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- Abhisar

I was teaching a kid about Elastic Head on collision and he was having some trouble deriving few relations so I am doing this post to help him and others looking for similar content.
\(\bigstar\) There is a typographical mistake below in Eq.3. The correct equation should be
\(\sf Va+{V_a}^{'}={V_b}^{'}\)
\(\bigstar\) There is one another typographical mistake in the statement just below Eq.1. The correct statement should be:
Also, in both elastic or inelastic collision momentum is conserved i.e. total initial momentum of the system is equal to the total final momentum of the system.

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- Abhisar

- jamiebookeater

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- Abhisar

Let's suppose that a body of mass \(\sf M_a\) is travelling with a velocity \(\sf V_a\) along a straight line and it collides with another stationary body of mass \(\sf M_b\). Consider the collision to be elastic in nature.

- Abhisar

\(\huge \bigstar\) Derive an equation for final velocities \(\sf V_a^{'}~and~V_b^{'}\) in terms of \(\sf M_a, M_b~and~V_a\).
Since, the collision is elastic we can say that final kinetic energy of the system is equal to the initial kinetic energy.
\(\sf \Rightarrow \frac{1}{2}{M_aV_a}^2 = \frac{1}{2}M_a{V_a^{'}}^2+\frac{1}{2}M_b{V_b^{'}}^2\)
\(\sf \Rightarrow {M_a(V_a}^{2}-{V_a^{'}}^2)=M_b{V_b^{'}}^2\)
\(\Rightarrow \sf M_a(V_a-V_a^{'})(V_a+V_a^{'})=M_b{V_b^{'}}^2\) ....Eq.1
Also, in both elastic or inelastic collision momentum is conserved i.e. total initial energy of the system is equal to the total final energy of the system.
\(\sf \Rightarrow M_aV_a=M_aV_a^{'}+M_bV_b^{'}\)
\(\sf \Rightarrow M_a(V_a-V_a^{'})=M_bV_b^{'}\) ...........Eq.2
Dividing Eq.1 with Eq.2 we get,
\(\sf V_a+{V_b}^{'}={V_b}^{'}\) .........Eq.3
Substituting this value in Eq.2 we get,
\(\boxed{\sf {V_a}^{'}=\frac{V_a(M_a-M_b)}{M_a+M_b}}\)
Substituting this value in Eq.3 we get,
\(\sf \boxed{{V_b}^{'}=\frac{2M_1V_1}{M_1+M_2}}\)

- IrishBoy123

https://gyazo.com/e1ca809cd98628a8b98808eff6b99869
just a typo

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- IrishBoy123

and the "ie" here is a non sequitur
https://gyazo.com/64cbadb3997368b8e712800f0163f5c7
because it confuses/conflates conservation of momentum with conservation of energy

- Abhisar

Thanks @irishboy123 , It should be,
\(\sf Va+{V_a}^{'}={V_b}^{'}\)

- arindameducationusc

yes, even I was wondering .... Thanks to @irishboy123
And Awesome derivation Abhisar,
It was very useful and got a good revision.
Thank you

- Abhisar

I am glad you found it helpful c:

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