a 40.0 kg bomb with a momentum of 640 kg m/s explodes into 3 pieces, the first one is 12.0 kg with a momentum of 380 kg m/s, while the second one is 15.0 kg with a momentum of 790 kg m/s (24.0 degrees West of South). Determine the velocity of the third piece after the explosion.
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initial mass( total mass) = 40kg. initial momentum=640kgm/s
mass of first piece=15kg. momentum of first piece= 380kgm/s
mass of second piece=15 kg. momentum of second piece = 790kgm/s
therefore mass of third piece would be= 40 - (12+15) = 40-27= 3kg
let momentum of third piece be x
by conservation of momentum
initial momentum = final momentum
640 = 380 + 790 + x
x= - 530kgm/s
negative sign indicates opposite direction
mass of first piece=15kg. momentum of first piece= 380kgm/s
mass of second piece=15 kg. momentum of second piece = 790kgm/s
therefore mass of third piece would be= 40 - (12+15) = 40-27= 3kg
let momentum of third piece be x
by conservation of momentum
initial momentum = final momentum
640 = 380 + 790 + x
x= - 530kgm/s
negative sign indicates opposite direction
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