Chapter 9: Linear Momentum & Collisions – Example Problems Study Notes

Chapter 9: Linear Momentum & Collisions

Example Problem 1: Air Bag Safety

This problem explores the physics of car collisions and the role of air bags in reducing injury by increasing the time over which a collision occurs, thereby reducing the force experienced by the driver.

• Key Concept: Impulse is the product of force and the time interval over which it acts. It is also equal to the change in momentum of an object.

• Impulse-Momentum Theorem: $F \Delta t = m \Delta v$

• Application: In a car crash, the change in velocity ($\Delta v$) is the same whether or not an air bag is present, but the time interval ($\Delta t$) over which the force acts is increased by the air bag.

• Force Ratio: The ratio of forces with and without the air bag is inversely proportional to the ratio of the time intervals: $\frac{F_{\text{no air bag}}}{F_{\text{air bag}}} = \frac{\Delta t_{\text{air bag}}}{\Delta t_{\text{no air bag}}}$

• Example: If a driver comes to rest in 0.02 s without an air bag and in 0.20 s with an air bag, the force experienced with the air bag is much less: $\frac{F_{\text{no air bag}}}{F_{\text{air bag}}} = \frac{0.20}{0.02} = 10$ This means the air bag reduces the force by a factor of 10.

Example Problem 2: Elastic Collision on a Frictionless Track

This problem involves two blocks on a frictionless track, where one block is released from a height and collides elastically with another block at rest. The goal is to find the velocities of both blocks after the collision.

• Key Concepts:

  • Conservation of Momentum: The total momentum before and after the collision is conserved. $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

  • Conservation of Kinetic Energy (Elastic Collision): The total kinetic energy is also conserved. $\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$

  • Solving for Final Velocities: For two objects in a one-dimensional elastic collision: $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} + \frac{2m_2}{m_1 + m_2} v_{2i}$ $v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i} + \frac{m_2 - m_1}{m_1 + m_2} v_{2i}$

  • Example: If $m_1 = 1.50\ \text{kg}$, $m_2 = 2.10\ \text{kg}$, and $h = 0.61\ \text{m}$:

    • First, find $v_{1i}$ using energy conservation: $mgh = \frac{1}{2} m v^2 \implies v = \sqrt{2gh}$

    • Then, use the above formulas to find $v_{1f}$ and $v_{2f}$.

Summary Table: Conservation Laws in Collisions

Type of Collision	Momentum Conserved?	Kinetic Energy Conserved?
Elastic	Yes	Yes
Inelastic	Yes	No
Perfectly Inelastic	Yes	No

Additional info: These problems illustrate the practical application of momentum and energy conservation principles in real-world and laboratory scenarios, such as vehicle safety and collision analysis.