Collisions and conservation laws
Collisions and conservation laws

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Collisions and conservation laws

3.1 Elastic collisions with a stationary target

We begin with an example of a one-dimensional elastic collision between two particles of identical mass, one of which is initially stationary. Our aim now is to find the final velocity of each particle after the collision.

Activity 1

A particle of mass m moves along the x -axis with velocity u subscript 1 x end and collides elastically with an identical particle at rest. What are the velocities of the two particles after the collision?

Answer

Let the final velocities be v subscript 1 x end and v subscript 2 x end. Conservation of momentum along the x -axis gives

m u subscript 1 x end = m v subscript 1 x end + m v subscript 2 x end
Equation label: (3)

and conservation of kinetic energy for this elastic collision gives

fraction 1 over 2 end m u squared subscript 1 x end = fraction 1 over 2 end m v squared subscript 1 x end + fraction 1 over 2 end m v squared subscript 2 x end.
Equation label: (4)

By eliminating common factors, Equation 3 can be simplified to give

u subscript 1 x end = v subscript 1 x end + v subscript 2 x end
i.e. v subscript 2 x end = u subscript 1 x end minus v subscript 1 x end
Equation label: (5)

and Equation 4 can be treated similarly to give

u squared subscript 1 x end = v squared subscript 1 x end + v squared subscript 2 x end.
Equation label: (6)

Rearranging Equation 6 gives

v squared subscript 2 x end = u squared subscript 1 x end minus v squared subscript 1 x end comma

the right-hand side of which may be rewritten using the general identity a squared minus b squared = open bracket a minus b close bracket open bracket a+ b close bracket, thus

v squared subscript 2 x end = open bracket u subscript 1 x end minus v subscript 1 x end close bracket open bracket u subscript 1 x end + v subscript 1 x end close bracket.

Dividing both sides of this last equation by v subscript 2 x end,

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

and using Equation 5 to simplify the resulting right-hand side gives

v subscript 2 x end = u subscript 1 x end + v subscript 1 x end.

Comparing this expression for v subscript 2 x end with that in Equation 5 shows that

v subscript 1 x end = 0 and thus v subscript 2 x end = u subscript 1 x end.

The result of Example 1 will be familiar to anyone who has seen the head-on collision of two bowls on a bowling green. The moving one stops, and the one that was initially stationary moves off with the original velocity of the first. In effect, the bowls exchange velocities.

Activity 2

Predict qualitatively (i.e. without calculation) what would happen when a body of mass m sub 1 collides with another body of mass m sub 2 that is initially at rest if:

(a) m sub 1 gg m sub 2 (The symbol gg should be read as ‘is very much greater than’.)

Answer

Experience should tell you that a high-mass projectile fired at a low-mass target would be essentially unaffected by the collision.

(b) m sub 2 gg m sub 1.

Answer

A low-mass projectile fired at a massive target would bounce back with unchanged speed.

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