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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.
```
7 years ago

## Answers : (2)

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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Anil Pannikar
IIT Bombay
```
7 years ago
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
```
7 years ago
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