How to understand Newton's 3rd Law

  • Thread starter ultrauser
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In summary: So, even though the car travels at constant speed, the car is accelerating because it's direction of travel is constantly changing. It's constantly "turning" because it's constantly following the curve of the earth.The first car (the one that gets hit), is not moving, so when it gets hit it's just like you getting hit by a car while standing still. Your body accelerates backwards. The car accelerates backwards.When I throw a rock I cause a force on it and a rock causes equal force on me. So if I would stay at frictionless surface would I move backward when throwing rock?Yes.
  • #36
The same example as above:

u-velocities before interaction
v-velocities after interaction

So I know
m1=100kg u1=0m/s
m2=10kg u2=10m/s


\begin{equation}

v_1 = \frac{u_1(m_1-m_2) + 2m_2u_2}{m_1+m_2};\\
v_2 = \frac{u_2(m_2-m_1) + 2m_1u_1}{m_1+m_2}; \end{equation}

\begin{equation}
v_1 = \frac{0(100-10) + 2*10*10}{100+10};\\
v_2 = \frac{10(10-100) + 2*100*0}{100+10}; \end{equation}

\begin{equation}
v_1= \frac{200}{110}=1.(81)\frac{m}{s}
\end{equation}

\begin{equation}
v_2= \frac{-900}{110}=-8.(18)\frac{m}{s}
\end{equation}

So does it mean first planetoid will now travel at 1.(81)m/s in left direction and second at -8.(18)m/s in right direction (I guess "-" mean it will reflect and travel into direction opposite to the direction before impact) ?

These are very different results from attempt I made before and I used the same data ( I can't get it why previous method gives such different results)
 
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  • #37
ultrauser said:
So does it mean first planetoid will now travel at 1.(81)m/s in left direction and second at -8.(18)m/s in right direction (I guess "-" mean it will reflect and travel into direction opposite to the direction before impact) ?
Exactly. The smaller mass bounces back.

These are very different results from attempt I made before and I used the same data ( I can't get it why previous method gives such different results)
The given data (masses and initial speeds) are insufficient to determine the final speeds. Here, we assumed an elastic collision and got the results consistent with that assumption. Earlier you assumed an impulse (force X time) during the collision that would stop the second planetoid and thus got results consistent with that assumption. (If you work out the speeds for that earlier case you will find that energy is not conserved.)
 
  • #38
Just for fun. For the elastic collision case, assume that the collision took exactly 1 ms (just like your previous example). Figure out the average force during the collision and compare that value to what you had earlier.
 
  • #39
Doc Al said:
Just for fun. For the elastic collision case, assume that the collision took exactly 1 ms (just like your previous example). Figure out the average force during the collision and compare that value to what you had earlier.

I don't know how to do it (what formulas should I use) ?

What I calculated earlier was average force needed to stop second planetoid (F= m*v/t ?) instead of force of actual impact?
 
  • #40
ultrauser said:
I don't know how to do it (what formulas should I use) ?
You'd find the average force using mΔv/Δt.

What I calculated earlier was average force needed to stop second planetoid (F= m*v/t ?) instead of force of actual impact?
Right. You calculated the average force assuming that the second planetoid was stopped.
 
  • #41
F=10*(10 - -8.18)/0.001 = 181.8/0.001= 181800N ? (probably not xP)
 
  • #42
ultrauser said:
F=10*(10 - -8.18)/0.001 = 181.8/0.001= 181800N ?
Right. When it bounces back elastically, the average force (with our simple assumption) is almost twice as much as before. (Of course there's no real reason to think the collision time would be the same, but it makes the point.)
 
  • #43
There is one more thing I would like to know, what conditions needs to be met so the small planetoid m=10kg would hit big planetoid m=100kg and mantain it's direction (instead of being reflected)
 

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