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Abhisar
 one year ago
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.
Abhisar
 one year ago
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.

This Question is Closed

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.8Let'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
 one year ago
Best ResponseYou've already chosen the best response.8\(\huge \bigstar\) 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. \(\sf \Rightarrow \frac{1}{2}{M_aV_a}^2 = \frac{1}{2}M_a{V_a^{'}}^2+\frac{1}{2}M_b{V_b^{'}}^2\) \(\sf \Rightarrow {M_a(V_a}^{2}{V_a^{'}}^2)=M_b{V_b^{'}}^2\) \(\Rightarrow \sf M_a(V_aV_a^{'})(V_a+V_a^{'})=M_b{V_b^{'}}^2\) ....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. \(\sf \Rightarrow M_aV_a=M_aV_a^{'}+M_bV_b^{'}\) \(\sf \Rightarrow M_a(V_aV_a^{'})=M_bV_b^{'}\) ...........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, \(\boxed{\sf {V_a}^{'}=\frac{V_a(M_aM_b)}{M_a+M_b}}\) Substituting this value in Eq.3 we get, \(\sf \boxed{{V_b}^{'}=\frac{2M_1V_1}{M_1+M_2}}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1https://gyazo.com/e1ca809cd98628a8b98808eff6b99869 just a typo

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1and the "ie" here is a non sequitur https://gyazo.com/64cbadb3997368b8e712800f0163f5c7 because it confuses/conflates conservation of momentum with conservation of energy

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.8Thanks @irishboy123 , It should be, \(\sf Va+{V_a}^{'}={V_b}^{'}\)

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0yes, even I was wondering .... Thanks to @irishboy123 And Awesome derivation Abhisar, It was very useful and got a good revision. Thank you

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.8I am glad you found it helpful c:
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