Author Topic: 1D collision : Conservation of Momentum  (Read 254379 times)

Fu-Kwun Hwang

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1D collision : Conservation of Momentum
« on: January 29, 2004, 05:14:42 pm »
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Newton's law of motion look the same to all observers in inertial frames of reference.
It is equally true that if momentum is conserved in one inertial reference frame, it is conserved in all inertial frames.

This java applet apply the above concept to one dimensional collision problem.
Two circular objects are confined to move in one diminution (between two gray blocks).
Press start button to run the animation.
    Circular objects will move with some predefined velocity (yellow arrow).

Click the mouse button to pause. Click it again to resume the animation.

While the animation is suspended:
    Click near the arrow of the velocity vector and drag it left/right to change the initial velocity.
    Click at the center of the circle to move it : drag the object left <--> right.
    Click within the circle to change the mass of that object.
      Click right mouse button to increase mass by one unit.
      Click left mouse button to decrease mass one unit.


Press Reset to reset most parameters to default values.
    eta is the coefficient of restitution
       eta = | relative velocity just after collision/relative velocity just before collision |
    for elastic collision eta=1., for perfectly inelastic collision eta=0.

You can select different frame of reference to view the relative motion of all the objects.
   lab is a laboratory inertial frame.
   $m_1, m_2$and CM are frame of reference with respect to left circular object $m_1$, right circular object $m_2$ and center of mass for $m_1$ and $m_2$.

The velocity of two objects after collision ($V'_1,V'_2$)can be calculated from velocity before collisions ($V_1,V_2$) and mass of two objects ($m_1,m_2$).

From conservation of momentum $m_1 V_1+m_2 V_2=$ $ m_1 V'_1+m_2 V'_2 $,
and conservation of energy $   frac{1}{2}m_1V_1^2+   frac{1}{2}m_2V_2^2=   frac{1}{2}m_1V_1'^2+   frac{1}{2}m_2V_2'^2$
So $m_1 (V_1-V_1')=m_2(V_2'-V_2)$
and $   frac{1}{2}m_1 (V_1^2-V_1'^2)=   frac{1}{2}m_2 (V_2'^2-V_2^2)$, which means $   frac{1}{2}m_1 (V_1-V_1')(V_1+V_1')=   frac{1}{2}m_2 (V_2'-V_2)(V_2'+V_2)$
So $V_1+V_1'=V_2'+V_2$

i.e. The equation need to be solved are
$m_1 V_1'+m_2 V_2'=$ $m_1 V_1+m_2V_2$ and $V_2'-V_1'=V_2-V_1$

The result is
$V'_1= frac{m_1-m_2}{m_1+m_2} V_1 +frac{2m_2}{m_1+m_2}V_2=V_{cm}+frac{m_2}{m_1+m_2}(V_2-V_1)=2V_{cm}-V_1$
and $V'_2=frac{2m_1}{m_1+m_2}V_1+frac{m_2-m_1}{m_2+m_1}V_2=V_{cm}+frac{m_1}{m_1+m_2}(V_1-V_2)=2V_{cm}-V_2$
where $V_{cm}=frac{m_1V_1+m_2V_2}{m_1+m_2}$

It means that $V'_1-V_{cm} = - (V_1-V_{cm})$ and $V'_2-V_{cm}= - (V_2-V_{cm})$
 or $V'_{1cm}= -V_{1cm}$ and $V'_{1cm}= -V_{1cm}$ where $V'_{1cm}=V'_1-V_{cm}$ ...etc.
From the point of center of mass coordinate system: both particles bounce back with the same speed (relative to center of mass).

-*-

You are welcomed to check out collision in 2 dimension.


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topic5
« Reply #1 on: January 30, 2004, 12:10:16 pm »
Subject: Conservation of Linear Momentum
Date: Tue, 28 Jul 1998 09:14:28 +0800
From: Lim Ai Phing <zuc@moe.edu.sg>
Reply-To: lap@pacific.net.sg
To: hwang@phy03.phy.ntnu.edu.tw
hi!!
I'm a physics teacher presently teaching my students dynamics.
Your collision java applet is really good, I'll ask my students to take a look here!!
But can I suggest that the colour of the numbers be changed
because yellow doesn't show up much against a gray background?
Aiping

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circles
« Reply #2 on: April 01, 2004, 11:04:35 pm »
The circles should be different colors to make it easier to differentiate between them.

Fu-Kwun Hwang

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Thank you for your suggestion
« Reply #3 on: April 10, 2004, 02:53:40 pm »
As you wish! I hope you like it better now!

andrewsmith

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topic5
« Reply #4 on: November 27, 2004, 01:59:44 am »
Could you use some visual indication of relative mass? Maybe increase the size of the masses or the thickness of the circle?

Javed

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Re: 1D collision : Conservation of Momentum
« Reply #5 on: June 13, 2007, 08:42:33 pm »
great program, can u please email me the source code, thanks

Fu-Kwun Hwang

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Re: 1D collision : Conservation of Momentum
« Reply #6 on: June 20, 2007, 12:57:55 pm »
You should have received the source code now.

Javed

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Re: 1D collision : Conservation of Momentum
« Reply #7 on: June 20, 2007, 04:31:06 pm »
thank you

afg

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Re: 1D collision : Conservation of Momentum
« Reply #8 on: December 22, 2008, 05:16:16 pm »
Hi,

Your program is great. Can I get the source code for it?

Thank you.

Fu-Kwun Hwang

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Re: 1D collision : Conservation of Momentum
« Reply #9 on: December 22, 2008, 05:18:01 pm »
Please check out the attachment under the first message.

ArdTraveller

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Re: 1D collision : Conservation of Momentum
« Reply #10 on: January 06, 2009, 07:30:14 pm »
Yeah it's a fantastic simulation can I get the source code too?THANKS

Fu-Kwun Hwang

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Re: 1D collision : Conservation of Momentum
« Reply #11 on: January 06, 2009, 09:26:13 pm »
The source code is available as an attachment at the first message. I do not know why you did not see it?  :o :o ::)

Question

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Re: 1D collision : Conservation of Momentum
« Reply #12 on: January 14, 2009, 01:45:51 pm »
Hello sir

since u gave the source code for this 1D collision

I would be happy if u could provide the source code for the Collision 2D
(the code that is not generated by EJS), coz it turns out that if i rewrite the code again in Netbeans i can't compile it, i want the output 2 be like the one for this 1D collision, meaning the source code of this 1D when i rewrote in Netbeans I could compile and run but not Collision 2D, thx 4 your help.

Fu-Kwun Hwang

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Re: 1D collision : Conservation of Momentum
« Reply #13 on: January 14, 2009, 05:43:57 pm »
You will find the source code for JDK1.0.2 version collision 2D applet in the downloaded ZIP file for that applet.

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Re: 1D collision : Conservation of Momentum
« Reply #14 on: February 16, 2009, 10:19:33 am »
Sir what does the greenish x that is always in between the circular objects signify, and at what code line that enables it to move with the objects sir?