A bomb exploded in mid air into two equal fragments .
What is angle between their directions of motion
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Let the bomb have a mass 2m,
Let the mass and velocity of the first fragment be m and v1 and of the 2nd be m and v2 respectively.
The initial velocity of the bomb be 0 assuming the bomb is at rest in air when it exploding
Intial momentum = final momentum
2m(0) = mv1 + mv2
=> mv1 + mv2 = 0
=> v1 = -v2
This means the two fragments will have equal but opposite velocities.
Thus direction of fragments opposite of each other.
Let the mass and velocity of the first fragment be m and v1 and of the 2nd be m and v2 respectively.
The initial velocity of the bomb be 0 assuming the bomb is at rest in air when it exploding
Intial momentum = final momentum
2m(0) = mv1 + mv2
=> mv1 + mv2 = 0
=> v1 = -v2
This means the two fragments will have equal but opposite velocities.
Thus direction of fragments opposite of each other.
Answered by
6
Hey !
_______________________
A bomb is exploded in mid air into two equal fragments
It means ,
Initial momentum = Final momentum
180° :+ The angle between their directions of motion .
As the two fragments would move in exactly opposite directions .
_________________________
_______________________
A bomb is exploded in mid air into two equal fragments
It means ,
Initial momentum = Final momentum
180° :+ The angle between their directions of motion .
As the two fragments would move in exactly opposite directions .
_________________________
Anonymous:
Thx ji ☺
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