Question

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg, roll toward each other and collide. The velocity is measured before and after each collision. The collected data is shown below.

A 5 column table with 3 rows. The first column is unlabeled with entries Trial 1, Trial 2, Trial 3, Trial 4. The second column is labeled Initial Velocity Ball A (meters per second) with entries positive 1, positive 0.5, positive 2, positive 0.5. The third column is labeled Initial Velocity Ball B (meters per second) with entries negative 2, negative 1.5, positive 1, negative 1. The third column is labeled Final Velocity Ball A (meters per second) with entries negative 2, negative 0.5, positive 1, positive 1.5. The fourth column is labeled Final Velocity Ball B (meters per second) with entries negative 1, negative 0.5, negative 2, negative 1.5.

Which trial shows the conservation of momentum in a closed system?

Trial 1
Trial 2
Trial 3
Trial 4

Answer 1

To determine which trial shows the conservation of momentum, we can use the principle of conservation of momentum which states that the total momentum before the collision must equal the total momentum after the collision.

The momentum \( p \) for a ball is calculated as:
\[ p = m \cdot v \]
where \( m \) is the mass and \( v \) is the velocity.

Given that both balls have a mass of 1.0 kg, the total momentum before and after the collision can be calculated for each trial.

Let's calculate the total momentum for each trial:

### Trial 1:
- **Initial momentum:**
- Ball A: \( p_A = 1.0 \cdot 1.0 = 1.0 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-2.0) = -2.0 \, \text{kg m/s} \)
- Total Initial Momentum: \( p_{initial} = 1.0 + (-2.0) = -1.0 \, \text{kg m/s} \)

- **Final momentum:**
- Ball A: \( p_A = 1.0 \cdot (-2.0) = -2.0 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-1.0) = -1.0 \, \text{kg m/s} \)
- Total Final Momentum: \( p_{final} = -2.0 + (-1.0) = -3.0 \, \text{kg m/s} \)

- **Momentum conservation:**
- \( p_{initial} \neq p_{final} \)

### Trial 2:
- **Initial momentum:**
- Ball A: \( p_A = 1.0 \cdot 0.5 = 0.5 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-1.5) = -1.5 \, \text{kg m/s} \)
- Total Initial Momentum: \( p_{initial} = 0.5 + (-1.5) = -1.0 \, \text{kg m/s} \)

- **Final momentum:**
- Ball A: \( p_A = 1.0 \cdot (-0.5) = -0.5 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-0.5) = -0.5 \, \text{kg m/s} \)
- Total Final Momentum: \( p_{final} = -0.5 + (-0.5) = -1.0 \, \text{kg m/s} \)

- **Momentum conservation:**
- \( p_{initial} = p_{final} \)

### Trial 3:
- **Initial momentum:**
- Ball A: \( p_A = 1.0 \cdot 2.0 = 2.0 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot 1.0 = 1.0 \, \text{kg m/s} \)
- Total Initial Momentum: \( p_{initial} = 2.0 + 1.0 = 3.0 \, \text{kg m/s} \)

- **Final momentum:**
- Ball A: \( p_A = 1.0 \cdot 1.0 = 1.0 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-2.0) = -2.0 \, \text{kg m/s} \)
- Total Final Momentum: \( p_{final} = 1.0 + (-2.0) = -1.0 \, \text{kg m/s} \)

- **Momentum conservation:**
- \( p_{initial} \neq p_{final} \)

### Trial 4:
- **Initial momentum:**
- Ball A: \( p_A = 1.0 \cdot 0.5 = 0.5 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-1.0) = -1.0 \, \text{kg m/s} \)
- Total Initial Momentum: \( p_{initial} = 0.5 + (-1.0) = -0.5 \, \text{kg m/s} \)

- **Final momentum:**
- Ball A: \( p_A = 1.0 \cdot 1.5 = 1.5 \, \text{kg m/s} \)
- Ball B: \( p_B = 1.0 \cdot (-1.5) = -1.5 \, \text{kg m/s} \)
- Total Final Momentum: \( p_{final} = 1.5 + (-1.5) = 0 \, \text{kg m/s} \)

- **Momentum conservation:**
- \( p_{initial} \neq p_{final} \)

### Conclusion:
Only Trial 2 shows conservation of momentum since \( p_{initial} = p_{final} \).

**Therefore, the answer is Trial 2.**