If a helium nucleus scatters to an angle of 120^\circ120 ∘ during an elastic collision with a gold nucleus, what is the final kinetic energy of the helium nucleus?

The answer to the question is;

The final kinetic energy of the helium nucleus is

7.53×10^(-13) J.

Explanation:

a) 0.5×m1×v1^2 = KE1, v1 = (2KE1/M1)^0.5 =

((2×8.0×10^(-13) J)/(6.68×10^(-27)) kg))^0.5

=1.548 × 10^7 m/s

Conservation of momentum along the x axis

m1v1 = m1V1'cos¤1 + m2V2'cos¤2... (1)

Conservation of momentum along the y axis

0 = m1V1'sin¤1 + m2V2'sin¤2.. (2)

Rearranging equation 1 and 2,

m1V1-m1v1 cos¤1 = m2v'2cos¤2 (ii)

-m1v'1sin(¤) = m2v'2 cos¤...(iii)

Squaring (ii) and (iii) gives

m2^2v2^2 = m1^2v1^2-2m1^2V1v1'cos¤+ m1^2v1^2

Solving for V2^2 and substituting into (i) we get

m1/m2×(V1^2-V1'^2) = m1^2/m2^2(V1^2+V1'^2-2V1V1'cos¤) so that

V1^2-V1'^2= m1/m2(V1^2+V1'^2-2V1V1'cos¤)

From where we have

m1 = 6.68×10(-27) kg, m2=3.29×10^(-23) kg,

v1 = 1.548×10^7 m/s

¤1 = 120°

We have

(1+m1/m2)V1'^2-((2m1/m2)v1cos¤1))V1'-(1-m1/m2)v1^2 = 0

From where

V1= -15323629.1 m/s or

= 15015578.5 m/s

and V2' = (m1/m2(v1^2-v1'^2))^0.5 =536246.981 m/s

Tan¤2 = -(v1'sin¤1)/(v1-v1'cos¤1)

= -0.56569

¤2 = tan^(-1)( -0.56569) = -29.5°

b) The final kinetic energy is

KEf =7.53×10^(-13)

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