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Mechanics

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16 Dec 2008 13:50:10 IST

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An easy one, yet everyone's invited.

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A right triangular wedge of mass $M$ rests on an absolutely smooth horizontal surface. A block with a mass $m$ is placed on the inclined surface of the wedge, the inclination being $\alpha$ with the horizontal. This system is released from rest. Determine the velocity of the wedge when the block lowers vertically through a height $h$ . Assume that there is no friction between the wedge and the block.

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VARUN RAJ

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16 Dec 2008 17:16:51 IST

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I THINK IT IS AN EASY SUM

FIRST WE KNOW THAT THE VEL OF THE BLOCK IS THE HORIZONTAL DIRECTION IS EQUAL

THATS THE EASY PART

V₁COS@=V2

THE HARD PART MAYBE I AM WRONG IN THIS PART

MGH=1/2M(VSIN@)^2

THIS IS THE 2ND STEP SO THE WHOLE SUM CAN BE CALCULATED

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Anant Kumar

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17 Dec 2008 10:13:43 IST

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I know how to solve it. What I want to know is, whether you guys can solve it. Anyway, what you suggested is entirely wrong (both the first part and the second part).

akki ~~ unlucky forever ~~

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17 Dec 2008 10:30:24 IST

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conserving momentum in horizontal direction ,

$Mv = m(ucos\alpha-v)$

energy conservation,

$mgh=\frac{Mv^2 }{2}+\frac{m({\sqrt{v^2+u^2 +2uvcos(\pi-\alpha) })}^2 }{2}$

using the two equation, we can easily get 'v'

is it correct...................

Ashutosh Sharma

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17 Dec 2008 10:40:10 IST

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Refer to last solved example in HC Verma...

Now, m desends by 'h' vertically, so, on the incline, it slips by h sin@

Now, from d example, we have acc._{m w,r,t, M} =

So, time taken = (Sorry, forgot g)

Now, acceleration of wedge =

Hence, velocity = x

Please check if correct...

Ashutosh Sharma

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17 Dec 2008 10:41:36 IST

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Looks like I have gone terribly wrong somewhere.... If that is the case... please correct....

akki ~~ unlucky forever ~~

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18 Dec 2008 20:10:02 IST

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sir what's the correct method,

pls reply.....................

Rajat Sen

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19 Dec 2008 21:21:06 IST

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Let $a_{2}$ be the acceleration of the incline and $a$ be that of the block relative to the wedge.

Suppose the normal reaction = $N$

so, we ghet the equations :

$Nsin \alpha = Ma_{2}$ .........1

$ma_{2} cos \alpha + mg sin \alpha = ma$ .....2

$N+ma_{2} sin \alpha = mg cos \alpha$ ....3

solving this we get :

$a_{2} = \frac{mg sin \alpha cos \alpha }{M + msin^{2} \alpha}$

$a = \frac{gsin \alpha (m+M)}{M+msin^{2} \alpha }$

now, by equating the time in both the cases we get :

$v_{wedge} = a_{2}\sqrt{\frac{2h}{asin \alpha}}$

So, we get the answer

shubham sunder

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20 Dec 2008 09:43:26 IST

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i am getting something like...

root (2mgh (1-sin@) / M )

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