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A bullet loses 1//20 of its velocity in...

A bullet loses ` 1//20` of its velocity in passing through a plank. What is the least number of plankd required to stop the bullet .

A

5

B

10

C

11

D

20

Text Solution

Verified by Experts

The correct Answer is:
C
`k_i=1/2 mv^2`
For one plank `rarr K =W`
`W=DeltakimpliesW= underset("Loss inK.E.")(underbrace(1/2mv^2-1/2m((19v)/(20))^2))`
`implies W=1/2 mv^2 [1-((19)/20)^2]`
n = Number of plank
`n=(1/2mv^2)/(W)=1/([1-((19)/20)^(2)])=((20)^2)/((20)^2-(19)^2)=400/(39)`
`n = 10.25 ~~ n = 11`
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