Collisions and conservation laws
Collisions and conservation laws

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Collisions and conservation laws

4 Elastic collisions in two or three dimensions

The laws of conservation of momentum and energy that we used to analyse elastic collisions in one dimension are also used to analyse elastic collisions in two or three dimensions. We simply treat the motions in each dimension as independent, and apply conservation of momentum separately along each Cartesian coordinate axis. Kinetic energy conservation continues to provide one additional equation relating the squares of the particle speeds. Since we have been careful to use vector notation throughout, this extension to two or three dimensions is easily made.

Consider the elastic collision of two identical bodies of mass m, one at rest and the other approaching with velocity bold u sub 1. The particles are no longer confined to move in one dimension, so our x -component equation (Equation 1), embodying conservation of momentum, becomes a full vector equation:

m bold u sub 1 = m bold v sub 1 + m bold v sub 2.

The law of conservation of energy (Equation 2) does not change, so

fraction 1 over 2 end m u sub 1 squared = fraction 1 over 2 end m v sub 1 squared + fraction 1 over 2 end m v sub 2 squared.

These can be simplified to:

bold u sub 1 = bold v sub 1 + bold v sub 2
Equation label: (10)

and

u sub 1 squared = v sub 1 squared + v sub 2 squared.
Equation label: (11)

These equations are most easily interpreted by a diagram. Figure 2 shows how the three vectors bold u sub 1, bold v sub 1 and bold v sub 2 are related to one another. Equation 10 tells us that all three velocity vectors must lie in a single plane , and that they must form a closed triangle. Equation 11 tells us that the triangle must be a right-angled triangle, since its sides obey Pythagoras’ theorem. The implication of this is striking, it means that the angle between bold v sub 1 and bold v sub 2 must be 90 super degrees.

Figure 2 (a) Elastic collision between particles of equal mass, with one at rest; (b) the corresponding vector triangle.

Following the elastic collision of two identical particles, one of which is initially at rest, the final velocities of the two particles will be at right-angles.

This is a simplifying feature of equal-mass collisions in two or three dimensions, analogous to the simple result of the exchange of velocities, which we found in one dimension.

You may have noticed that this result does not tell us exactly where the bodies go after the collision. Any pair of final velocities which can be represented by Figure 2 will be equally satisfactory, and there are an infinite number of these. The reason for this is that we have said nothing about the shape or size of the bodies, or just how they collide. We usually need to have additional information of this kind if we are to determine unique final velocities in such cases. Figure 3 shows the outcome of a particular collision in which spherical bodies make contact at a specific point. The location of this point is the sort of additional information required to determine unique values for v sub 1 and v sub 2.

Figure 3 When ball 1 strikes ball 2, the reaction forces at the contact point ensure that ball 2 is propelled away along the line of centres, as in snooker.

Activity 7

For the case illustrated in Figure 2 (two bodies of equal mass, one of which is initially at rest), if the moving body has an initial speed of 10 mathrm mathrm superscript minus 1 end, and is deflected through 20 degrees in the collision, find the magnitudes and directions of the velocities bold v sub 1 and bold v sub 2.

Answer

We draw a vector triangle like the one shown in Figure 2b

Figure 4 Vector triangle

We can now see that

v sub 1 =10 cosine 20 degrees mathrm mathrm superscript minus 1 end = 9.4 mathrm mathrm superscript minus 1 end

and is at the given angle of 20 degrees to the x -axis; bold v sub 2 has a magnitude

v sub 2 = 10 sine 20 degrees mathrm mathrm superscript minus 1 end = 3.4 mathrm mathrm superscript minus 1 end

and must be at 70 degrees to the x -axis so that the two angles add up to 90 degrees.

Activity 8

In the same situation (Figure 2), if, instead of the outcome specified in Activity 7, the speed of the moving body is reduced from 10.0 mathrm mathrm superscript minus 1 end to 6.0 mathrm mathrm superscript minus 1 end by the collision, find the final velocities.

Answer

Using the triangle in Figure 2 Pythagoras’ theorem tells us that v sub 1 squared + v sub 2 squared = u sub 1 squared, so

v sub 2 squared = u sub 1 squared minus v sub 1 squared = open bracket 10.0 squared minus 6.0 squared close bracket mathrm squared mathrm superscript minus 2 end = 64 mathrm squared mathrm superscript minus 2 end

so that v sub 2 = 8.0 mathrm mathrm superscript minus 1 end.

Now bold v sub 1 is at an angle arccos open bracket 6.0/10.0 close bracket = 53 super degrees and bold v sub 2 at an angle arccos open bracket 8.0/10.0 close bracket = 37 degrees to the x -axis. You will observe that the two angles add up to 90 degrees, as they should.

When the masses of the two colliding particles are unequal the algebraic manipulations required to solve elastic collision problems become rather complicated, but no new physics is involved in the solution so we will not pursue such problems here.

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