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The velocity of the bullet becomes one third after it penetrates $4 \mathrm{~cm}$ 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 $(4+x) \mathrm{cm}$ inside the block. The value of $\mathrm{x}$ is :
Option: 1
$2.0$

Option: 2
$1.0$

Option: 3
$0.5$

Option: 4
$1.5$

Answers (1)

best_answer

$\mathrm{V_{1}=\frac{u}{3}\left ( Given \right )}$

$\mathrm{V_{2}=0}$

* For constant retardation by the equations of motion

$\mathrm{V_1^2=u^2+2 a s }$

$\mathrm{\frac{u^2}{9}-u^2=2 a\left(4 \times 10^{-2}\right) }$

$\mathrm{-\frac{4 u^2}{9 \times 4 \times 10^{-2}}=a }$

$\mathrm{a=\frac{-100 u^2}{9} \rightarrow(1) }$

$\mathrm{v_2^2=u^2+2 a s }$

$\mathrm{0=u^2+2\left(-\frac{100 u^2}{9} \times(4+x) \times 10^{-2}\right) }$

$\mathrm{u^2=\frac{2 u^2(4+x)}{9} }$

$\mathrm{x=0.5 \mathrm{~cm}}$

Hence 3 is correct option

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