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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 19:10:27 IST
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a small block of mass m is placed over a wedge of mass M at a ht.H initially from the ground.Find the acceleration of wedge,acc.of block perpendicular to plane AB and velocity of wedge when m reaches B.no external force acts on the wedge i cannot draw the diagram guys plz experts.angle of inclination of plane is theta.
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10 Aug 2007 19:36:50 IST
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reply yaar plz solve this wrt the ground frame and plz explain how to solve sums related to wedge constraints n all
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10 Aug 2007 19:40:43 IST
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try 2 conserve momentum nd energy...question is not really clear....just conserve the stuff nd u shd get it!!! plz....rate me!!!!!
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10 Aug 2007 19:41:36 IST
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i was wrong the first time. did it orally and in a hurry jatinroxx corrected me. wen we conserve energy, we also take the energy attained by the wedge. and then write the 2 eqns and then solve them
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ameya s,
1 yr
iit bombay.
if you ever want to help me clean my room, go STUDY NOW!!! |
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10 Aug 2007 19:47:54 IST
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how u took v1cos theta
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10 Aug 2007 19:49:36 IST
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mgh=1/2mv^2 + 1/2 mV^2 mvcos@= (M+m)V solving these two, we get the required speed
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ameya s,
1 yr
iit bombay.
if you ever want to help me clean my room, go STUDY NOW!!! |
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19 Aug 2007 20:34:34 IST
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Let the acceleration of the block m wrt the wedge be a and that of the wedge wrt thee ground be A.
The block is observed wrt the wedge and hence its FBD has a pseudo force mA directed opposite to the direction of motion of the wedge.
mg sin  + mAcos  = ma ---------------(1)
N + mA sin = mg cos ----------------(2)
For the wedge we resolve parallel and perpendicular to the ground.
N' = Mg + N cos ---------------(3)
N sin = MA ----------------------(4)
From (1), (2) and (3),
A = mg cos sin/(M + msin2) ---------------(5)
From (5) and (1)
a = (m+M)gsin/(M + msin2) -----------------(6)
The acceleration block m is the vector sum of a and A.
a is directed along the inclined plane and hence has no component perpendicular to the plane.
Hence acceleration of the block perpendicular to the plane
aper=A sin = mg cos sin2/(M + msin2)
Let the velocity of the block relative to the wedge be v and that of the wedge relative to the ground be V. Then by momentum and energy conservation,
m(vcos - V) - MV = 0
and
(1/2)m(v2+V2- 2vVcos)+ (1/2)mV2 = mgh
Solve for V
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19 Aug 2007 21:34:02 IST
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i think applying conservation of energy with velocity relative to the block will yield wrong answer.......you need absolute velocity of the sliding block....that will be somewhat comlex, but solvable..
pl rate me if you think it is a god suggestion
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