Answer:
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)
