Collision between 3 masses, with spring

[Karol]

Homework Statement

The masses C, of magnitude 2m, and B, of magnitude m, move with velocity V and hit a stationary mass A of magnitude m. the collision lasts a very short time and is plastic but the masses don't stick together. the spring with constant k is ideal and the surface is smooth.
What's the max acceleration of C during the contact between A and B
Box A continues, after it separates, with constant velocity, what's it.
What's the max contraction of the spring after A and B part?
How long did the contact between A and B take place?

Homework Equations

Kinetic energy, potential energy of a spring: ##E_p=\frac{1}{2}kx^2##
Impulse-momentum: ##mv=Ft##
Conservation of momentum: ##m_1v_1+m_2v_2=m_1v_1'+m_2v_2'##

The Attempt at a Solution

I divide into 2 collisions. the first is plastic between boxes A and B in which the members cling to one another. the final velocity of the boxes A and B is ##\frac{V}{2}## and i find it from conservation of momentum between box B and both boxes after the collision:
$$mV=2mv\;\rightarrow\; v=\frac{V}{2}$$
Now these 2 boxes collide elastically with C. the spring contracts, expands and reaches, again, the initial relaxed length L₀. then and there the boxes A and B part and A continues alone.
One system consists of the boxes A and B and the other system is box C alone. conservation of energy+momentum:
$$\begin{cases}\frac{1}{2}(2m)V^2+\frac{1}{2}(2m)\left( \frac{V}{2} \right)^2=\frac{1}{2}(2m)v_1^2+\frac{1}{2}(2m)v_2^2 \\ 2mV+2m\frac{V}{2}=2mv_1+2mv_2 \end{cases}\;\Rightarrow\; v_1=\frac{V}{2},\; v_2=V$$
This is also A's velocity after it detaches, which happens at distance L₀ again.
At some point, when the spring contracts to it's max, A and B halt momentarily. from the point of view of an observer on C all A and B's kinetic energy gets into the spring. A and B's velocity relative to C is ##\frac{V}{2}##:
$$\frac{1}{2}(2m)\left( \frac{V}{2} \right)^2=\frac{1}{2}kx^2\;\rightarrow\; x=\sqrt{\frac{m}{2k}}V$$
C's max acceleration:
$$ma=F=kx\;\rightarrow\;2m\cdot a=\sqrt{\frac{m}{2k}}V\;\rightarrow\;a=\sqrt{\frac{mk}{2}}\frac{V}{2m}$$
After A has left, B stretches the string and returns. this is another collision:

$$\begin{cases}\frac{1}{2}mV^2+\frac{1}{2}(2m)\left( \frac{V}{2} \right)^2=\frac{1}{2}(2m)v_C^2+\frac{1}{2}mv_B^2 \\ mV+2m\frac{V}{2}=2mv_C+mv_B \end{cases}\;\Rightarrow\; v_C=\frac{5}{6}v,\; v_B=\frac{V}{3}$$
Now it compresses the spring, how much?
The distance between B and C is x. i want to find x as a function of t: ##x=f(t)## so that i will be able to differentiate it and find the maximum contraction. i can't. i use a simplified setting: a spring attached to a wall.

Lets say x is at distance x₀ from the wall, the velocity there is v₀ and it advances distance dx. the acceleration is ##F=kx=ma\;\rightarrow\; a=\frac{k\cdot x_0}{m}##. the acceleration in the interval dx is considered constant, and the mean velocity in dx is:
$$v-\frac{k\cdot x_0}{m}dt$$
from kinematics:
$$dx=\left( v-\frac{k\cdot x_0}{2m}dt \right)dt\;\rightarrow\;\frac{dx}{dt}=v_0-\frac{k\cdot x_0}{2m}dt$$
$$\int \frac{dx}{dt} dt=\int dx=x=\int v_0 dt-\int \frac{k\cdot x_0}{2m}dt^2$$
I don't know to handle dt², if everything is correct till there, of course.

 

[haruspex]

A and B halt momentarily. from the point of view of an observer on C

I assume that full stop should not be there, that you are saying A and B halt momentarily from the point of view of an observer on C

At some point, when the spring contracts to it's max, A and B halt momentarily. from the point of view of an observer on C all A and B's kinetic energy gets into the spring. A and B's velocity relative to C is V2V2\frac{V}{2}:

That argument worries me. You are using a non-inertial frame. But KE is different in different inertial frames, so a non-inertial frame means energy is not conserved.
Consider the common mass centre and speeds relative to it. I think you'll find the spring PE is half what you calculated.

For the motion after A and B part, it must be SHM relative to the common mass centre, no?

 

[Karol]

With A and B attached, The velocity of COM:
$$v_{cm}=\frac{2mV+2m\frac{V}{2}}{4m}=\frac{3}{4}V$$
In the COM frame the masses approach and move away from the COM at the same velocity:

So there is a momentary halt. the total kinetic energy:
$$E_k=2\cdot 2m\cdot\frac{V^2}{16}=\frac{1}{4}V^2$$
This is the compressed spring's energy:
$$\frac{1}{4}V^2=\frac{1}{2}kx^2\;\rightarrow\; x=\frac{V}{\sqrt{2k}}$$
$$F=kx\;\rightarrow\; 2ma=k\frac{V}{\sqrt{2k}}\;\rightarrow\;a=\frac{\sqrt{2k}V}{4m}$$
I understand intuitively that there is SHM, but why? is it because the masses move symmetrically in the COM or just because the restoring force is F=kx?
Harmonic motion:
$$\dot x=-\omega A\sin(\omega t+\theta),\; \omega=\sqrt{\frac{k}{m}}=\frac{2\pi}{T}$$
In the COM system at t=0 the spring is loose, ##\dot x=\frac{V}{4}## and ##\theta=\frac{3}{2}\pi##
The COM is one third the distance between B and C, and there the spring stationary while aside that point it contracts\expands. looking at B from that point gives:
$$\omega=\sqrt{\frac{\frac{3}{2}k}{m}}=\sqrt{\frac{3k}{2m}}$$
$$\frac{V}{4}=\omega A=\sqrt{\frac{k}{m}}A\;\rightarrow\;A=\frac{V}{4}\sqrt{\frac{2m}{3k}}$$
The time A and B are in contact is found also from the second collision harmonic motion conditions:
$$\omega=\sqrt{\frac{2k}{2m}}=\sqrt{\frac{k}{m}}=\frac{2\pi}{T}\;\rightarrow\;T=2\pi\frac{m}{k}$$
The time A and B are in contact is T.

 

[haruspex]

In the COM frame the masses approach and move away from the COM at the same velocity:

But the masses are different now. In the frame of the common mass centre, momentum is zero.

 

[Karol]

The velocity of COM ##\frac{3}{4}V##, the velocities of C and B in the COM are ##\frac{1}{4}V##
The reduced mass:
$$\overline{m}=\frac{m_1m_2}{m_1+m_2}=\frac{2m^2}{3m}=\frac{2}{3}m$$
$$\omega=\sqrt{\frac{k}{\overline{m}}}=\sqrt{\frac{3k}{2m}}$$
$$\omega=\frac{2\pi}{T}\;\rightarrow\; T=\frac{2\pi}{\omega}$$
T, the period, is independent of coordinate systems, so why does it change, since i use ##\overline{m}##?
The mass is an inherent property of C and B. i know i can express momentum with ##\overline{m}##: ##P=\overline{m}V_{12}##, but what else?

 

[haruspex]

why does it change, since i use ##¯\overline{m}##?

Change? I did not see a different value, except via a line of reasoning that I regarded as wrong.
Another way to get to this is to observe that the mass m is effectively attached to the common mass centre by a spring of constant 3k/2, and 2m by a spring of constant 3k. Both lead to a frequency ##\sqrt{\frac{3k}{2m}}##

 

[Karol]

So the result:
$$\frac{V}{4}=\omega A=\sqrt{\frac{k}{m}}A\;\rightarrow\;A=\frac{V}{4}\sqrt{\frac{2m}{3k}}$$
Is correct, right? and also correct is the time A and B were in contact::
$$T=2\pi\frac{m}{k}$$

 

[haruspex]

So the result:
$$\frac{V}{4}=\omega A=\sqrt{\frac{k}{m}}A\;\rightarrow\;A=\frac{V}{4}\sqrt{\frac{2m}{3k}}$$
Is correct, right? and also correct is the time A and B were in contact::
$$T=2\pi\frac{m}{k}$$

The max compression is for after separation, right? The speeds cannot both be V/4 relative to the mass centre then.

For the time in contact, they are only in contact for half a cycle, no?

 

[Karol]

After separation ##v_{cm}=\frac{2}{3}V##. the relative velocity between C and B ##v_{rel}=\frac{V}{3}##
If i want to use the reduced mass ##\overline{m}=\frac{2}{3}m## then:
$$\frac{V}{3}=\omega A=\sqrt{\frac{3k}{2m}}\cdot A\;\rightarrow\; A=\frac{V}{3}\sqrt{\frac{2m}{3k}}$$
If i use the partial spring method, there is 1/3 spring between COM and C, i get the same A.
For the time they are in contact it's half cycle because B collides A when the spring is loose, at L₀, and they separate at L₀ again.

 

[haruspex]

After separation ##v_{cm}=\frac{2}{3}V##. the relative velocity between C and B ##v_{rel}=\frac{V}{3}##

How do you get that relative velocity? Why does separation suddenly change it?

 

[Karol]

How do you get that relative velocity? Why does separation suddenly change it?

Correction, ##v_{rel}=\frac{V}{2}##, but that doesn't change the result for A.

 

[Karol]

The kinetic energy in the COM frame can be calculated using the reduced mass:
$$E_k=\frac{1}{2}\overline{m}V_{12}^2=\frac{1}{2}\frac{2}{3}m\frac{V^2}{4}=\frac{1}{12}mV^2$$
And it differs from the KE i found in post #3, where i also had a mistake:
$$E_k=2\frac{1}{2}\cdot 2m\cdot\frac{V^2}{16}=\frac{1}{8}V^2$$

 

[haruspex]

The kinetic energy in the COM frame can be calculated using the reduced mass:
$$E_k=\frac{1}{2}\overline{m}V_{12}^2=\frac{1}{2}\frac{2}{3}m\frac{V^2}{4}=\frac{1}{12}mV^2$$

Yes.
But to get the maximum contraction after separation, I feel that working in the COM frame only complicates things. You can just consider the loss in KE in going from point of separation to when B and C are at the same velocities.

 

[Karol]

You can just consider the loss in KE in going from point of separation to when B and C are at the same velocities.

How do i find the common velocity of C and B after A parts?

 

[haruspex]

How do i find the common velocity of C and B after A parts?

Conservation of momentum on B+C.

 

[Karol]

Conservation of momentum on B+C.

$$2m\frac{V}{2}+mV=3mv\;\rightarrow\; v=\frac{2}{3}V$$
$$E_{k\ initial}=\frac{mV^2}{2}\left( \frac{1}{4}+1 \right)=\frac{5}{8}mV^2$$
$$E_{k\ final}=\frac{mV^2}{2}\frac{4}{9}=\frac{2}{9}mV^2$$
$$\Delta E_k=\frac{13}{72}mV^2$$
In post #3 i got:
$$\Delta E_k=\frac{1}{4}mV^2$$

 

[haruspex]

##E_{k final}=\frac{mV^2}{2}\frac 49##

The mass is 3m here. Similarly, in the E_initial equation, one mass is 2m.

 

[Karol]

$$E_{k\ initial}=\frac{mV^2}{2}\left(2 \frac{1}{4}+1 \right)=\frac{3}{4}mV^2$$
$$E_{k\ final}=\frac{3mV^2}{2}\frac{4}{9}=\frac{2}{3}mV^2$$
$$\Delta E_k=\frac{1}{12}mV^2$$
Still different than in #3

 

[haruspex]

$$E_{k\ initial}=\frac{mV^2}{2}\left(2 \frac{1}{4}+1 \right)=\frac{3}{4}mV^2$$
$$E_{k\ final}=\frac{3mV^2}{2}\frac{4}{9}=\frac{2}{3}mV^2$$
$$\Delta E_k=\frac{1}{12}mV^2$$
Still different than in #3

I was never sure what the KE calculated as mV²/8 in #3, and corrected to mV²/12 in #12, referred to. I gather it was KE of A+B+C in "the com frame", but which com frame?
The mV²/12 in #12 is for the KE that goes into compression of the spring after collision. That agrees with the lost KE (of B and C) computed in #18.

 

[Karol]

There is still a problem with the amplitude A i got in #9
$$A=\frac{V}{3}\sqrt{\frac{2m}{3k}}=\frac{2}{3}\frac{1}{\sqrt{6}}V\sqrt{\frac{m}{k}}$$
While with the KE method:
$$\Delta E_k=\frac{1}{12}mV^2=\frac{1}{2}kx^2\;\rightarrow\; x=\frac{1}{\sqrt{6}}V\sqrt{\frac{m}{k}}$$

 

[haruspex]

There is still a problem with the amplitude A i got in #9
$$A=\frac{V}{3}\sqrt{\frac{2m}{3k}}=\frac{2}{3}\frac{1}{\sqrt{6}}V\sqrt{\frac{m}{k}}$$
While with the KE method:
$$\Delta E_k=\frac{1}{12}mV^2=\frac{1}{2}kx^2\;\rightarrow\; x=\frac{1}{\sqrt{6}}V\sqrt{\frac{m}{k}}$$

In your post #11 you agreed that the relative velocity V/3 you quoted in post #9 should have been V/2, but maintained that it did not change the result for A. I did not check that at the time, but looking at it now I believe it does change the result for A and leads to agreement with the KE method.

 

[Karol]

So to solve for 2 masses with a spring must i use the reduced mass ##\over{m}## only?
Is there also an other method?