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Discussion Response Post to: Solve this problem....

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Rohan (1468)

Blazing goIITian

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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

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

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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