Three blocks start off at rest on a frictionless floor and touching each other. The block on the left has a mass of 3.85 kg. The block in the middle has a mass of 2.15 kg. The block on the right has a mass of 8.51 kg. Suddenly an explosion splits up for left block from the other two, which remain together for the time being, sending it to the left at a speed of 3.3 m/s relative to the other two blocks. 0.71 seconds later , A second explosion splits up the right block from the middle block sending the right block to the right with a speed of 3.03 m/s relative to the middle block.
A.) after the second explosion, what is the velocity of the middle block?
B.) assuming that the middle block started out at position of 0 m, what is the position 3.96 seconds after the initial explosion?
Three blocks start off at rest on a frictionless floor and touching each other. The block on the left has a mass of 3.85 kg. The block in the middle has a mass of 2.15 kg. The block on the right has a mass of 8.51 kg. Suddenly an explosion splits up for left block from the other two, which remain together for the time being, sending it to the left at a speed of 3.3 m/s relative to the other two blocks. 0.71 seconds later , A second explosion splits up the right block from the middle block sending the right block to the right with a speed of 3.03 m/s relative to the middle block.
A.) after the second explosion, what is the velocity of the middle block?
B.) assuming that the middle block started out at position of 0 m, what is the position 3.96 seconds after the initial explosion?
A.) after the second explosion, what is the velocity of the middle block?
B.) assuming that the middle block started out at position of 0 m, what is the position 3.96 seconds after the initial explosion?
A) The momentum should be conserved in a collision.
The momentum with which the right block moves is p=m*v, i.e., p = (8.51*3.03) = 25.7853 kg*m/s.
So the momentum of the middle block should also be the same.
Therefore the velocity of the middle block is v=p/m.i.e., v = (25.7853 / 2.15) = 11.9931 m/s
B) The distance travelled by the middle block after the initial explosion can be given by distance =( speed * time) =(speed * 3.96)
So first we have to find the speed with which the middle block moves.
During the initial collision also momentum should be conserved.
The momentum of the left object is p= m * v.i.e. p = 3.85 * 3.3 = 12.705 kg*m/s.
As the middle block and the right block are touching each other, both of them experience the same momentum.
So the mass is taken as the sum of middle block and right block.i.e. total mass= 2.15 + 8.51 = 10.66 kg
The velocity of the middle and the right block is v = p / m.i.e v = 12.705 / 10.66 = 1.191 m / s.
therefore the distance travelled by the middle block after the initial collision is distance =(speed * time ) i.e.
(1.191 * 3.96 ) = 4.716 m
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