| Contents for this page | Related topics |  |
|---|---|---|
|
Isolated systems Collisions Linear collisions Perpendicular collisions Collisions at oblique angles Additional questions |
Projectiles Energy Work Power |
 Data Glossary |
| Learning Outcomes | ||
| After studying this section, you will understand the difference between elastic, inelastic and perfectly elastic collisions. | ||
| Helpful background knowledge | ||
| Introduction to vectors | Addition of vectors | Equations of motion |
We have come across this in the Grade 11 Mechanics section of the Electronic Science Tutor, but we might as well mention it again! It is important to realise that the Law of Conservation of Momentum (
) only applies to ISOLATED SYSTEMS, that is, systems where the only forces that are acting are those between the interacting bodies. In the sort of problems that you might encounter at Grades 11 and 12 levels, you may assume that you are dealing with isolated systems, unless told otherwise. An example of a system that is not isolated would be two railway trucks colliding. Here, apart from the forces between the colliding trucks, there is the force of friction between the wheels of the trucks and the rails. Normally, such frictional forces are ignored.
A COLLISION may be defined as an isolated event during which two or more bodies exchange forces for a relatively short time interval. Normally, we think of collisions as implying that the colliding bodies come into actual contact, but this is not necessarily so (). As a result of a collision, the colliding bodies undergo a change in velocity. Familiar events are the motion of the balls in a billiard game, and, sadly, traffic accidents. On a microscopic scale, gas and liquid molecules are continuously undergoing collisions between one another and the walls of the containing vessel.
Collisions may be ELASTIC, INELASTIC or PERFECTLY INELASTIC.
It follows that if the total kinetic energy of the colliding bodies is the same after and before the collision, then the collision was elastic. If the total kinetic energy after the collision is less than the total kinetic energy before the collision, then the collision is inelastic. It need not be perfectly inelastic, the case when the colliding objects stick together after colliding. In the real macroscopic world, all collisions are inelastic.
When the kinetic energy of the moving objects is not conserved, what happens to it? Remember that energy cannot be created or destroyed, so we must consider the following options:
This is a revison and extension of the topic which is dealt with in Grade 11.
Linear collisions are those in which the colliding objects both move along the same straight line (). The objects may move in the same or in the opposite direction. The examples below are taken from the Grade 11 section of the Electronic Science Tutor, and are included here for the purpose of revision.
If one of the bodies is at rest, and the moving object strikes it "head-on", we get two useful equations (see their derivations):
We can examine three cases:
From equation (1), v1 = 0, and from equation (2), v2 = u1. The first object comes to rest, and the struck object moves off with the same velocity as that of the striking object before the collision. There is a simple exchange of velocities. Those of you who are into billiards and snooker will be familiar with this!
From equation (1), v1 ≅ u1, and from equation (2), v2 ≅ 2u1. The heavier object does not appreciably change its velocity, and the lighter object moves at twice the same velocity with which it was struck.
From equation (1), v1 ≅ -u1, and from equation (2), v2 ≅ 0. The striking object effectively bounces back from the heavier object, which remains at rest. This is the approximate situation found when a falling ball bounces off a hard floor.
Here we are interested in collisions where the initial velocities of the colliding objects are at 90º to each other. We should remember that momentum is a vector quantity, but kinetic energy is a scalar quantity.
Consider two objects of masses m1 and m2. The mass m1 moves from left to right in a horizontal direction with a velocity u1, while the mass m2 moves upwards in a vertical direction, with a velocity u2. After colliding, the masses stick together, and move with a final velocity v in a direction θ ° with the horizontal. The principle underlying this situation is that the final momentum, a vector quantity, can be resolved into an x- and a y-component. Initially, since m1 moves in a horizontal direction, its momentum only has an x-component, while the momentum of m2, which moves only in a vertical direction, only has a y-component. These x- and the y-components are separately conserved, and their vector sum is the resultant momentum of the combined masses (m1 + m2) after the collision. We can draw up a table:
| Objects | x-Momentum before | y-Momentum before | x-Momentum after | y-Momentum after |
|---|---|---|---|---|
| m1 | m1u1 | 0 | - | - |
| m2 | 0 | m2u2 | - | - |
| (m1 + m2) | - | - | (m1 + m2) vcosθ | (m1 + m2) vsinθ |
The ratio of the two momentums before the collision readily gives the value of the angle θ:
While the magnitude, |v|, of the resultant velocity is given by applying Pythagoras' theorem:
If you have trouble understanding this, you might want to refresh your memory by looking at the method of components in resolving vectors.
The next level of difficulty involves collisions at glancing angles, that is, collisions taking place at an angle that is less than 90°. Typically, knowing the momentums of the two colliding objects, we would want to calculate the momentum of the combined objects after the collision.
Consider two masses A and B moving towards each other at angles α and β to the horizontal, as shown at the left in the diagram below.
Before the collision, let the momentum of A be pa and the momentum of B be pb. After the collision, the two bodies coalesce, and move at an angle θ to the horizontal, with a momentum pc, as shown at the right in the above diagram. As before, let us draw up a table:
| Objects | x-Momentum before | y-Momentum before | x-Momentum after | y-Momentum after |
|---|---|---|---|---|
| A | pacosα | pasinα | - | - |
| B | pbcos(360 - β) () |
pbsin(360 - β) | - | - |
| (A + B) | - | - | pcx = pccosθ | pcy = pcsinθ |
The x-components are conserved, so pacosα + pbcos(360 - β) = pccosθ.
Similarly, we equate the y-components: pasinα + pbsin(360 - β) = pcsinθ.
These two equations enable us to calculate the value of the final momentum pc and the angle θ. The magnitude |pc| will be given by |pc| = (pcx2 + pcy2)½, while tanθ = pcy/pcx.
m1 = m2 = m. The conservation of momentum equation reduces to mu1 = mv1 + mv2, while the conservation of kinetic energy equation reduces to mu1² = mv1² + mv2². The resultant momentum vector mv is equal to the initial momentum vector mu1. From the triangle ABC, applying the converse of Pythagoras' theorem, we see that the final momentum vectors must be at right angles to each other.
If the mass m2 is at rest, then v1 = 0.
| Conservation of momentum: | m1u1 = m1v1 + m2v2, rearranging to m1(u1 - v1) = m2v2 (A) |
| Conservation of kinetic energy (collision is elastic): |
½m1u1² = ½m1v1² + ½m2v2², |
| which gives | m1(u1² - v1² ) = m1(u1 - v1)(u1 + v1) = m2v2² (B) |
| Dividing (B) by (A), we get | v2 = u1 + v1. |
| Substituting this value of v2 in (A) and rearranging gives | v1 = u1(m1 - m2)/(m1 + m2). |
| Using this value of v1 in (B) gives, after rearranging, | v2 = 2u1m1/(m1 + m2). |
