A bomb is thrown in a horizontal direction with a velocity of 50m/s it explode into two part of mass 6kg and 3kg.The heavier part continue to move in the same direction with a velocity of 80m/s . Calculate the velocity
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Question is incomplete.
At the end it should be “Calculate the velocity of lighter part”
Initial momentum of system
= (6kg + 3kg) × 50 m/s
= 450 kgm/s
Final momentum of system
= (6kg × 80m/s) + 3v
= 480 + 3v
According to law of conservation of momentum
Initial momentum of system = Final momentum of the system
450 = 480 + 3v
3v = -30
v = -10 m/s
Velocity of lighter part is -10 m/s
Here -ve sign indicates that it will travel opposite to the motion of heavier part.
At the end it should be “Calculate the velocity of lighter part”
Initial momentum of system
= (6kg + 3kg) × 50 m/s
= 450 kgm/s
Final momentum of system
= (6kg × 80m/s) + 3v
= 480 + 3v
According to law of conservation of momentum
Initial momentum of system = Final momentum of the system
450 = 480 + 3v
3v = -30
v = -10 m/s
Velocity of lighter part is -10 m/s
Here -ve sign indicates that it will travel opposite to the motion of heavier part.
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Answer:
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