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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:

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 v1 and object B is stationary (V1= 0) with respect to the observer.

>>> 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

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

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 V1= 0, the total momentum in the initial configuration is:

P1 = mv1

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

P2 = mv2 + MV2

By the Law of Conservation of Momentum:

P1 = P2

mv1 = mv2 + MV2

Solve for v1:

v1 = v2 + (M/m)V2

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

v1 = v2 + kV2

To make it easy to compare with energy, square v1:

v12 = v22 + 2kv2V2 + k2V22

Total kinetic energy

The total kinetic energy before the collision is:

E1 = mv12/2

The total energy after the collision is:

E2 = mv22/2 + MV22/2

By the Law of Conservation of Energy:

E1 = E2

mv12/2 = mv22/2 + MV22/2

Solve for v12:

v12 = v22 + kV22

Relationship between v2 and V2

In order to determine the relationship between the velocity of A after the collision (v2) and the velocity of B after the collision (V2), you compare the equations for v12 for both total momentum and total energy. Then, equating the solutions for v12, you can find the relationship between v2 and V2.

The momentum relationship is:

v12 = v22 + 2kv2V2 + k2V22

and the energy relationship is:

v12 = v22 + kV22

You can combine the equations to get:

v22 + 2kv2V2 + k2V22 = v22 + kV22

Simplify the equation:

2kv2V2 + k2V22 = kV22

2v2+ kV2= V2

Solve for v2:

2v2 = V2 − kV2

v2 = V2(1 − k)/2

Rearrange factors to solve for V2:

V2 = 2v2/(1 − k)

This is the relationship between the velocities after the collision, v2 and V2 as a function of the ratio k = M/m.

Relationship with the initial velocity

Starting with the total momentum equation, you can determine the values of v2 and V2 in terms of v1.r

v2 as a function of v1

To state v2 as a function of v1, start with the total momentum equation:

v1 = v2 + kV2

Substitute V2 = 2v2/(1 − k):

v1 = v2 + 2kv2/(1 − k)

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

v1(1 − k) = v2(1 − k) + 2kv2

v1(1 − k) = v2− kv2 + 2kv2

v1(1 − k) = v2 +2kv2

v1(1 − k) = v2(1 + k)

Solve for v2:

v2 = v1(1 − k)/(1 + k)

V2 as a function of v1

Again, start with:

v1 = v2 + kV2

Substitute v2 = V2(1 − k)/2:

v1 = V2(1 − k)/2 + kV2

2v1 = V2(1 − k) + 2kV2

2v1 = V2(1 + k)

Solve for V2

V2 = 2v1/(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 v2 or V2 are:

v2 = V2(1 − k)/2

V2 = 2v2/(1 − k)

The relationships of v2 and V2 as a function of v1 are:

v2 = v1(1 − k)/(1 + k)

V2 = 2v1/(1 + k)


Use your skills to examine situations


Resources and references

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