Understanding Elastic Collisions
In physics, an elastic collision is a collision in which both total momentum and total kinetic energy are conserved. For two objects colliding on a frictionless track, we use the following two fundamental laws:
- Conservation of Momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
- Conservation of Kinetic Energy: $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$
Alternatively, for elastic collisions, the relative velocity of approach equals the relative velocity of separation: $v_{1i} - v_{2i} = v_{2f} - v_{1f}$
The Problem Setup
Given:
- Glider 1: $m_1 = 0.15\text{ kg}$, $v_{1i} = +0.80\text{ m/s}$ (right is positive)
- Glider 2: $m_2 = 0.30\text{ kg}$, $v_{2i} = -2.2\text{ m/s}$ (left is negative)
Step-by-Step Solution
Step 1: Conservation of Momentum
$(0.15)(0.80) + (0.30)(-2.2) = 0.15v_{1f} + 0.30v_{2f}$ $0.12 - 0.66 = 0.15v_{1f} + 0.30v_{2f}$ $-0.54 = 0.15v_{1f} + 0.30v_{2f}$ Dividing by 0.15 simplifies this to: $-3.6 = v_{1f} + 2v_{2f} \quad \text{(Equation A)}
Step 2: Relative Velocity Equation
$v_{1i} - v_{2i} = v_{2f} - v_{1f}$ $0.80 - (-2.2) = v_{2f} - v_{1f}$ $3.0 = v_{2f} - v_{1f} \implies v_{2f} = 3.0 + v_{1f} \quad \text{(Equation B)}
Step 3: Solve the System
Substitute (B) into (A): $-3.6 = v_{1f} + 2(3.0 + v_{1f})$ $-3.6 = v_{1f} + 6.0 + 2v_{1f}$ $-9.6 = 3v_{1f}$ $v_{1f} = -3.2\text{ m/s}$
Now solve for $v_{2f}$ using (B): $v_{2f} = 3.0 + (-3.2) = -0.2\text{ m/s}$
Conclusion
- Glider 1 final velocity: $3.2\text{ m/s}$ to the left.
- Glider 2 final velocity: $0.2\text{ m/s}$ to the left.