A bomb explodes into 3 equal fragments with the velocities of two of them being 9i and 9j.find the magnitude and direction of the velocity of the third?
Asked by umapushkala | 6th Mar, 2012, 04:44: PM
Before explosion, the total momentum of the bomb is zero.
Let p1, p2 and p3 be the momentum of the three fragments.
Let v1, v2 and v3 be theire respective velocities.
If m be the mass of each fragment, then
p1 = mv1 p2 = mv2, p3 = mv3
v1 =9i, v2 =9j
resultant of p1 and p2 = sqrt( p1^2 + p2^2)
= m ( sqrt( 9i^2 + 9j^2) kgm/s
Total momentum should be conserved
therefore, p + p3 =0
In magnitude p =p3
v3 = p3/m = mv3/m = m ( sqrt( 9i^2 + 9j^2)/m = sqrt( 9i^2 + 9j^2)m/s
Answered by | 7th Mar, 2012, 12:22: PM
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