Two helium nuclei are heading directly towards each other with the same speed, s

Question

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 = 2.7x106 m/s when they are a distance di = 2.1x10-12 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

Mass of one helium nucleus = 6.644*10-27 kg

As both are Helium nucleus moving with same speed towards each other initially at 2.7*106 m/s

Change in Potential Energy of helium nuclei seperated at distance initially(di) = 2.1*10-12 m and final df

Uf - Ui = Gm1m2[1/di-1/df] = 6.67*10-11*(6.644*10-27)2/[1/(2.1*10-12) - 1/df]

Change in Kinetic Energy = 1/2 m (vf - vi)2

Hence 1/2 m (vf - vi)2 = 6.67*10-11*(6.644*10-27)2/[1/(2.1*10-12) - 1/df]

1/2*6.644*10-27(0 - 2.7*106) = 6.67*10-11*(6.644*10-27)2/[1/(2.1*10-12) - 1/df]

By Calculating df = 3.28671*10-24 m

Hence di/df = 6.38936*1011

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