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A bullet enters a wooden block with velocity $120 \mathrm{~m} / \mathrm{s}$. The bullet travels $1.5 \mathrm{~s}$ in the block before its velocity reduces to zero due to resistance force which is proportional to the square root of the velocity. The distance travelled by the bullet in the wooden block is
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The correct answer is:
$90 \mathrm{~m}$
Given, velocity of bullet, $u=120 \mathrm{~m} / \mathrm{s}$
and $t=1.5 \mathrm{~s}$
If deceleration of bullet inside the block is $a$.
Then from the first equation of motion, $v=u-a t$
$\begin{aligned} & & 0 & =120-a \times 1.5 \\ \Rightarrow & & a & =\frac{120}{1.5}=80 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
If bullet travels a distance of $s$ meter before it comes to rest, from third equation of the motion,
$\begin{aligned} v^2 & =u^2-2 a s \\ 0 & =120^2-2 \times 80 \times s \\ \Rightarrow \quad s & =\frac{120 \times 120}{2 \times 80}=90 \mathrm{~m}\end{aligned}$
Hence, no option is correct.
and $t=1.5 \mathrm{~s}$
If deceleration of bullet inside the block is $a$.
Then from the first equation of motion, $v=u-a t$
$\begin{aligned} & & 0 & =120-a \times 1.5 \\ \Rightarrow & & a & =\frac{120}{1.5}=80 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
If bullet travels a distance of $s$ meter before it comes to rest, from third equation of the motion,
$\begin{aligned} v^2 & =u^2-2 a s \\ 0 & =120^2-2 \times 80 \times s \\ \Rightarrow \quad s & =\frac{120 \times 120}{2 \times 80}=90 \mathrm{~m}\end{aligned}$
Hence, no option is correct.
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