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2 Jul 2007 11:34:17 IST
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let A bomb of mass m is thrown vertically in the air ... it explodes into two equal parts when its velocity is 21m/sec. If one part moves upto the height of 40m from the pt. of explosion then what is the position of second part fom pt. of explosion.
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momentum before explosion and momentum after explosion must be same bcoz no external force is acting on the system... Intial velocity of first part can be found by: v^2 - u^2 =2as where v=0 a=10 s=40 therefore, u=20root2 applying law of conservation of linear momentum let x be the velocity of second part M21=m20root2 + mx as M=2m therefore the equation becomes 42=20root2 + x x=42 - 20root2=11.72 now,u know the velocity of second part so position can be found by applying same formula u get s= -9.76 minus sign here means that other part is going downwards rate my efforts if you find them useful
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2 Jul 2007 12:23:30 IST
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numericaly your ans is correct but i think that it should be with positive sign...
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2 Jul 2007 12:34:31 IST
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sorry that was a mistake you are right the answer is with positive sign
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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2 Jul 2007 13:24:18 IST
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Good work maverick
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All men dream but not equally. Those who dream by night in the dusty recesses of their minds wake in the day to find that it was vanity; but the dreamers of the day are dangerous men, for they may act their dream with open eyes to make it possible. |
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2 Jul 2007 14:27:40 IST
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After the explosion , the part of the mass which moves up to a height of 40m needn't go directly upwards. It can also take a parabolic path and reach a maximum height of 40m from the point of explosion. The question is incomplete in itself, it should be mentioned that the part of the bomb goes DIRECTLY upwards.P.S this can also be solved in much less time using conservation of momentum. If the time of explosion is infinetely small then dp becomes almost zero and hence momentum is conserved just befor and after the collision.
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2 Jul 2007 16:04:43 IST
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momentum before collision=momentum aftr collisison.
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2 Jul 2007 17:30:29 IST
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mv= m/2v1 + m/2v2
v1=root(2g40)
on solving get v2
put v2sq=2gh
to get h
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devesh
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2 Jul 2007 17:33:02 IST
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thanx everbody you all are right i will RATE YOU ALL
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2 Jul 2007 17:34:59 IST
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I would like to add a few things
System: The particle
The only external force acting on the system before and after the explosion are the gravitational forces. Now, assuming the time taken for the explosion to be small, the impulse generated by these gravitational forces is negligible. Hence linear momentum can be assumed to be constant in the vertical and horizontal directions.
In the horizontal direction,
0 = (m/2)v1x + (m/2)v2x
v1x = -v2x-----------(1)
As there is no acceleration along the X axis, the pieces move equal distances from the point of explosion, in equal time intervals along the X axis.
In the vertical direction,
mv0 = (m/2)v1y + (m/2)v2y
2v0 = v1y + v2y -----------(2)
Also,
v1y2/2g = h1
v1y =  2gh1 ---------(3)
From (2) and (3)
v2y = 2v0 -  2gh1
Height to which the second piece rises
h2 = v2y2/2g
h2 = (2v0 - 2gh1)2/2g
This is the height to which the second piece rises, and horizontally, it is at the same distance from the projection point as the first piece. This completely determines the position of the first piece.
Substitute
v0 = 21 m/s
h1 = 40 m
to get the numerical answer.
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All men dream but not equally. Those who dream by night in the dusty recesses of their minds wake in the day to find that it was vanity; but the dreamers of the day are dangerous men, for they may act their dream with open eyes to make it possible. |
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2 Jul 2007 18:06:26 IST
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Good explanation elessar...........
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