Consider the \"collision\" between two helium nuclei (\"alpha particles\"). Thes
ID: 1487590 • Letter: C
Question
Consider the "collision" between two helium nuclei ("alpha particles"). These particles have a mass equal to four protons and a charge equal to two protons.
Two helium nuclei are heading directly towards each other with the same speed, so v1 = - v2 where v1 and v2 are the velocities of the particles. A detector determines that the speed of the particles is vi = 1.3x106 m/s when they are a distance di = 2.5x10-12 m apart.
If df is the closest distance the particles approach each other, what is the ratio of the initial to final distances, di/df?
Give your answer to at least three significant digits.
Remember: Momentum conservation requires that v1 = - v2 during the entire "collision". (Total momentum is equal to zero.)
Explanation / Answer
energy conservation
1/2 m vi^2 + 1/2 m vi^2 + k q^2 /di = k q^2 / df
m vi^2 + k (4 e)^2 / di = k (4 e^2) / df
(4 x 1.6726 x 10-27) x (1.3 x 10^6)^2 + 9 x 10^9 x (4 x 1.6 x 10^-19)^2 / 2.5 x10^-12 = 9 x 10^9 x (4 x 1.6 x 10^-19)^2 / df
1.1306776 x 10^-14 + 1.47456 x 10^-39 = 3.6864 x 10^-27 / df
df = 3.26034583 x 10^-41 m
di/df = 2.5x10^-12 / 3.26034583 x 10^-41
di / df = 7.66789822 x 10^-54