Collisions and conservation laws
Collisions and conservation laws

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Collisions and conservation laws

3.2 Elastic collisions in general

We now consider the general one-dimensional elastic collision between particles of mass m sub 1 and m sub 2 which move with initial velocities u subscript 1 x end and u subscript 2 x end before the collision and final velocities v subscript 1 x end and v subscript 2 x end after the collision. As stated earlier, the outcome of collisions of this kind is determined by Equation 1 and Equation 2. We shall not write down the details (though you might like to work them out for yourself) but by arguments similar to those used in Example 1, the following result can be obtained.

If the initial velocity of particle 1 relative to particle 2 is taken to be

u subscript 12 x end = u subscript 1 x end minus u subscript 2 x end

and if the final velocity of particle 1 relative to particle 2 is taken to be

v subscript 12 x end = v subscript 1 x end minus v subscript 2 x end comma

then, as a result of an elastic collision

v subscript 12 x end = minus u subscript 12 x end.

In other words:

In a one-dimensional elastic collision between two particles the relative velocity of approach is the negative of the relative velocity of separation

v subscript 1 x end minus v subscript 2 x end = minus open bracket u subscript 1 x end minus u subscript 2 x end close bracket.
Equation label: (7)

Combining this result (which incorporates the conservation of kinetic energy) with Equation 1 (which expresses conservation of momentum), leads to the following expressions for the final velocities:

v subscript 1 x end = fraction u subscript 1 x end open bracket m sub 1 minus m sub 2 close bracket + 2 m sub 2 u subscript 2 x end over m sub 1 + m sub 2 end open bracket elastic close bracket
Equation label: (8)
v subscript 2 x end = fraction u subscript 2 x end open bracket m sub 2 minus m sub 1 close bracket + 2 m sub 1 u subscript 1 x end over m sub 1 + m sub 2 end open bracket elastic close bracket.
Equation label: (9)

In the next two exercises you can use Equations 8 and 9. However, these equations are complicated so you are not expected to memorise them. You should be able to solve this type of question starting from the equations of conservation of momentum (Equation 1) and kinetic energy (Equation 2).

Activity 3

A neutron of mass m rebounds elastically in a head-on collision with a gold nucleus of mass 197 m that is initially at rest. What fraction of the neutron’s initial kinetic energy is transferred to the recoiling gold nucleus? Repeat this calculation when the target is a carbon nucleus at rest and of mass 12 m.

Answer

The initial kinetic energy is

uppercase E sub trans = fraction 1 over 2 end m u subscript 1 x end squared

and the final kinetic energy is

uppercase E sub trans = fraction 1 over 2 end m v subscript 1 x end squared comma

therefore the loss in energy is

Delta uppercase E sub trans = fraction 1 over 2 end m u subscript 1 x end squared minus fraction 1 over 2 end m v subscript 1 x end squared = fraction 1 over 2 end m open bracket u subscript 1 x end squared minus v subscript 1 x end squared close bracket

and the fractional loss is

fraction Delta uppercase E sub trans over uppercase E sub trans end = fraction u subscript 1 x end squared minus v subscript 1 x end squared over u subscript 1 x end squared end = 1 minus open bracket fraction v subscript 1 x end squared over u subscript 1 x end squared end close bracket.

With u subscript 2 x end = 0 in Equation 8,

v subscript 1 x end = u subscript 1 x end fraction m sub 1 minus m sub 2 over m sub 1 + m sub 2 end comma

we can write the fractional loss as

fraction Delta uppercase E sub trans over uppercase E sub trans end = 1 minus open bracket fraction v subscript 1 x end squared over u subscript 1 x end squared end close bracket = 1 minus open bracket fraction m sub 1 minus m sub 2 over m sub 1 + m sub 2 end close bracket squared.

For gold

fraction Delta uppercase E sub trans over uppercase E sub trans end = 1 minus open bracket fraction 196 over 198 end close bracket squared = 0.02 comma mathrm.

For carbon

fraction Delta uppercase E sub trans over uppercase E sub trans end = 1 minus open bracket fraction 11 over 13 end close bracket squared = 0.28 comma mathrm.

So, a low-mass nucleus is much more effective than a more massive nucleus when it comes to slowing down fast neutrons by elastic collisions. It is because of this fact that carbon is used in a nuclear reactor for just this purpose.

Activity 4

A tennis player returns a service in the direction of the server. The ball of mass 50 mathrm arrives at the racket of mass 350 mathrm with a speed of 45 mathrm mathrm superscript minus 1 end and the racket is travelling at 10 mathrm mathrm superscript minus 1 end at impact. Calculate the velocity of the returning ball, assuming elastic conditions.

Answer

We designate the ball as particle 1 and the racket as particle 2, with the ball initially travelling along the positive x -direction. From Equation 8

v subscript 1 x end = fraction u subscript 1 x end open bracket m sub 1 minus m sub 2 close bracket + 2 m sub 2 u subscript 2 x end over open bracket m sub 1 + m sub 2 close bracket end

so

v subscript 1 x end = fraction [45 mathrm mathrm superscript minus 1 end times open bracket minus 0.30 mathrm close bracket + 2 times 0.350 mathrm times open bracket minus 10 mathrm mathrm superscript minus 1 end close bracket ] over 0.400 mathrm end

i.e.

v subscript 1 x end = fraction open bracket minus 13.5 mathrm mathrm mathrm superscript minus 1 end minus 7.0 mathrm mathrm mathrm superscript minus 1 end close bracket over 0.4 mathrm end = minus 51 mathrm mathrm superscript minus 1 end.
S217_1

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