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If a body looses half of its velocity on penetrating $3 \mathrm{~cm}$ in a wooden block, then how much will it penetrate more before coming to rest
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Verified Answer
The correct answer is:
1 cm
For first condition
Initial velocity $=u$, Final velocity $=u / 2, s=3 \mathrm{~cm}$
From $v^2=u^2-2 a s_{\Rightarrow}\left(\frac{u}{2}\right)^2=u^2-2 a s \quad a=\frac{3 u^2}{8 s}$
Second condition
Initial velocity $=u / 2$, Final velocity $=0$
$\begin{aligned} & \text { From } v^2=u^2-2 a x_{\Rightarrow} 0=\frac{u^2}{4}-2 a x \\ & \quad x=\frac{u^2}{4 \times 2 a}=\frac{u^2 \times 8 s}{4 \times 2 \times 3 u^2}=s / 3=1 \mathrm{~cm}\end{aligned}$
Initial velocity $=u$, Final velocity $=u / 2, s=3 \mathrm{~cm}$
From $v^2=u^2-2 a s_{\Rightarrow}\left(\frac{u}{2}\right)^2=u^2-2 a s \quad a=\frac{3 u^2}{8 s}$
Second condition
Initial velocity $=u / 2$, Final velocity $=0$
$\begin{aligned} & \text { From } v^2=u^2-2 a x_{\Rightarrow} 0=\frac{u^2}{4}-2 a x \\ & \quad x=\frac{u^2}{4 \times 2 a}=\frac{u^2 \times 8 s}{4 \times 2 \times 3 u^2}=s / 3=1 \mathrm{~cm}\end{aligned}$
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