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.
http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html

https://www.pathwayz.org/Tree/Plain/CONSERVATION+OF+MOMENTUM+-+1D