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Nitin Gupta Grade: 11
        


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)

Anil Pannikar AskiitiansExpert-IITB
85 Points
										

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
vikas askiitian expert
510 Points
										

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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