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A block of mass 'M' is moving on rough horizontal surface with momentum 'P'. The coefficient of friction between the block and surface is ' $\mu$ '. The distance covered by block before it stops is $[\mathrm{g}=$ acceleration due to gravity $]$
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Verified Answer
The correct answer is:
$\frac{\mathrm{P}^{2}}{2 \mu \mathrm{M}^{2} \mathrm{~g}}$
Using kinematic relation,
$v^{2}=u^{2}-2 a s$
The momentum pis given by
$\begin{array}{l}
p=M u \\
u=\frac{P}{M}
\end{array}$
and acceleration is given as
$a=\mu g$
Substituting values in Eq. (i), we get
$\begin{array}{l}
0=\mathrm{u} 2-2 \mathrm{as}(\because f \in \text { alvelocity }, v=0) \\
0=\left(\frac{\mathrm{p}}{\mathrm{M}}\right)^{2}-2 \mu \mathrm{gs} \\
\left(\frac{\mathrm{p}}{\mathrm{M}}\right)^{2}=2 \mu \mathrm{gs} \\
\Rightarrow \mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}^{2} \mu \mathrm{g}}
\end{array}$
$v^{2}=u^{2}-2 a s$
The momentum pis given by
$\begin{array}{l}
p=M u \\
u=\frac{P}{M}
\end{array}$
and acceleration is given as
$a=\mu g$
Substituting values in Eq. (i), we get
$\begin{array}{l}
0=\mathrm{u} 2-2 \mathrm{as}(\because f \in \text { alvelocity }, v=0) \\
0=\left(\frac{\mathrm{p}}{\mathrm{M}}\right)^{2}-2 \mu \mathrm{gs} \\
\left(\frac{\mathrm{p}}{\mathrm{M}}\right)^{2}=2 \mu \mathrm{gs} \\
\Rightarrow \mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}^{2} \mu \mathrm{g}}
\end{array}$
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