A 0.015kg marble moving to the right at 0.225m/s makes an elastic head-on collision with a 0.030kg at 0.180m/s. After a collision, the smaller marble moves to the left at 0.315m/s. assume that neither marble rotates before or after the collision and both marbles are moving on a frictionless surface. What is the velocity of the 0.030kg marble after the collision?
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Expert's answer
2020-04-08T10:42:30-0400
Solution: We will use the law of momentum conservation. The initial momentum of first (small) marble is P1=m1⋅v1 . That is for second marble P2=m2⋅v2. After elastic collision the momentum of smaller marble is P1new=m1⋅V1 and greater one P2new=m2⋅V2 . Due to the fact that the collision is elastic head-on and without rotation, all vectors are aligned relative to the same direction and differ only in the sign. The law of momentum conservation state
(1)P1+P2=P1new+P2new substituting formulas for momentums we have
(2) m1⋅v1−m2⋅v2=−m1⋅V1+m2⋅V2 . Here we have taken into account the + sign for right moving and the - sign for left moving . We solve this equation with respect to an unknown speed V2
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