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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?

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2 Answers 2

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Normally you solve an elastic collision with just momentum and energy conservation, because you really don't know what happens at impact. The formulas are given in this answer. For equal masses in one dimension the velocities are exchanged.

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.

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  • $\begingroup$ It is strange that whatever time of impact the velocity change for both objects is always equal to velocity of second object before impact (it would mean second object would always stop after hitting the first regardless of velocity or time) $\endgroup$
    – user82115
    Commented Jun 28, 2014 at 0:26
  • $\begingroup$ But it is correct if they are the same mass and the impact is on center (which makes it 1D). When you use energy/momentum conservation you ignore the details of the interaction-you just assume it doesn't dissipate any energy. In your example, the duration could double, the force cut in half, and you would come out the same. $\endgroup$
    – Ross Millikan
    Commented Jun 28, 2014 at 1:29
  • $\begingroup$ But according to that if I would hit object with greater or smaller mass I would always have 0 velocity after impact :| (when I threw rock at massive wall it is reflected) $\endgroup$
    – user82115
    Commented Jun 28, 2014 at 1:53
  • $\begingroup$ No, if the objects are different masses, after impact they will both be moving. If the heavy one hits the light, they both move in the direction the heavy one was going. If the light hits the heavy, it rebounds. You assumed a time of impact, then calculated the force to stop the incoming rock. That cannot be correct if the masses don't match. $\endgroup$
    – Ross Millikan
    Commented Jun 28, 2014 at 2:14
  • $\begingroup$ Whatever the time of impact I assume I always get zero velocity for second body after impact. What is wrong here? $\endgroup$
    – user82115
    Commented Jun 28, 2014 at 4:57
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If the masses are the same, then as stated above, the initially moving mass stops and the other one acquires the velocity of the initially moving one. This is the mechaism behind the so called "Newton's cradle". BTW, the collision has to be head on, ie take place in one dimension.

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