Solution1:
The ball collides with the ground with a velocity v=sqrt{2g(L/2)}. Due to inelastic nature of the collision v2 is lost and only v1 is left.
v1=v(1/2)....the angle is 30*.
Now the ball continues to move left till the string becomes taut again. The angle is again 30*. Only the tangential component of velocity ie. v3 survives the jerk given by the string.
v3=v(1/2)
Now you can apply conservation of eng. of h=v3*v3/2g;
(v3 comes out to be sqrt(gL)/4.)
So:h=L/32.
Is this a previous subjective JEE question?
ReplyDeletetheres smthin im unsure of
ReplyDelete"the only loss of energy occurs during the inelastic collision and is equal to 3mgl/8.
therefore max height achieved should be l/8 ?"
no kunal. and no digvijay.
ReplyDeletewhen the ball reaches the left extreme of its path it has a velocity component parallel to the string.
If this velocity is sustained any longer, the string will break. It has to go.. and it goes due to a sudden jerk by the string.
Eng. is lost due to this also.
I agree it was tough and maybe beyond JEE.
Still it was a cute one..clearing many concepts in the way. Don't you agree?
Any suggestions?
Yea.. the problem is mast 1dum ... :)
ReplyDeleteI got the same but i took 4 in underroot ... a big blunder :(
n whats the origin of the ques ?
right anuj...jst wanted 2 confirm..sure ws a cute one...waise where did u find these?irodov mein to nai hain...
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ReplyDeleteu can do it by ur method also digvijay. only error is that the total loss in energy is 3mgl/8 in the first collision and 3mgl/32 by the impulse generated from the string. net loss is 15mgl/32
ReplyDelete[v={sqrt(gl)}/2, vel. after first collision, {1/2 m(vcos30)^2 = 3mgl/32} energy lost b4 it leaves ground.
initial energy is mgl/2 (and not mgl)
loss in energy is 15mgl/32
remaining energy = mgl/32
nice question
that's right.
ReplyDelete