Double collision of spring mass system

The velocity of COM = V₂2 = V₄ ........( let )

Now , the kinetic energy of the balls must be converted totally into the potential energy of the spring after everything has stopped : )

So , 12 K A ² = 12 m V₄²

Or , A = V₀ Mω ( M + m )

Now , the motion of the ball 2 can be completely described as : -

x₂ ( t ) = x₂ ( t ) [ In COM frame ] + x_COM ( t )

= V₀ Mω ( M + m ) sin ( ω t ) + V₀ M tM + m

= V₀ M tM + m [ 1 + sin ( ω t )ω t ]

whereas , ........ x₁ ( t ) = V₀ t ( M - mM + m )

For collision to occur between balls 1 and 2 , we obviously need to have

x₁ = x₂

Sir ,

Mm = - sin xx = ξ ...............( 1 )

How is -

M = mξ _Min. ?

Later , I found out the mistake -

mM = - sin xx ...................instead of ( 1 )

Thanks Sir .

Once you get the equation
sin x x= - ξ
and you are looking for the min of ξ so that a solution may exist, just look for the absolute minimum of the function sin xx. This is easier said than done.
One way is to differentiate w.r.t. x and equate the derivative to zero. This gives the transcendental equation
tan x = -x.
There is no algebraic solution, but graphical means give x ≈ 2.02875 (which is very roughly 2/3π).
From here we get M = m/ξ_min ≈ 10 kg. (This again you taken wrongly, ricky. You found M = m ξ_min which won't be 10kg)

Well yes, I overlooked the interval you specified (and the last statement was a blind guess).
But your assertion that the min is 2/3Ï€ is not correct. If you see the answer I gave was only approximate.

fantastic work ricky......

& an excellent ques from anant sir,,,

EDIT - POST NUMBER 10 : LINE NUMBER 5 SHOULD BE READ AS : -

M V₀² = M V₁² + m V₂²

Cancelling appropriate terms , we obtain -

Mm = - sin ( ω t )ω t = ξ ......... ( let )

Now we need to minimise " ξ " .

Jordan ' s inequality states that , for any " x E ( 0 , π2 ) " ;

sin xx ≥ 2π

Just changing the interval suitably , we can derive the following : - For any " x E ( π , 3 π2 ) " ;

- sin xx ≥ 23 π

Hence , ξ _Min. = 23 π

Consequently , M = m ξ _Min. = 2 m3 π ≈ 10 Kg

approach is okay... however the graphs are not the ones you are looking for......
and
M\approx \dfrac{2}{3\pi}m

Here , ω = √ K √ m , where " K = New spring constant " .

Immediately , one would ask , " why is there a new spring constant " ? It is because the effective length of the spring has been halved in COM frame . Accordingly , since the spring constant is inversely proportional to the length of the spring , hence ,

K = 2 k

Now the quest is to find the amplitude .

Let V₁ and V₂ be the velocities of the respective balls ( 1 and 2 ) after the first collision .

Of course , after the first collision , the speeds of balls 2 and 3 must be equal .

Momentum and energy conservation gives us -

M V₀ = m V₂ + M V₁

M V₀² = m V₂² + M V₀² )

Easily solving for the assumed velocities , we find -

V₁ = V₀ ( M - mM + m )

V₂ = V₀ ( 2 MM + m )

If we now fix the reference frame to the center of mass of the two connected balls , it is easy to realize that the masses are just contracting and expanding in that frame , i . e , they are performing S . H . M .

Now , it has to be noted that the equation of the S . H . M . has to be a " sine " curve , since , at " t = 0 " , there was no S . H . M .

Let the equation of the motion with respect to COM be - x ( t ) = A sin ( ω t )

Is this going to end in the archives?

Thanks yaar , but now how to proceed ?

I am really very sorry if I hurt your emotions , very sorry . But in fact , I just wanted to know how to bring in that " sin x " , that ' s all . I know that it ' ll execute SHM with a reduced mass frequency , but where does " sin x " come in ? Please , this is my doubt .

I am weak in Physics [2][2] , so again sorry if I hurt you .

can u post the solution sir?