Help on energy conservation?

Momentum and Conservation of Energy 2D Help?

• Show that if a neutron is scattered through 90 degrees in an elastic collision with an initially stationary deuteron, the neutron loses 2/3 of its initial kinetic energy to the deuteron.(mass of neutron=1 atomic mass unit and mass of deuteron=2 atomic mass units) I know that the conservation of momentum and energy applies, but the 90 degrees is messing with my brains.If the neutron was initially moving in the positive x direction then after collision it will have a velocity in the y component only right? But what about the deuteron after collision? It will have an x and y component so how do I incorporate that into the equations and derive the proof? Ideas anyone?

• Answer:

  Suppose the neutron is moving towards the positive x-axis and hits a deuteron. After the collison, neutron moves towards +ve y-direction and the deuteron at an angle θ below x-axis. If m = mass of neutron, 2m = mass of deuteron Let u = initial velocity of neutron and u' = its final velocity and v = velocity after collision of the deteron Momentum balance along x and y-axis gives mu = (2m) * vcosθ => u = 2vcosθ ... ( 1 ) mu' = (2m) * v sinθ => u' = 2vsinθ ... ( 2 ) Kinetic energy balance gives (1/2) mu^2 = (1/2)mu'^2 + (1/2) (2m) v^2 => u^2 = u'^2 + 2v'^2 ... ( 3 ) Eliminating θ from eqns. ( 1 ) and ( 2 ), u^2 + u'^2 = 4v^2 ... ( 4 ) Eliminating v from eqns. ( 3 ) and ( 4 ), 2u^2 = 2u'^2 + u^2 + u'^2 => u^2 = 3u'^2 => Loss of KE of the neutron = (1/2) mu^2 - (1/2)mu'^2 = (1/2) mu^2 - (1/2) m * (1/3)u^2 = (2/3) * (1/2) mu^2 = (2/3) od the original KE of the neutron. [Note: KE balance is valid only assuming elastic collision.]