How to understand Newton's 3rd Law

[Doc Al]

But using this method velocity after impact of second planetoid would always be zero regardless of how massive or light (smaller mass) is the object it hits

Why do you think that? What do you mean by "using this method"? All we are doing is applying conservation of momentum.

If you would like to see the second bounce off of the first, make the second very light.

 

[ultrauser]

Why do you think that? What do you mean by "using this method"?

I am calculating it like that (different masses example):
planetoid 1 - standing ; planetoid 2 -constant velocity(<-- left direction)
m1=100kg; m2 = 10kg
v1=0m/s ; v2=10m/s
time of impact t=0.001s

F=m*v/t; F=10kg*10m/s / 0.001s = 100 000N

a2=F/m2 = 100 000/10 = 10 000 m/s2 velocity change 10 000m/s2 * 1/1000s = 10 m/s so now planetoid 2 speed (constant speed - velocity change = 10-10=0)

a1=F/m2 = 100 000/100 = 1000m/s2 velocity change 1000m/s2 * 1/1000s = 1m/s planetoid 1 speed

whatever data I always get velocity change equal to constant velocity for second planetoid (resulting in 0m/s)

 

[Doc Al]

I am calculating it like that (different masses example):
planetoid 1 - standing ; planetoid 2 -constant velocity(<-- left direction)
m1=100kg; m2 = 10kg
v1=0m/s ; v2=10m/s
time of impact t=0.001s

Note that the impact force and resulting acceleration is unknown.

F=m*v/t; F=10kg*10m/s / 0.001s = 100 000N

You are merely assuming that the force brings planetoid #2 to a stop. Don't do that.

whatever data I always get velocity change equal to constant velocity for second planetoid (resulting in 0m/s)

That's because you are just assuming that, whether you realize it or not. So of course you get that same answer every time.

Do this instead. Assume a perfectly elastic collision. Then you have enough information to properly calculate the final velocities of both objects without having to make assumptions about the forces involved.

 

[ultrauser]

Do this instead. Assume a perfectly elastic collision. Then you have enough information to properly calculate the final velocities of both objects without having to make assumptions about the forces involved.

Isn't it what I did in that example? What do you mean and how to do it?

You are merely assuming that the force brings planetoid #2 to a stop. Don't do that.

What do you mean by that? I thought if I have data and calculate force than calculation isn't assumption (and force is accurate)

 

[Doc Al]

Isn't it what I did in that example? What do you mean and how to do it?


What do you mean by that? I thought if I have data and calculate force than calculation isn't assumption (and force is accurate)

You merely assumed that the force was 10kg*10m/s / 0.001s = 100,000 N. That's one possibility, but it is not required. Why not calculate the final speed of the first planetoid, given this assumption?

If you make that assumption then of course the second object comes to rest. That's what you assumed to calculate the force.

To calculate final speeds for the case of a perfectly elastic collision, apply both momentum conservation and energy conservation.

 

[ultrauser]

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)

 

[Doc Al]

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.)

 

[Doc Al]

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.

 

[ultrauser]

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?

 

[Doc Al]

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.

 

[ultrauser]

F=10*(10 - -8.18)/0.001 = 181.8/0.001= 181800N ? (probably not xP)

 

[Doc Al]

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.)

 

[ultrauser]

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)