A bullet moving with a speed of 100 ms -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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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 $v^{2} = u^{2} - 2as$ ⇒ ⇒ $\theta = u^{2} = 2 \alpha s$
$S = - \frac{u^{2}}{2 \alpha}$ $s \propto u^{2}$ [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. $s_2 = 4(s_1) = 4(2s)$ ⇒ ⇒ $S_{2} = 8 \sigma$
So, number of planks required = 8
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