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For an inelastic impact situation where two bodies traveling in opposite directions (vehicles say) collide and coalesce (perfectly inelastic collision) , one can derive the following expression for equal and opposite transfer of momentum. On the basis that if impact forces are equal and opposite, so too is the transfer of momentum.
Δp =m1m2(v2-v1) /(m1+m2)
In passing it's interesting to note that the quantity:
m1m2 /(m1+m2)
also occurs in the calculation of reduced mass.
But that's an aside - my question is can one derive a similar expression for momentum transfer in the case of a simple elastic collision? In terms of the pre-collision parameters.
Δp =m1m2(v2-v1) /(m1+m2)
In passing it's interesting to note that the quantity:
m1m2 /(m1+m2)
also occurs in the calculation of reduced mass.
But that's an aside - my question is can one derive a similar expression for momentum transfer in the case of a simple elastic collision? In terms of the pre-collision parameters.