# Applications of Collision Equations

by Ron Kurtus (revised 2 January 2014)

The general equations for the collision of a moving object **A** with a stationary object **B** was determined in *Derivation of a Simple Collision* and are stated below.

Applications of the equation include when the mass of **A < B**, **A = B**, and **A > B**.

Some questions you may have include:

- What is the collision equation?
- What is the solution when mass of
**A**is less than**B**? - What is the solution when mass of
**A**equals that of**B**? - What is the solution when mass of
**A**is greater than**B**?

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

## Collision equations

The relationships between the velocities when object **A** collides with object **B** are:

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)

Between **v _{2}** or

**V**:

_{2}

v_{2}= V_{2}(1 − k)/2

V_{2}= 2v_{2}/(1 − k)

where

**v**is the initial velocity of object_{1}**A****v**is the resulting velocity of object_{2}**A****V**is the resulting velocity of object_{2}**B****k**is the ratio of the masses of the objects:**k = M/m****m**is the mass of object**A****M**is the mass of object**B**

## Mass of A is less than B

Consider the situation when the mass of **A < B**. Suppose **M = 2m**. Then **k = 2**.

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

v_{2}= v_{1}(1 − k)/(1 + k)

v_{2}= v_{1}(1 − 2)/(1 + 2)

v_{2}= −v_{1}/3

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

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

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

V_{2}= 2v_{1}/3

### Relationship of v_{2} and V_{2}

v_{2}= V_{2}(1 − k)/2

v_{2}= V_{2}(1 − 2)/2

v_{2}= −V_{2}/2

This means that for a case when **M = 2m**, object **A** would move in the opposite direction in a velocity of 1/2 the velocity of object **B**.

Resulting motion when mass of **A** less than **B**

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

## Mass of A equals B

Suppose the mass of object **A** is the same as that of object **B**. Then **k = 1**.

v_{2}= V_{2}(1 − k)/2

v_{2}= V_{2}(1 − 1)/2

v_{2}= 0

However, to maintain the conservation of momentum and energy, you can use the momentum equation:

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

v_{1}= V_{2}

Thus, the collision sequence looks like:

Collision of equal mass objects

This means that when the masses of **A** and **B** are equal, the collision results in object **A** becomeing stational and object **B** moving foward at the same velocity as **v _{1}**. This effect can be seen in Newton's Cradle.

## Mass of A is greater than B

Suppose **m = 2M** (or **M = m/2**). Then **k = 1/2**.

v_{2}= V_{2}(1 − k)/2

v_{2}= V_{2}(1 − 1/2)/2

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

v_{2}= V_{2}/4

Also, since

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

V_{2}= 2v_{1}/(1 + ½)

V_{2}= 2v_{1}/(3/2)

V_{2}= 4v_{1}/3

and

v_{2}= v_{1}/3

The collision scenario is:

Resulting motion when mass of A greater than B

This means that for a case when **m = 2M**, object **A** would continue to move in the same direction after the collision at a velocity of **1/4** the velocity of object **B**.

## Summary

By inserting the value of **k** in the collsion equations, you can see the resulting motion of the objects.

Test your ideas

## Resources and references

### Websites

### Books

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

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## Where are you now?

## Applications of Collision Equations