Applications of Collision Equations
About 3 min read
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 v2 and V2 as a function of v1 are:
v2 = v1(1 − k)/(1 + k)
V2 = 2v1/(1 + k)
Between v2 or V2:
v2 = V2(1 − k)/2
V2 = 2v2/(1 − k)
where
- v1 is the initial velocity of object A
- v2 is the resulting velocity of object A
- V2 is the resulting velocity of object 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.
v2 as a function of v1
v2 = v1(1 − k)/(1 + k)
v2 = v1(1 − 2)/(1 + 2)
v2 = −v1/3
V2 as a function of v1
V2 = 2v1/(1 + k)
V2 = 2v1/(1 +2)
V2 = 2v1/3
Relationship of v2 and V2
v2 = V2(1 − k)/2
v2 = V2(1 − 2)/2
v2 = −V2/2
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.
Mass of A equals B
Suppose the mass of object A is the same as that of object B. Then k = 1.
v2 = V2(1 − k)/2
v2 = V2(1 − 1)/2
v2 = 0
However, to maintain the conservation of momentum and energy, you can use the momentum equation:
v1 = v2 + kV2
v1 = V2
Thus, the collision sequence looks like:
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 v1. 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.
v2 = V2(1 − k)/2
v2 = V2(1 − 1/2)/2
v2 = V2(1/2)/2
v2 = V2/4
Also, since
V2 = 2v1/(1 + k)
V2 = 2v1/(1 + ½)
V2 = 2v1/(3/2)
V2 = 4v1/3
and
v2 = v1/3
The collision scenario is:
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
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