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A body of mass 9 kg explodes into three parts of mass 2 kg, 3 kg and 4 kg. The part...
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The body with mass 9 kg undergoes an explosion leading to the formation of 3 smaller parts with mass 2 kg, 3 kg and 4 kg. The part with mass 2 kg moves along the positive x-axis with speed 18 m/s and the part with mass 4 kg moves along the positive y-direction with speed 15 m/s.
Initially, the momentum of the body is zero. The law of conservation of momentum can be used here and we know that after the explosion, the total momentum of the three parts that are created should also be equal to zero.
The momentum of the part with mass 2 kg moving along the positive x-axis is equal to 2*18 = 36 kg*m/s. And the momentum of the part with mass 4 kg moving along the positive y-axis is 4*15 = 60 kg*m/s. The sum of these two vectors is one with magnitude `sqrt(36^2 + 60^2) = 12*sqrt 34` . And the angle of the vector `theta = tan^-1(60/36) ~~ 59.06` degrees anti-clockwise with the positive x-axis.
The momentum of the third part is a vector with the same magnitude as the sum determined earlier and in the opposite direction.
The velocity of the part with mass 3 kg has a magnitude of `4*sqrt 34` m/s and it is at an angle 59.06 degrees in the anticlockwise direction with the negative x-axis.
Posted by justaguide on November 4, 2013 at 3:38 PM (Answer #1)
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