# A particle of mass m elastically collides head on with a stationary particle of mass M. What does the mass of M need to be to minimize the incoming particles kinetic energy?

Use conservation of linear momentum.

`P_i=P_f`

`mv_(m,i)=m v_(m,f)+M v_(M,f)`

(1) `m(v_(m,i)-v_(m,f))=M v_(M,f)`

Use conservation of kinetic energy since this collision is fully elastic.

`KE_i=KE_f`

`1/2 m v_(m,i)^2=1/2 m v_(m,f)^2+1/2 M v_(M,f)^2`

Solve for `v_(m,i)`

`m (v_(m,i)^2- v_(m,f)^2)=M v_(M,f)^2`

`m (v_(m,i)- v_(m,f))(v_(m,i)+ v_(m,f))=M v_(M,f)^2`

Plug in equation `(1)` .

`M v_(M,f)(v_(m,i)+ v_(m,f))=M v_(M,f)^2`

`v_(m,i)+ v_(m,f)=v_(M,f)`

(2)    `v_(m,i)=v_(M,f)- v_(m,f)`

Now we know the kinetic energy of the incoming particle will be minimized if `v_(m,f)^2` is at a minimum. So now lets again use the momentum conservation equation.

`mv_(m,i)=m v_(m,f)+M v_(M,f)`

`mv_(m,i)-M v_(M,f)=m v_(m,f)`

Plug in `(2)` to eliminate `v_(m,i)` .

`m(v_(M,f)- v_(m,f))-M v_(M,f)=m v_(m,f)`

`mv_(M,f)-M v_(M,f)=2m v_(m,f)`

`(m-M)v_(M,f)=2m v_(m,f)`

`1/2 (1-M/m)v_(M,f)=v_(m,f)`

Square both sides to make it proportional to `KE_i` .

`1/4 (1-M/m)^2 v_(M,f)^2=v_(m,f)^2`

Here you can see the minimum occurs when M=m. When this happens the incoming particle completely stops.

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