# Derivation of a Simple Elastic Collision

by Ron Kurtus (2 August 2013)

An **elastic collision** of two objects is one where both the total momentum and kinetic energy of the pair is conserved. Typically, they are point objects. It is an ideal situation that does not really occur in Nature.

By a simple collision, I mean when an object directly collides with a stationary object along a line through their centers, such that the resulting motion is not at some angle. This is called a one-dimensional collision. Also, there is a requirement that the objects are perfectly stiff and do not lose kinetic energy in the collision.

You can determine the resulting velocities of the objects is by combining the laws of *Conservation of Momentum* and *Conservation of Energy* before and after the collision. The Law of Conservation of Momentum states that in a closed system the momentum is conserved in any direction. Conservation of Energy states that in a closed system, the energy remains constant.

Some questions you may have include:

- What are the initial considerations?
- What is the relationship between the final velocities?
- What are the relationships with the initial velocity?

This lesson will answer those questions. Useful tool: Units Conversion

## Initial considerations

Suppose an object **A** of mass **m** is moving toward object **B** of mass **M**, along a line through their centers. Object **A** is moving at velocity **v _{1}** and object

**B**is stationary (

**V**) with respect to the observer.

_{1}= 0**>>> NOTE: I could view it as both moving relative to each other toward their center of mass.** **NOTE: derivation is too complicated.**

Object **A** moving toward stationary object **B**

### After collision

After object **A** collides with object **B**, the motion of both **A** and **B** change according to the ratio of their masses and conservation or momentum and energy, provided each object is perfectly stiff and will not lose kinetic energy through deformation in the collision.

Example of motion of objects after collision

**Note**: Exactly what happens at the point of collision is not considered in this derivation.

### Total momentum

The total momentum of two objects is:

P = mv + MV

Since **V _{1}= 0**, the total momentum in the initial configuration is:

P_{1}= mv_{1}

After object **A** collides with **B**, the total momentum is:

P_{2}= mv_{2}+ MV_{2}

By the Law of Conservation of Momentum:

P_{1}= P_{2}

mv_{1}= mv_{2}+ MV_{2}_{}

Solve for **v _{1}**:

v_{1}= v_{2}+ (M/m)V_{2}

For the sale of convenience, set **k = M/m**. Thus:

v_{1}= v_{2}+ kV_{2}

To make it easy to compare with energy, square **v _{1}**:

v_{1}^{2}= v_{2}^{2}+ 2kv_{2}V_{2}+ k^{2}V_{2}^{2}

### Total kinetic energy

The total kinetic energy before the collision is:

E_{1}= mv_{1}^{2}/2

The total energy after the collision is:

E_{2}= mv_{2}^{2}/2 + MV_{2}^{2}/2

By the Law of Conservation of Energy:

E_{1}= E_{2}

mv_{1}^{2}/2= mv_{2}^{2}/2 + MV_{2}^{2}/2

Solve for **v _{1}^{2}**:

v_{1}^{2}= v_{2}^{2}+ kV_{2}^{2}

## Relationship between **v**_{2} and V_{2}

_{2}

In order to determine the relationship between the velocity of **A** after the collision (**v _{2}**) and the velocity of

**B**after the collision (

**V**), you compare the equations for

_{2}**v**for both total momentum and total energy. Then, equating the solutions for

_{1}^{2}**v**, you can find the relationship between

_{1}^{2}**v**and

_{2}**V**.

_{2}The momentum relationship is:

v_{1}^{2}= v_{2}^{2}+ 2kv_{2}V_{2}+ k^{2}V_{2}^{2}

and the energy relationship is:

v_{1}^{2}= v_{2}^{2}+ kV_{2}^{2}

You can combine the equations to get:

v_{2}^{2}+ 2kv_{2}V_{2}+ k^{2}V_{2}^{2}= v_{2}^{2}+ kV_{2}^{2}

Simplify the equation:

2kv_{2}V_{2}+ k^{2}V_{2}^{2}= kV_{2}^{2}

2v_{2}+ kV_{2}= V_{2}

Solve for **v _{2}**:

2v_{2}= V_{2}− kV_{2}

v_{2}= V_{2}(1 − k)/2

Rearrange factors to solve for **V _{2}**:

V_{2}= 2v_{2}/(1 − k)

This is the relationship between the velocities after the collision, ** v _{2}** and

**V**as a function of the ratio

_{2}**k = M/m**.

## Relationship with the initial velocity

Starting with the total momentum equation, you can determine the values of **v _{2}** and

**V**in terms of

_{2}**v**.r

_{1}### v_{2} as a function of v_{1}

To state **v _{2}** as a function of

**v**, start with the total momentum equation:

_{1}

v_{1}= v_{2}+ kV_{2}

Substitute** V _{2} = 2v_{2}/(1 − k)**:

v_{1}= v_{2}+ 2kv_{2}/(1 − k)

Multiply both sides of equation by **(1 − k)** and simplify:

v_{1}(1 − k) = v_{2}(1 − k) + 2kv_{2}

v_{1}(1 − k) = v_{2}− kv_{2}+ 2kv_{2}

v_{1}(1 − k) = v_{2}+2kv_{2}

v_{1}(1 − k) = v_{2}(1 + k)

Solve for **v _{2}**:

v_{2}= v_{1}(1 − k)/(1 + k)

### V_{2} as a function of v_{1}

Again, start with:

v_{1}= v_{2}+ kV_{2}

Substitute **v _{2} = V_{2}(1 − k)/2**:

v_{1}= V_{2}(1 − k)/2 + kV_{2}

2v_{1}= V_{2}(1 − k) + 2kV_{2}

2v_{1}= V_{2}(1 + k)

Solve for **V _{2}**

V_{2}= 2v_{1}/(1 + k)

## Summary

If an object directly collides with another, such that the resulting motion of the objects is along a line through their centers, the resulting velocities of the objects is determined by the laws of *Conservation of Momentum* and *Conservation of Energy.*

By applying the laws before and after the collision, you can determine that the resulting velocities are related to the ratio of the masses of the objects.

The resulting relationships between **v _{2}** or

**V**are:

_{2 }

v_{2}= V_{2}(1 − k)/2

V_{2}= 2v_{2}/(1 − k)

The relationships of **v _{2}** and

**V**as a function of

_{2}**v**are:

_{1}

v_{2}= v_{1}(1 − k)/(1 + k)

V_{2}= 2v_{1}/(1 + k)

Use your skills to examine situations

## Resources and references

### Websites

### Books

**Top-rated books on the Physics of Motion**

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## Derivation of a Simple Collision