Question

elastic collision physics problem solution

Answer 1

Certainly! An elastic collision is a collision in which both momentum and kinetic energy are conserved. To solve an elastic collision problem, we generally need the following equations:

1. **Conservation of Momentum:**
\[
m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}
\]
where:
- \(m_1\) and \(m_2\) are the masses of the colliding objects,
- \(v_{1i}\) and \(v_{2i}\) are the initial velocities of the objects,
- \(v_{1f}\) and \(v_{2f}\) are the final velocities of the objects after the collision.

2. **Conservation of Kinetic Energy:**
\[
\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2
\]

### Example Problem:

**Problem:** A 2 kg cart moving at 3 m/s collides elastically with a 1 kg cart at rest. What are their velocities after the collision?

### Solution Steps:

1. **Identify knowns and assign variables:**
- \(m_1 = 2 \, \text{kg}\), \(v_{1i} = 3 \, \text{m/s}\) (initial velocity of cart 1)
- \(m_2 = 1 \, \text{kg}\), \(v_{2i} = 0 \, \text{m/s}\) (initial velocity of cart 2)
- Let \(v_{1f}\) and \(v_{2f}\) be the final velocities of cart 1 and cart 2, respectively.

2. **Set up momentum conservation equation:**
\[
2 \cdot 3 + 1 \cdot 0 = 2 v_{1f} + 1 v_{2f}
\]
Simplifying gives:
\[
6 = 2 v_{1f} + v_{2f} \quad \text{(Equation 1)}
\]

3. **Set up kinetic energy conservation equation:**
\[
\frac{1}{2} \cdot 2 \cdot 3^2 + \frac{1}{2} \cdot 1 \cdot 0^2 = \frac{1}{2} \cdot 2 \cdot v_{1f}^2 + \frac{1}{2} \cdot 1 \cdot v_{2f}^2
\]
This becomes:
\[
9 = v_{1f}^2 + \frac{1}{2} v_{2f}^2 \quad \text{(Equation 2)}
\]

4. **Solve the equations:**

From Equation 1, we can express \(v_{2f}\) in terms of \(v_{1f}\):
\[
v_{2f} = 6 - 2 v_{1f}
\]

Substituting \(v_{2f}\) into Equation 2:
\[
9 = v_{1f}^2 + \frac{1}{2}(6 - 2v_{1f})^2
\]
Expanding the square:
\[
(6 - 2v_{1f})^2 = 36 - 24v_{1f} + 4v_{1f}^2
\]
So we get:
\[
9 = v_{1f}^2 + \frac{1}{2}(36 - 24v_{1f} + 4v_{1f}^2)
\]
Multiplying through by 2 to eliminate the fraction:
\[
18 = 2v_{1f}^2 + 36 - 24v_{1f} + 4v_{1f}^2
\]
Combine like terms:
\[
0 = 6v_{1f}^2 - 24v_{1f} + 18
\]
Dividing everything by 6 gives:
\[
0 = v_{1f}^2 - 4v_{1f} + 3
\]
Factoring this quadratic:
\[
0 = (v_{1f} - 3)(v_{1f} - 1)
\]
So, \(v_{1f} = 3 \text{ or } 1\). Since cart 1 must slow down, we take \(v_{1f} = 1 \text{ m/s}\).

5. **Determine \(v_{2f}\):**
Using \(v_{1f} = 1\) in Equation 1:
\[
6 = 2(1) + v_{2f} \implies v_{2f} = 4 \, \text{m/s}
\]

### Final Result:
- Final velocity of cart 1, \(v_{1f} = 1 \, \text{m/s}\)
- Final velocity of cart 2, \(v_{2f} = 4 \, \text{m/s}\)

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