The Formula for Elastic Collision: An In-Depth Look

Elastic collisions are a common phenomenon in physics, encountered both in everyday life and in various scientific applications. These collisions occur when two objects collide and then separate without any loss in kinetic energy. The mathematical description of these collisions is based on two fundamental principles: the conservation of momentum and the conservation of kinetic energy.

Conservation of Momentum

The conservation of momentum states that the total momentum of a closed system is constant if no external forces are acting upon it. Mathematically, this can be expressed as:

\[

m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f}

\]

Where \( m_1 \) and \( m_2 \) are the masses of the two objects, \( v_{1i} \) and \( v_{2i} \) are their initial velocities, and \( v_{1f} \) and \( v_{2f} \) are their final velocities after the collision.

Conservation of Kinetic Energy

The conservation of kinetic energy states that the total kinetic energy of the system remains constant during an elastic collision. This can be expressed as:

\[

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

\]

The Formula for Elastic Collision

Combining the conservation of momentum and kinetic energy principles, we can derive the formula for the final velocities \( v_{1f} \) and \( v_{2f} \) in an elastic collision:

\[

v_{1f} = \frac{(m_1 – m_2)}{(m_1 + m_2)} \cdot v_{1i} + \frac{(2 \cdot m_2)}{(m_1 + m_2)} \cdot v_{2i}

\]

\[

v_{2f} = \frac{(2 \cdot m_1)}{(m_1 + m_2)} \cdot v_{1i} – \frac{(m_1 – m_2)}{(m_1 + m_2)} \cdot v_{2i}

\]

Applications

Physics Experiments

Researchers and scientists use the formula for elastic collision to design experiments that explore the basic properties of matter and energy.

Billiards and Sports

In games like billiards, understanding the principles of elastic collision can help players make more accurate shots.

Engineering and Transport

In vehicle design, engineers use these equations to understand how materials will respond to impacts, thus improving safety measures.

Astrophysics

Elastic collision equations are also applied in astrophysics, for example, in the study of celestial bodies and phenomena like gravitational slingshots.

Conclusion

Understanding the formula for elastic collision is essential in many fields of science and engineering. It provides the mathematical framework that describes how objects will behave when they collide under elastic conditions, which is vital for everything from game design to space exploration.

By delving into the conservation of momentum and kinetic energy, we can gain a comprehensive understanding of how elastic collisions work, empowering us to predict and interpret a wide range of physical phenomena.

**PROBLEMS AND SOLUTIONS**

Conceptual Questions

1. What is an elastic collision?

Solution: An elastic collision is a type of collision in which both kinetic energy and momentum are conserved.

2. Is it possible to have an elastic collision where one object is stationary?

Solution: Yes, it is possible. In this case, the stationary object will start moving, and the moving object’s velocity will change in such a way that kinetic energy and momentum are conserved.

3. What are the conserved quantities in an elastic collision?

Solution: In an elastic collision, both momentum and kinetic energy are conserved.

4. Can kinetic energy be lost in an elastic collision?

Solution: No, kinetic energy is conserved in an elastic collision.

5. How is the elastic collision formula useful in sports like billiards?

Solution: The formula helps in predicting the path and velocity of balls after the collision, which can improve game strategy.

Calculation Questions

6. Two objects with masses \(m_1 = 2 \, \text{kg}\) and \(m_2 = 3 \, \text{kg}\) collide elastically. Their initial velocities are \(v_{1i} = 5 \, \text{m/s}\) and \(v_{2i} = -2 \, \text{m/s}\). What are their final velocities?

Solution:

Using the formula for elastic collision,

\[

v_{1f} = \frac{(2 – 3)}{(2 + 3)} \times 5 + \frac{(2 \times 3)}{(2 + 3)} \times (-2) = \frac{-1}{5} \times 5 + \frac{6}{5} \times (-2) = -1 + (-1.2) = -2.2 \, \text{m/s}

\]

\[

v_{2f} = \frac{(2 \times 2)}{(2 + 3)} \times 5 – \frac{(2 – 3)}{(2 + 3)} \times (-2) = \frac{4}{5} \times 5 + \frac{1}{5} \times 2 = 4 + 0.4 = 4.4 \, \text{m/s}

\]

7. Two identical balls (\(m_1 = m_2\)) collide elastically. If one is stationary (\(v_{2i} = 0\)) and the other is moving at \(10 \, \text{m/s}\), what will be their velocities after the collision?

Solution:

\[

v_{1f} = \frac{(m_1 – m_2)}{(m_1 + m_2)} \times 10 + \frac{(2 \times m_2)}{(m_1 + m_2)} \times 0 = 0

\]

\[

v_{2f} = \frac{(2 \times m_1)}{(m_1 + m_2)} \times 10 – \frac{(m_1 – m_2)}{(m_1 + m_2)} \times 0 = 10

\]

8. What would be the final velocities if both balls in question 7 were moving toward each other at \(10 \, \text{m/s}\)?

Solution:

\(v_{1f} = -10 \, \text{m/s}\) and \(v_{2f} = 10 \, \text{m/s}\), as they will simply switch velocities.

9. In an elastic collision, \(m_1 = 5 \, \text{kg}\), \(m_2 = 5 \, \text{kg}\), \(v_{1i} = 5 \, \text{m/s}\), \(v_{2i} = -5 \, \text{m/s}\). What are \(v_{1f}\) and \(v_{2f}\)?

Solution:

\(v_{1f} = -5 \, \text{m/s}\) and \(v_{2f} = 5 \, \text{m/s}\), they will switch velocities.

10. An object with \(m_1 = 1 \, \text{kg}\) and \(v_{1i} = 5 \, \text{m/s}\) collides with a stationary object \(m_2 = 4 \, \text{kg}\). What are the final velocities?

Solution:

\[

v_{1f} = \frac{(1 – 4)}{(1 + 4)} \times 5 + \frac{(2 \times 4)}{(1 + 4)} \times 0 = \frac{-3}{5} \times 5 + 0 = -3 \, \text{m/s}

\]

\[

v_{2f} = \frac{(2 \times 1)}{(1 + 4)} \times 5 – \frac{(1 – 4)}{(1 + 4)} \times 0 = \frac{2}{5} \times 5 = 2 \, \text{m/s}

\]

11. Is it possible for two objects to stick together after an elastic collision?

Solution: No, in an elastic collision, objects will separate after the collision and kinetic energy is conserved.

12. Can the final velocities of both objects be the same after an elastic collision?

Solution: Yes, it’s possible, particularly if both objects have the same mass and opposite initial velocities.

13. If \(m_1 = 2 \, \text{kg}\) and \(m_2 = 4 \, \text{kg}\) and both are initially at rest, what are their final velocities after an elastic collision?

Solution:

Both will remain at rest as there is no initial velocity to transfer. \(v_{1f} = 0 \, \text{m/s}\), \(v_{2f} = 0 \, \text{m/s}\).

14. Two objects with \(m_1 = 3 \, \text{kg}\) and \(m_2 = 6 \, \text{kg}\) have initial velocities of \(v_{1i} = 4 \, \text{m/s}\) and \(v_{2i} = -3 \, \text{m/s}\). Calculate their final velocities.

Solution:

\[

v_{1f} = \frac{(3 – 6)}{(3 + 6)} \times 4 + \frac{(2 \times 6)}{(3 + 6)} \times (-3) = \frac{-3}{9} \times 4 + \frac{12}{9} \times (-3) = -\frac{4}{3} + \frac{-12}{3} = -\frac{16}{3} \, \text{m/s}

\]

\[

v_{2f} = \frac{(2 \times 3)}{(3 + 6)} \times 4 – \frac{(3 – 6)}{(3 + 6)} \times (-3) = \frac{6}{9} \times 4 + \frac{3}{9} \times 3 = \frac{24}{9} + \frac{9}{9} = \frac{33}{9} \, \text{m/s} = \frac{11}{3} \, \text{m/s}

\]

15. If \(m_1 = 2 \, \text{kg}\) and \(m_2 = 2 \, \text{kg}\), and \(v_{1i} = 3 \, \text{m/s}\), \(v_{2i} = -1 \, \text{m/s}\), find the final velocities.

Solution:

\(v_{1f} = -1 \, \text{m/s}\), \(v_{2f} = 3 \, \text{m/s}\), they will switch velocities.

16. In an elastic collision, \(m_1 = 4 \, \text{kg}\), \(m_2 = 2 \, \text{kg}\), \(v_{1i} = 2 \, \text{m/s}\), \(v_{2i} = 4 \, \text{m/s}\). What are the final velocities?

Solution:

\[

v_{1f} = \frac{(4 – 2)}{(4 + 2)} \times 2 + \frac{(2 \times 2)}{(4 + 2)} \times 4 = \frac{2}{6} \times 2 + \frac{4}{6} \times 4 = \frac{2}{3} + \frac{16}{9} = \frac{6}{9} + \frac{16}{9} = \frac{22}{9} \, \text{m/s}

\]

\[

v_{2f} = \frac{(2 \times 4)}{(4 + 2)} \times 2 – \frac{(4 – 2)}{(4 + 2)} \times 4 = \frac{8}{6} \times 2 – \frac{2}{6} \times 4 = \frac{16}{9} – \frac{8}{9} = \frac{8}{9} \, \text{m/s}

\]

17. What will be the final velocities if \(m_1 = 3 \, \text{kg}\) and \(m_2 = 5 \, \text{kg}\) have the same initial velocities of \(v_{1i} = v_{2i} = 4 \, \text{m/s}\)?

Solution:

\(v_{1f} = 4 \, \text{m/s}\), \(v_{2f} = 4 \, \text{m/s}\), their velocities will remain the same.

18. An object \(m_1 = 1 \, \text{kg}\) moving at \(v_{1i} = -6 \, \text{m/s}\) collides with another object \(m_2 = 1 \, \text{kg}\) moving at \(v_{2i} = 6 \, \text{m/s}\). Calculate the final velocities.

Solution:

\(v_{1f} = 6 \, \text{m/s}\), \(v_{2f} = -6 \, \text{m/s}\), they will switch velocities.

19. Two objects \(m_1 = 5 \, \text{kg}\) and \(m_2 = 7 \, \text{kg}\) have initial velocities \(v_{1i} = 10 \, \text{m/s}\) and \(v_{2i} = -10 \, \text{m/s}\). What are the final velocities?

Solution:

\(v_{1f} = -10 \, \text{m/s}\) and \(v_{2f} = 10 \, \text{m/s}\), they will switch velocities.

20. If \(m_1 = 4 \, \text{kg}\) and \(m_2 = 4 \, \text{kg}\), and \(v_{1i} = 3 \, \text{m/s}\) and \(v_{2i} = -3 \, \text{m/s}\), find the final velocities.

Solution:

\(v_{1f} = -3 \, \text{m/s}\), \(v_{2f} = 3 \, \text{m/s}\), they will switch velocities.