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

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

Text Solution

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The correct Answer is:

Let the thickness of one plank is s
if bullet enters with velocity u then it leaves with velocity
`v=(u-(u)/(20))=(19)/(20)u`
from `v^(2)=u^(2)-2as`

`rArr ((19)/(20)u)^(2)=u^(2)-2as rArr (400)/(39)=(u^(2))/(2as)`
Now if the n planks are arranged just to stop the bullet then again from `v^(2)=u^(2)-2as`
`0=u^(2)=2` ans
`rArr n = (u^(2))/(2as)=(500)/(39)`
`rArr n = 10.25`
As the planks are more than 10 so we can consider n = II

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

` 8`

`7`

` 11`

` 14`

A bullet losses 1/6 of its velocity in passing through a plank. What is the least number of planks required to stop the bullet ?

A bullet looses ((1)/(n))^(th) of its velocity passing through one plank.The number of such planks that are required to stop the bullet can be:

`(n^(2))/(2n-1)`

`(2n^(2))/(n-1)`

infinite