) after the collision

### Analysis:
In an elastic collision, both momentum and kinetic energy are conserved. We can use these principles to set up our equations.

1. **Conservation of Momentum**:
\[
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
\]

Substituting the known values:
\[
(4.0 \, \text{kg})(5.0 \, \text{m/s}) + (2.0 \, \text{kg})(-4.0 \, \text{m/s}) = (4.0 \, \text{kg}) v_1 + (2.0 \, \text{kg}) v_2
\]
\[
20.0 \, \text{kg·m/s} - 8.0 \, \text{kg·m/s} = 4.0 v_1 + 2.0 v_2
\]
\[
12.0 \, \text{kg·m/s} = 4.0 v_1 + 2.0 v_2 \, \, \, \, \text{(1)}
\]

2. **Conservation of Kinetic Energy**:
\[
\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2
\]

Substituting the known values:
\[
\frac{1}{2} (4.0)(5.0^2) + \frac{1}{2} (2.0)(-4.0^2) = \frac{1}{2} (4.0) v_1^2 + \frac{1}{2} (2.0) v_2^2
\]
\[
\frac{1}{2} (4.0)(25) + \frac{1}{2} (2.0)(16) = 2.0 v_1^2 + 1.0 v_2^2
\]
\[
50.0 + 16.0 = 2.0 v_1^2 + 1.0 v_2^2
\]
\[
66.0 = 2.0 v_1^2 + 1.0 v_2^2 \, \, \, \, \text{(2)}
\]

Now we have two equations (1) and (2) with two unknowns (v1 and v2).

### Solution:
We can solve equation (1) for one variable and substitute it into equation (2). Let’s solve equation (1) for \(v_2\):

From equation (1):
\[
2.0 v_2 = 12.0 - 4.0 v_1
\]
\[
v_2 = 6.0 - 2.0 v_1 \, \, \, \, \text{(3)}
\]

Now substitute equation (3) into equation (2):

\[
66.0 = 2.0 v_1^2 + 1.0 (6.0 - 2.0 v_1)^2
\]
Expanding \( (6.0 - 2.0 v_1)^2 \):
\[
(6.0 - 2.0 v_1)(6.0 - 2.0 v_1) = 36.0 - 24.0 v_1 + 4.0 v_1^2
\]
Now substituting this into the equation:
\[
66.0 = 2.0 v_1^2 + (36.0 - 24.0 v_1 + 4.0 v_1^2)
\]
Combining terms:
\[
66.0 = 6.0 v_1^2 - 24.0 v_1 + 36.0
\]
Rearranging gives us:
\[
6.0 v_1^2 - 24.0 v_1 - 30.0 = 0
\]
Dividing the entire equation by 6:
\[
v_1^2 - 4.0 v_1 - 5.0 = 0
\]

Now we can use the quadratic formula:
\[
v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 1\), \(b = -4\), and \(c = -5\):
\[
v_1 = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}
\]
\[
v_1 = \frac{4 \pm \sqrt{16 + 20}}{2}
\]
\[
v_1 = \frac{4 \pm \sqrt{36}}{2}
\]
\[
v_1 = \frac{4 \pm 6}{2}
\]
This gives us two solutions:
1. \(v_1 = \frac{10}{2} = 5 \, \text{m/s}\)
2. \(v_1 = \frac{-2}{2} = -1 \, \text{m/s}\)

Since the balls collide, we use the second result:
\[
v_1 = -1 \, \text{m/s}
\]

Now substitute \(v_1\) back into equation (3) to find \(v_2\):
\[
v_2 = 6.0 - 2.0(-1) = 6.0 + 2.0 = 8.0 \, \text{m/s}
\]

### Statement:
After the elastic collision:
- The velocity of the 4.0 kg ball (ball 1) is \(-1.0 \, \text{m/s}\) (to the left).
- The velocity of the 2.0 kg ball (ball 2) is \(8.0 \, \text{m/s}\) (to the right).