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.

Question

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 · Answer

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 · Answer

$\huge \bigstar$ 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 · Answer

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

irishboy123 · Answer

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

abhisar · Answer

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

arindameducationusc · Answer

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

abhisar · Answer

I am glad you found it helpful c: