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A marble going at a speed of 2 ms^(-1) ...

A marble going at a speed of `2 ms^(-1)` hits another marble of equal mass at rest. If the collision is perfectly elastic, then the velocity of the first marble after collision is

A

`4 ms^(-1)`

B

`0 ms^(-1)`

C

`2 ms^(-1)`

D

`3 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`vecv_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))vecu_(1)+((2m_(2))/(m_(1)+m_(2))) vecu_(2)`.
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Knowledge Check

  • In , if the collision were perfectly elastic, what would be the speed of deuteron after the collision ?

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