Distance of closest approach for proton and alpha particle collision

[seto6]

Homework Statement

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?

Homework Equations

potential energy and electric potential energy.

The Attempt at a Solution

following is a simplified attempt at solution

k(Q_p)(Q_a)/r=(.5(M_p)V^2)+(.5(M_a)V^2)
and solved for r and got 1.7X10^-13 m
but i keep getting it wrong could some 1 tell me where i am going wrong
the answer is 1.3X10^-13m

 

[Andrew Mason]

Homework Statement

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?

Homework Equations

potential energy and electric potential energy.

The Attempt at a Solution

following is a simplified attempt at solution

k(Q_p)(Q_a)/r=(.5(M_p)V^2)+(.5(M_a)V^2)
and solved for r and got 1.7X10^-13 m
but i keep getting it wrong could some 1 tell me where i am going wrong
the answer is 1.3X10^-13m

The problem is the frame of reference you are using.

At the point of closest approach, are the two particles stopped in the lab frame? Why, or why not?

AM