A bullet moving with a speed of 100 m{s^{ - 1 | vedclass

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Question

(b)Let the thickness of each plank is $s$. If the initial speed of bullet is $100 m/s$ then it stops by covering a distance $2s$ By applying ${v^2} =

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(b)Let the thickness of each plank is $s$. If the initial speed of bullet is $100 m/s$ then it stops by covering a distance $2s$ By applying ${v^2} = {u^2} - 2as$

$&rArr;$ $0 = {u^2} - 2as$ $s = \frac{{{u^2}}}{{2a}}$ 

$s \propto {u^2}$ [If retardation is constant] 

If the speed of the bullet is double then bullet will cover four times distance before coming to rest

i.e. ${s_2} = 4({s_1}) = 4(2s)$ 

$&rArr;$ ${s_2} = 8s$ 

So number of planks required $ = 8$

A bullet moving with a speed of $100$ $m{s^{ - 1}}$can just penetrate two planks of equal thickness. Then the number of such planks penetrated by the same bullet when the speed is doubled will be

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