1. A neutron moving at 8.0 X 107 m/s makes a head-on elastic collision with a he

Question

1. A neutron moving at 8.0 X 107 m/s makes a head-on elastic collision with a helium nucleus. (Mass of the helium = 4 X mass of the neutron.)
a) Find the speed of the neutron after the collision.
b) Which direction is the neutron moving (forward or back)?
c) Let f = (final kinetic energy of the neutron)/(initial kinetic energy of the neutron). Show that f = 0.36.
d) Suppose that for many elastic collisions, not necessarily head-on, the average value of f = 0.2. How many collisions are needed, on the average, to reduce the neutron’s kinetic energy by a factor of 106?

Explanation / Answer

Pi = mv = Mn * 8.0 X 107
By conservation of Momentum ,
Mn * 8.0 X 107 = 4Mn Vhe + Mn Vf

By conservation of energy 

1/2 mVi2 = 1/2 4mVhe2 + 1/2 mVf2

Velocity of helium =  32 X 10^6 m/s .

Velocity of neutron =  4.8 X 10^7 m/s . (Backwards)

F = (1/2 mVf2)/1/2 mVi2

 = (4.8/8)^2 = 0.36

(d) Average value of f = 0.2

 To reduce by a factor of 10^6

Here I did n't understand the question Properly can u elabrate on it . 

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