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A bullet moving with a speed of $100 \mathrm{~ms}^{-1}$ 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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The correct answer is:
$8$
Given, $u_{1}=100 \mathrm{~ms}^{-1}$
$$
\Rightarrow u_{2}=2 \times u_{1}=2 \times 100=200 \mathrm{~ms}^{-1}
$$
Using equation of motion,
$$
v^{2}=u^{2}-2 a s \quad(\because \text { acceleration is negative })
$$
As, final velocity, $v=0$
$\Rightarrow \quad u^{2}=2 a s$ or $s=\frac{u^{2}}{2 a} \Rightarrow s \propto u^{2}$ $\therefore \quad \frac{s_{1}}{s_{2}}=\frac{u_{1}^{2}}{u_{2}^{2}}=\frac{(100)^{2}}{(200)^{2}}=\frac{1}{4}$ $\Rightarrow \quad s_{2}=4 s_{1}=4 \times 2$ planks $=8$ planks
$$
\Rightarrow u_{2}=2 \times u_{1}=2 \times 100=200 \mathrm{~ms}^{-1}
$$
Using equation of motion,
$$
v^{2}=u^{2}-2 a s \quad(\because \text { acceleration is negative })
$$
As, final velocity, $v=0$
$\Rightarrow \quad u^{2}=2 a s$ or $s=\frac{u^{2}}{2 a} \Rightarrow s \propto u^{2}$ $\therefore \quad \frac{s_{1}}{s_{2}}=\frac{u_{1}^{2}}{u_{2}^{2}}=\frac{(100)^{2}}{(200)^{2}}=\frac{1}{4}$ $\Rightarrow \quad s_{2}=4 s_{1}=4 \times 2$ planks $=8$ planks
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