Asked by Cheesecake
I've got a question:
Imagine this scenario where someone releases a marble down an inclined ramp and at the end of the ramp a block is fixed. So the marble will roll down the ramp and it will collide with the block.
Since the block is fixed, m(marble)u(marble) + m(block)u(block) = m(marble)v(marble) + m(block)v(block) by conservation of momentum. Since u(block) and v(block) = 0, u(marble) = v(marble).
Yet, KE of the marble is lost and when it rebounds it will not go back up to its original height. How can momentum be conserved in this case?
Imagine this scenario where someone releases a marble down an inclined ramp and at the end of the ramp a block is fixed. So the marble will roll down the ramp and it will collide with the block.
Since the block is fixed, m(marble)u(marble) + m(block)u(block) = m(marble)v(marble) + m(block)v(block) by conservation of momentum. Since u(block) and v(block) = 0, u(marble) = v(marble).
Yet, KE of the marble is lost and when it rebounds it will not go back up to its original height. How can momentum be conserved in this case?
Answers
Answered by
Cheesecake
To clarify, KE of the marble is lost through heat for eg. due to this collision being inelastic. If KE of the marble is lost, then how does u(marble) = v(marble)?
Answered by
Damon
The earth moved under its feet.
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