a bomb initially at rest explodes into 3 fragments of equal masses . 2 fragments fly off at right angles to each other at velocities 9m/s and 12m/s respectively. calculate the speed of the third fragment ?

expl in detail please

Question

Vikas TU · Answer

A bomb is initially at rest means its velocity of centre of mass = zero.
Hence,

MVcom = (m1v1 + m2v2 + m3v3)/(m1 + m2 + m3)
0 = [m(v1)i + m(v2)j + m(v3)]/3m
v3 = -v1i – v2j
put v1 and v2
=> -9i – 12j m/s

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a bomb initially at rest explodes into 3 fragments of equal masses . 2 fragments fly off at right angles to each other at velocities 9m/s and 12m/s respectively. calculate the speed of the third fragment ? expl in detail please

a bomb initially at rest explodes into 3 fragments of equal masses . 2 fragments fly off at right angles to each other at velocities 9m/s and 12m/s respectively. calculate the speed of the third fragment ?

expl in detail please

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a bomb initially at rest explodes into 3 fragments of equal masses . 2 fragments fly off at right angles to each other at velocities 9m/s and 12m/s respectively. calculate the speed of the third fragment ? expl in detail please

a bomb initially at rest explodes into 3 fragments of equal masses . 2 fragments fly off at right angles to each other at velocities 9m/s and 12m/s respectively. calculate the speed of the third fragment ?expl in detail please

1 Answers

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a bomb initially at rest explodes into 3 fragments of equal masses . 2 fragments fly off at right angles to each other at velocities 9m/s and 12m/s respectively. calculate the speed of the third fragment ?

expl in detail please