A body of mass \(M\) collides elastically with another stationary body of mass \(m\) \((M>m)\) . After impact body were moving at an angle of \(\alpha \) and \(\beta \) with original direction of \(M\). Find Maximum value of \(\alpha \) ?

Please Try and Post a Solution (Not only answers)..
**Please State All Possible Approaches , I believe there are more than 2**

Help Me! Thanks...

Karan

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## Comments

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TopNewestIs the answer \( tan^{-1}\left(\dfrac{m}{\sqrt{M^{2}-m^{2}}}\right) \)

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Yes absolutly...

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yes it is

@Karan Shekhawat - i think the slickest way is to go to COM frame and then return back, thats how i handle all collision problems

in Com frame, angle can vary freely from 0 to pi/2, since the only constraint is that net momentum be 0 and this just requires particles to move in opposite directionsso tan(a)

(a is angle in com frame)= \( \dfrac {v_y}{v_x} \) ,now in COM frame, the speeds should not change after collision(you can easily check) (when it is elastic)(direction may change)now in ground frame

tan(b) = \( \dfrac {v_y}{v_{x1}} \) (because the vertical component is same in both frames)

b is angle in ground frame, also \(v_{x1} \) is horizontal velocity in ground framenow \(v_{x1} = v_x + v_{com} \)

further let v=u (initial speed in ground frame)

so we have \( v_x = \dfrac {mucos(a)}{m+m} \quad and \quad v_c = \dfrac {Mu}{M+m} \)

substituting we have

\( tan(b) = \dfrac {sin(a)}{cos(a)+\dfrac {M}{m}} \) which is maximum at cos(a) = \( \dfrac {-m}{M} \) (the minus sign is crucial)

the answer is

sin(b)= \( \dfrac {m}{M} \)as ronak wrote (except in tan)

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Ahh .... it's also cool ...Thank you saketh bro ... :)

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This is a very good problem that I had done a long time ago. The method that I know of is straightforward and there is no 'trick' involved. I don't see the point in typing out the whole thing again as this problem has been extensively discussed in a few online forums.

This has the required result and a full derivation. Hope this helps :)

@Karan Shekhawat

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Ahhh.... Squaring Part is Tricky ... I was Fed up when I did not able to seprate \(\alpha \) ... But Yes It should be come in my mind that I should square the momentum equation ....

Thanks again and again :)

Btw what is That site " Fouram " ? Is it just like Brilliant.org ?

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No no :P ...a forum is actually any place or website where people discuss ideas ...It's a general term and doesn't refer to a particular website...examples include math-stack exchange, physics-stack exchange, AOPS etc.

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Btw I have one Solution .. which I didn't understand , Can You please Help me in that also ?

Method is C-frame ... And Then They used COLM and Using relative velocity and state maximum velocity occure when V1/cm is perpendicular to V1 or Vcm .... Something like that ... !

But I didn't understand that .... Please Help me in that also

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Ahh, so there is another method...I'll try and find an argument along the lines you've mentioned...If I get any idea, I'll get back to you on it. :)

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Isn't the answer 180 degree.

Just do a head on collision.

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It is not possible for \(M>m\).

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The mass that comes and hits has greater mass and thus a recoil is not possible in this case. (this case would infact give the least value of deviation,\(\alpha=0\))

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@Shashwat Shukla @Raghav Vaidyanathan @Mvs Saketh @Ronak Agarwal @Krishna Sharma

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