this question is from my previous assignment which was due 2 weeks back. just revising for mid-term not sure how to solve it. if someone explains it thoroughly it would be grate(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton.

A proton and an alpha particle headed directly toward each other, had each initial speed of 3.9×10−3 c when they were far away.

Here, as is customary when describing processes involving nuclear targets, the speed is expressed as a fraction of c, the speed of light.

What is the distance of closest approach between the proton and the alpha particle?

Hint: There are two conserved quantities. Make use of both.

2. Relevant equations

U=Eqd or KqQ/r

K=1/2*(mv^2)

3. The attempt at a solution

.5m(v_i)^2 + Eq(d_i)= .5m(v_f)^2 + Eq(d_f)

where (V_f)=0 then i solved for (d_f)

and got the 1.87*10^-27 but the answer should be d_f=1.3*10^-13

could some 1 explain this to me and the reasoning!! please

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# Homework Help: Colliding Particles using energy

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