Consider the \"collision\" between two helium nuclei (\"alpha particles\"). Thes
ID: 1488170 • 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 = 4.1x106 m/s when they are a distance di = 1.5x10-12 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
conservation of energy
when the helium molecules are heading towards each other the kinetic energy of the molecules vanishes due to repulsive force between the two nuclei, after that they move away from each other
say v1-------> <------------ v2 (+ve in + x direction, -Ve in -v e direction)
after collision potential energy = change in kinetic energy
k*q1*q2 / df = 1/2 m(v2^2+v1^2)
df = 1/2 m(v2^2+v1^2)/ k*q1*q2
= (0.5*6.644*10^-27(4.1*106 ^2+4.1*106 ^2))/(9*10^9*4*1.6*10^-19)
df = 5.3*10^-14 m
ratio of df/di = 5.3*10^-14 m/1.5*10^-12m
= 0.0353m