#### Answer

The final speed of the proton is $1.7\times 10^7~m/s$ to the left.
The final speed of the carbon atom is $3.1\times 10^6~m/s$ to the right.

#### Work Step by Step

Let $m_A$ be the mass of the proton and let $m_B$ be the mass of the carbon atom.
Let $v_A$ and $v_B$ be the initial velocity of each particle.
Let $v_A'$ and $v_B'$ be the final velocity of each particle.
We can use conservation of momentum to set up an equation.
$m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can set up another equation.
$v_A – v_B = v_B' - v_A'$
$v_A - 0 = v_B' - v_A'$
$v_A' = v_B' - v_A$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$
$v_B' = \frac{2m_A~v_A}{m_A+m_B}$
$v_B' = \frac{2m_A~v_A}{m_A+12m_A}$
$v_B' = \frac{2~v_A}{13}$
$v_B' = \frac{(2)(2.0\times 10^7~m/s)}{13}$
$v_B' = 3.1\times 10^6~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A$
$v_A' = 3.1\times 10^6~m/s - 2.0\times 10^7~m/s$
$v_A' = -1.7\times 10^7~m/s$
The final speed of the proton is $1.7\times 10^7~m/s$ to the left.
The final speed of the carbon atom is $3.1\times 10^6~m/s$ to the right.