# Physics Challange

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

Note by Karan Shekhawat
4 years, 10 months ago

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

- 4 years, 10 months ago

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 directions

so 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 frame

now $v_{x1} = v_x + v_{com}$

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)

- 4 years, 10 months ago

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

- 4 years, 10 months ago

Yes absolutly...

- 4 years, 10 months ago

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 :)

- 4 years, 10 months ago

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 ?

- 4 years, 10 months ago

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.

- 4 years, 10 months ago

oopes Proved myself Foolish. .. :) But I didn't Know about them earliear ...

- 4 years, 10 months ago

That's quite alright :)

- 4 years, 10 months ago

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 ... !

- 4 years, 10 months ago

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. :)

- 4 years, 10 months ago

Yes Please .. Thanks ! Please I have not tell exact correct lines ... In fact there may be some error in my lines .. I have tell you just rough idea as far as I remmeber .. So Please Don't blame me if You found anything wrong in my lines ... :)

- 4 years, 10 months ago

No worries... the keywords would only be 'relative frame' and 'maximum when perpendicular'....The rest I don't care about :D

- 4 years, 10 months ago

Thanks ... Take your time , ... I 'am waiting !

- 4 years, 10 months ago

- 4 years, 10 months ago

Just do a head on collision.

- 4 years, 10 months ago

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,$\alpha=0$)

- 4 years, 10 months ago

It is not possible for $M>m$.

- 4 years, 10 months ago