How Do Proton and Alpha-Particle Speeds Affect Their Closest Approach Distance?

[erinec]

Homework Statement

A proton and an alpha-particle are fired directly toward each
other from far away each with an initial speed of 0.01c.
Determine their distance of closest approach measured between
their centres.

Homework Equations

See below.

The Attempt at a Solution

What I did was..

Kinetic Energy of the system = Potential Energy of the system
(.5*m_p*v²) + (.5*m_He*v²) = k*q_p*q_He / r
and then I solved for r, which gave me an incorrect answer.

What am I doing wrong?

I am just wondering.. could you please try this question and tell me what you get?

I get 1.16*10⁵...
which is not the correct answer..

The correct answer is 19.2 fm.;;;

 

[dark adonis]

Homework Statement

A proton and an alpha-particle are fired directly toward each
other from far away each with an initial speed of 0.01c.
Determine their distance of closest approach measured between
their centres.

Homework Equations

See below.

The Attempt at a Solution

What I did was..

Kinetic Energy of the system = Potential Energy of the system
(.5*m_p*v²) + (.5*m_He*v²) = k*q_p*q_He / r
and then I solved for r, which gave me an incorrect answer.

What am I doing wrong?

I am just wondering.. could you please try this question and tell me what you get?

I get 1.16*10⁵...
which is not the correct answer..

The correct answer is 19.2 fm.;;;

Let's start by giving the relevant masses:
$$m_{He} = 6.64 \times 10^{-27} kg$$
$$m_{p} = 1.67 \times 10^{-27} kg$$
Charge:
$$q_{He} = 3.2 \times 10^{-19}C$$
$$q_p = 1.6 \times 10^{-19} C$$
Physical Constants:
$$k =9\times 10^9 \frac{N m^2}{C^2}$$
$$c = 3\times 10^8 m/s$$
Doing some algebra and plugging in these values I get something on the order of 10^-16 m. Are you sure you plugged everything in correct?