The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completel

Question

The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at 4 + x cm inside the block. The value of x is

TestHub · Accepted Answer

According to the given information for the first $4 cm$ , $u = v_{0}$ and $v = \frac{v_{0}}{3}$ . Applying equation of motion for constant acceleration,
$v^{2} = u^{2} + 2 a s$
$\Rightarrow {(\frac{v_{0}}{3})}^{2} = v^{2} + 2 a (4) \Rightarrow a = - \frac{8 {v_{0}}^{2}}{9 (8)} = - \frac{{v_{0}}^{2}}{9}$
When the body stops, $v = 0$
$\Rightarrow 0 = {v_{0}}^{2} + 2 a (4 + x)$
$\Rightarrow {v_{0}}^{2} = 2 (\frac{{v_{0}}^{2}}{9}) (4 + x)$
$\Rightarrow 4.5 = 4 + x$
$\Rightarrow x = 0.5 cm$

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