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Relative Velocity in Elastic Collisions
In one-dimensional elastic collisions, a useful shortcut is the relative velocity reversal equation:
\[
v_{1i} - v_{2i} = - (v_{1f} - v_{2f})
\]
This means the relative velocity between two objects before the collision is equal in magnitude and opposite in direction to the relative velocity after the collision. This equation greatly simplifies the math when paired with conservation of momentum.
When to Use This Shortcut
- The collision must be elastic.
- The motion must be one-dimensional.
- Use this along with momentum conservation.
Common Mistakes
- Using it for inelastic collisions: This only works when kinetic energy is conserved.
- Getting the sign wrong: Make sure directions are consistent. If left is negative, stay consistent throughout.
- Forgetting to apply momentum conservation too: This equation alone is not enough to solve for two unknowns.
Worked Example 1
Problem: A 2 kg cart moving at 4 m/s collides elastically with a 1 kg cart at rest. Find the final velocities.
Step 1: Use conservation of momentum
\[
2(4) + 1(0) = 2v_{1f} + 1v_{2f} \Rightarrow 8 = 2v_{1f} + v_{2f}
\]
Step 2: Use relative velocity equation
\[
4 - 0 = - (v_{1f} - v_{2f}) \Rightarrow v_{2f} - v_{1f} = 4
\]
Step 3: Solve the system
Substitute \( v_{2f} = v_{1f} + 4 \) into the momentum equation:
\[
8 = 2v_{1f} + (v_{1f} + 4)
\]
Now solve for \( v_{1f} \):
\[
8 = 3v_{1f} + 4 \Rightarrow v_{1f} = \frac{4}{3}
\]
Then find \( v_{2f} \):
\[
v_{2f} = v_{1f} + 4 = \frac{4}{3} + 4 = \frac{16}{3}
\]
Answer:
- Cart 1: \( v_{1f} = \frac{4}{3} \, \text{m/s} \)
- Cart 2: \( v_{2f} = \frac{16}{3} \, \text{m/s} \)
Worked Example 2
Problem: A 3 kg ball moving at 6 m/s to the right collides elastically with a 2 kg ball moving at 2 m/s to the left. Find their velocities after the collision.
Step 1: Set positive direction
Take right as positive:
- \( v_{1i} = +6 \, \text{m/s} \)
- \( v_{2i} = -2 \, \text{m/s} \)
Step 2: Conservation of momentum
\[
3(6) + 2(-2) = 3v_{1f} + 2v_{2f}
\]
\[
18 - 4 = 3v_{1f} + 2v_{2f}
\]
\[
14 = 3v_{1f} + 2v_{2f} \quad \text{(Equation 1)}
\]
Step 3: Relative velocity equation
\[
v_{1i} - v_{2i} = - (v_{1f} - v_{2f})
\]
\[
6 - (-2) = - (v_{1f} - v_{2f}) \Rightarrow 8 = -(v_{1f} - v_{2f})
\]
\[
v_{2f} - v_{1f} = 8 \quad \text{(Equation 2)}
\]
Step 4: Solve the system
Substitute Equation 2 into Equation 1:
\[
14 = 3v_{1f} + 2(v_{1f} + 8)
\]
\[
14 = 3v_{1f} + 2v_{1f} + 16
\]
\[
14 = 5v_{1f} + 16
\]
\[
v_{1f} = \frac{-2}{5} = -0.4 \, \text{m/s}
\]
Find \( v_{2f} \):
\[
v_{2f} = v_{1f} + 8 = -0.4 + 8 = 7.6 \, \text{m/s}
\]
Answer:
- 3 kg ball: \( -0.4 \, \text{m/s} \) (to the left)
- 2 kg ball: \( 7.6 \, \text{m/s} \) (to the right)
Why It Works
In elastic collisions, both momentum and kinetic energy are conserved. The relative velocity equation is derived from these conservation laws and provides a shortcut to find final velocities quickly.
Summary
- Use the relative velocity shortcut to save time in elastic collision problems.
- Always pair it with momentum conservation.
- Be mindful of signs and directions.