Question

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 nuclei (197Au). The energy of the incoming helium nucleus was 7.58 ✕ 10−13 J, and the masses of the helium and gold nuclei were 6.68 ✕ 10−27 kg and 3.29 ✕ 10−25 kg, respectively (note that their mass ratio is 4 to 197. Assume that the helium nucleus travels in the +x-direction before the collision.)
(a)
If a helium nucleus scatters to an angle of 121° during an elastic collision with a gold nucleus, calculate the helium nucleus' final speed (in m/s) and the final velocity (magnitude in m/s and direction counterclockwise from the +x-axis) of the gold nucleus.
(b)
What is the final kinetic energy (in J) of the helium nucleus?

Answer 1

a) Final speed of the helium nucleus: 1.14 x 10^5 m/s
Final velocity of the gold nucleus: 1.14 x 10^5 m/s, 121° counterclockwise from the +x-axis

b) Final kinetic energy of the helium nucleus: 5.76 x 10^-13 J