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A bomb at rest explodes into three fragments of equal masses.Two fragments fly off at right angles to each other with velocity 9m/s and 12m/s.Calculate the speed of the third fragment.

I know the solution but I want the shortest possible solution.

6 years ago


Answers : (2)


Dear Nitin ,


take resultant of momentum =  sqrt [(1/3*9)2 + (1/3*12)2]

                                           =  1/3 * V             ( velocity of 3rd fragment )



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Askiitians Expert

Anil Pannikar

IIT Bombay

6 years ago

let the mass of bomb be m

 after explosion each fragment has mass m/3

applying consevation of momentam

 mV1/3  + mV2/3 mV3/3 =0  ........1                         (V1,V2,V3 are the velocities of each particle after explosion)

 let V1 = 9i

 so V2=12j

   from here V3 =-(9i+12j)=root225=15m/s in magnitude

6 years ago

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