# Calculating velocity change after impact?

Let's say there is no gravity here and objects won't crush. We have 2 rocks with $m=10\text{ kg}$. First rock has velocity $v_1=0\text{ m/s}$ and second $v_2=10\text{ m/s}$ (flying in leftward direction). The second rock hit first causing force for a time $t=0.001\text{ s}$.

$F=ma \rightarrow F=\frac{mv}{t} \rightarrow F=\frac{10*10}{0.001} = 100000\text{ N}$

$a=F/m= 100 000/10 = 10 000\text{ m/s}^2$

So acceleration is $10 000\text{ m/s}^2$ in time $0.001\text{ s}$ meaning change in velocity would be $10\text{ m/s}$? It would mean first rock would travel at velocity $10\text{ m/s}$ (flying in leftward direction) and the second would stop $(10-10=0)$ but it seems against all logic.

I probably messed up something here so anyone can help?

It turns out your interaction time and acceleration multiply to get the correct velocity change, but since you didn't insist on momentum and energy conservation it might not. When you calculated the acceleration as $\frac {mv}t$, you assumed that the first rock would stop. That is correct here, but if the second rock had a different mass your calculation would look just the same.