A bullet moving with a speed of 100 m/s can | vedclass

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(B) Let the thickness of each plank be $s$. If the initial speed of the bullet is $u_1 = 100 \ m/s$,it stops after covering a total distance of $2s$.

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the thickness of each plank be $s$. If the initial speed of the bullet is $u_1 = 100 \ m/s$,it stops after covering a total distance of $2s$. Using the equation of motion $v^2 = u^2 - 2as$,where $v=0$ is the final velocity and $a$ is the constant retardation:$0 = u_1^2 - 2a(2s) \implies 2a = \frac{u_1^2}{2s}$.Since the retardation $a$ is constant,the total distance $S$ covered by the bullet is proportional to the square of the initial speed $(S \propto u^2)$.If the speed is doubled,$u_2 = 2u_1$. The new distance covered $S_2$ will be:$S_2 = \left(\frac{u_2}{u_1}ight)^2 S_1 = (2)^2 (2s) = 4 	imes 2s = 8s$.Since each plank has thickness $s$,the number of planks penetrated is $8s / s = 8$.

$A$ bullet moving with a speed of $100 \ m/s$ 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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