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Relative speed, center of mass, kinetic energy

  • Thread starter alaix
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Homework Statement


Let m1, m2 and m3, the mass of 3 particles and v12, v23, v13 their respective relative velocities.

a) Show that the total kinetic energy of the system in relation to the center of mass is:

( m1m2v12^2 + m2m3v23^2 + m1m3v13^2 ) / (m1+m2+m3)

b) generalize this result


Homework Equations





The Attempt at a Solution




I DID attempt a lot of stuff but I feel like a headless chicken running around here....
I know I won't have any answer before I post any try on here, but a tip on how/where to start would be greatly appreciated
 

Answers and Replies

  • #2
ehild
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I DID attempt a lot of stuff but I feel like a headless chicken running around here....
I tried to imagine you as a headless chicken running around :rofl: ....But DO show some of that lot of stuff you attempted.
First, what is the meaning of those relative velocities v12, v23, v13? How can you express the relative velocities of pairs of particles with their velocities with respect to the CM?

ehild
 
  • #3
ehild
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I think the expression is not correct, it is twice the KE.

But it is a nice tricky problem, so I step further.
The velocities are vectors,

[tex]\vec v_{ij}=\vec v_i-\vec v_j[/tex]

[tex]m_i m_j \vec v_{ij}^2=m_i m_j(\vec v_i^2+\vec v_j^2-2 \vec v_i \vec v_j)[/tex]
The numerator of the original expression is
[tex]N=m_1 m_2\vec v_{12}^2+m_2 m_3\vec v_{23}^2+m_3 m_1\vec v_{31}^2=m_1 m_2(\vec v_1^2+\vec v_2^2-2 \vec v_1 \vec v_2)+m_2 m_3(\vec v_2^2+\vec v_3^2-2 \vec v_2\vec v_3)+m_3 m_1(\vec v_3^2+\vec v_1^2-2 \vec v_3\vec v_1)[/tex]

Upon rearranging ,

[tex]m_1v_1^2 (m_2 +m_3 )+m_2 v_2^2 (m_1+m_3 )+m_3 v_3^2 (m_1+m_2 )-2(m_1 m_2\vec v_1\vec v_2+m_2 m_3 \vec v_2\vec v_3+m_3 m_1\vec v_3\vec v_1)[/tex]

Here comes the trick:

[tex]m_1v_1^2 (m_2 +m_3 )=m_1v_1^2 (m_1+m_2 +m_3 )-m_1^2v_1^2
[/tex]

so

[tex]N=(m_1v_1^2 +m_2v_2^2+m_3v_3^2)(m_1+m_2 +m_3)-(m_1^2 v_1^2+m_2^2 v_2^2+m_3^2 v_3^2+2 m_1m_2 \vec v_1 \vec v_2+2 m_2 m_3 \vec v_2 \vec v_3+2m_3 m_1 \vec v_3\vec v_1)
[/tex]

Let be the total mass M.
The first term is 2M KE. The second term is the square of the total momentum, (Ʃmivi)^2=P^2, and we know that the KE of the CM is KECM=P^2/(2M)

The original expression is

[tex]\frac{m_1m_2 \vec v_{12}^2+m_2 m_3 \vec v_{23}^2+m_3 m_1 \vec v_{31}^2}{M}=2KE-2KE_{CM}[/tex]

twice the KE with respect to the CM. The factor 2 is missing from the original expression.

ehild
 
Last edited:

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