1V = \frac{m\sqrt{2gh}.cos\alpha}{M+m}.............!!!
Q. 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 α being 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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12 Answers
1Mv + m(√2ghcotθ -v) = 0 -> initial vel of com
=> v= m(√2ghcotθ)/(m+M)
1well, sir... i'm daring to ask a silly question... i'm unable to get my mistake in energy conservation equation... plzzz sort out d error...
here is my energy conservation for d entire process...
mgd = 1/2.(M+m).V2 + mg(d-h) + 1/2.m.(√2.gsinα.hcosecα)2 ....... (here 'd' is initial vertical height of body 'm')
LHS = [initial potential energy of mass 'm']...
RHS = [combined kinetic energy of (M+m)] + [final potential energy of mass 'm'] + [kinetic energy of mass 'm' after descending through a vertical height 'h' along d inclined]...
solving dis i'm getting V=0... [11] [2]
plzzz temme what's d error...!!!???
66The kinetic energy term for 'm' has been written wrongly. Try to figure out why that's wrong.
1mgh=1/2MV2+(1/2)m(vx2+vy2)
mvx=MV
Vsinα=vycosα-vxsinα
3 equation 3 variable...
directions of velocities are:
<------ V
-----> vx
|
|
|
\/ vy
incline's direction:
*
*
*
*
*
_____ *
21I guess these 3 eqtns by GHOSTY wud clinch it :
mgh=1/2MV2+(1/2)m(vx2+vy2)
mvx=MV
Vsinα=vycosα-vxsinα
33Yes i think ghosty eqns are right....
Rohan ka me v1 v2 kya hai pata nahi.. but the same steps in both method..
momentum conservation in horizontal
energy conservation
velocity in common normal direction is same...
1yes ghostly's equations are right..
id didnt see that already someone had posted it ..
posted i a hurry after reading prob.
it will be better if i delete it .
