To solve this problem, we use the principles of conservation of momentum and conservation of kinetic energy, since the collision is elastic.

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Given data:

• Mass of cart 1 ($m_1$): $0.150 \, \text{kg}$
• Initial velocity of cart 1 ($v_{1i}$): $+0.80 \, \text{m/s}$ (positive direction is to the right)
• Mass of cart 2 ($m_2$): $0.300 \, \text{kg}$
• Initial velocity of cart 2 ($v_{2i}$): $-2.20 \, \text{m/s}$ (negative direction is to the left)

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Equations:

1. Conservation of Momentum:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$ where $v_{1f}$ and $v_{2f}$ are the final velocities of cart 1 and cart 2, respectively.

2. Conservation of Kinetic Energy:

$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$

3. Relative Velocity Condition (for Elastic Collisions):

$v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$ This simplifies calculations significantly.

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Step 1: Apply the relative velocity condition

$v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$ Substitute the initial velocities: $0.80 - (-2.20) = -(v_{1f} - v_{2f})$ $3.00 = -(v_{1f} - v_{2f})$ v_{1f} - v_{2f} = -3.00 \tag{1}

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Step 2: Apply conservation of momentum

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$ Substitute the given values: $(0.150)(0.80) + (0.300)(-2.20) = (0.150)v_{1f} + (0.300)v_{2f}$ $0.120 - 0.660 = 0.150v_{1f} + 0.300v_{2f}$ -0.540 = 0.150v_{1f} + 0.300v_{2f} \tag{2}

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Step 3: Solve the system of equations

We now solve the system of equations:

1. $v_{1f} - v_{2f} = -3.00$
2. $0.150v_{1f} + 0.300v_{2f} = -0.540$

From equation (1):

v_{1f} = v_{2f} - 3.00 \tag{3}

Substitute $v_{1f}$ from equation (3) into equation (2):

$0.150(v_{2f} - 3.00) + 0.300v_{2f} = -0.540$ $0.150v_{2f} - 0.450 + 0.300v_{2f} = -0.540$ $0.450v_{2f} - 0.450 = -0.540$ $0.450v_{2f} = -0.090$ $v_{2f} = -0.200 \, \text{m/s}$

Solve for $v_{1f}$ using equation (3):

$v_{1f} = v_{2f} - 3.00$ $v_{1f} = -0.200 - 3.00$ $v_{1f} = -3.20 \, \text{m/s}$

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Final Velocities:

• Final velocity of cart 1: $v_{1f} = -3.20 \, \text{m/s}$ (to the left)
• Final velocity of cart 2: $v_{2f} = -0.200 \, \text{m/s}$ (to the left)

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Verification:

Conservation of Momentum:

$0.120 - 0.660 = 0.150(-3.20) + 0.300(-0.200)$ $-0.540 = -0.480 - 0.060$ $-0.540 = -0.540 \quad \text{(verified!)}$

Conservation of Kinetic Energy:

$\frac{1}{2}(0.150)(0.80)^2 + \frac{1}{2}(0.300)(2.20)^2 = \frac{1}{2}(0.150)(-3.20)^2 + \frac{1}{2}(0.300)(-0.200)^2$ $0.048 + 0.726 = 0.768 + 0.006$ $0.774 = 0.774 \quad \text{(verified!)}$

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Key Results:

1. Final velocity of cart 1: $v_{1f} = -3.20 \, \text{m/s}$ (to the left)
2. Final velocity of cart 2: $v_{2f} = -0.200 \, \text{m/s}$ (to the left)

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Would you like a deeper explanation of any part? 😊

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Relative Questions:

1. How does the mass ratio affect the velocities after an elastic collision?
2. What happens if the collision were perfectly inelastic instead?
3. How does the relative velocity condition simplify elastic collision problems?
4. Can we solve this using only energy conservation? Why or why not?
5. How do the results change if the initial velocity of one cart were zero?

Tip: Always verify both momentum and kinetic energy conservation when solving elastic collision problems to avoid calculation errors!