anonymous
  • anonymous
A particle of mass 4m explodes into 3 pieces of masses m,m,2m. The equal masses move along X & Y axes with velocities 4m/s & 6m/s respectively. Find the magnitude of velocity the heavier mass (Choices --Under root 17,2 x under root 13,under root 13 or under root 13/2 m/s
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1350120916624:dw| since the momentum of the system was zero at the beginning after the explode it remains zreo. now we have for the two masses : speed = sqrt(6^2+4^2) = sqrt(52) = 2sqrt(13) so the magnitude of the momentum of the two masses is m*2sqrt(13) this should be equal to 2m * |V| where |V| is the magnitude of the velocity of the larger mass 2m*|V| = m * 2sqrt(13) |V| = sqrt(13)
anonymous
  • anonymous
I fail to understand the logic " since the momentum of the system was zero at the beginning after the explode it remains zero"--Can this be elaborated? Many thanks for the answer.. .
anonymous
  • anonymous
conservation of momentum .. no external forces acted upon the system so the total momentum has not changed

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