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Blazing goIITian

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9 Mar 2009 21:17:23 IST

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A ball falls vertically on an inclined plane of inclination alpha , with speed v and makes a perfec

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A ball falls vertically on an inclined plane of inclination alpha , with speed v and makes a perfectly elastic collision. Where will it hit da plane again ?

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ABINASH

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Joined: 20 Apr 2007

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9 Mar 2009 23:39:07 IST

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U can use momentum conservation along the x axis of the plane.... i.e x along the plane and y perpendicular to it. So, vel_y changes in the opposite direction with the same magnitude and vel_x remains the same as there's no impulse in that direction. So, the ball strikes @ a distance of 2v^2sin(alpha)/g..... along the plane from the point of collision.

ramyani chakrabarty

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9 Mar 2009 23:41:30 IST

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plz post detail soln.

ABINASH

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10 Mar 2009 00:43:02 IST

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See.. we solve the problem tilting the coordinate axes by alpha..... i.e x now moves along the plane and y perpendicular to it. So, when the ball strikes the plane, its velocity along this x vel_x is vsin(alpha) and vel_y is -vcos(alpha). After the collision, vel_x remains the same and vel_y becomes vcos(alpha). Now, its a projectile, with vel_y as the upward component of the velocity and vel_x as the horiznontal and the gravity value is also gcos(alpha) . So, when it again strikes the plane, the vertical displacement is zero. calculating this time, we get t = 2v/g.

Hence, range is vel_x*t = 2v^2sin(alpha)/g.... dats it.

Hope this helps....

Anant Kumar

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Joined: 10 Jul 2008

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10 Mar 2009 01:06:10 IST

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Take the coordinate axes as shown. The initial velocities along the two axes are $v_{0x}=vsinalpha$ and $v_{0y}=vcos alpha$ . The (constant) accelerations along the axes are $a_x=gsinalpha$ and $a_y = -gcosalpha$ . The $y$ coordinate during the flight is
$y=(vcosalpha)t -dfrac{1}{2}(gcosalpha) t^2$ and the $x$ coordinate $x=(vsinalpha) t + dfrac{1}{2} (gsinalpha) t^2$ .
When the object returns to the plane, $y=0$ which gives $t=dfrac{2v}{g}$ . For this $t$ , the distance of the point where the object strikes the plane from the initial point, is
$x=vsinalphaleft(dfrac{2v}{g}ight) + dfrac{1}{2}gsinalpha left(dfrac{4v^2}{g^2}ight)=dfrac{4v^2sinalpha}{g}$

ABINASH

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10 Mar 2009 01:10:09 IST

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Sir is rite..... I apologize..... I 4gt to take the accn along the x direction wich is gsin(alpha)...

So, the displacement along this x is vsin(alpha)*t + 1/2gsin(alpha)*t^2.....

Thnx sir..... n apologies to Ramyani...

ramyani chakrabarty

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10 Mar 2009 12:09:29 IST

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thank u sir !

I m so stressed.

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