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A solid cylinder of mass $3\, kg$ is rolling on a horizontal surface with velocity $4\, m s^{- 1}$. It collides with a horizontal spring of force constant $200 \,N m^{-1}$. The maximum compression produced in the spring will be ............... $\mathrm{m}$
$0.5$
$0.6$
$0.2$
$0.7$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,At\,{\rm{maximum}}\,compression\,the\\
solid\,cylinder\,will\,stop.\\
According\,to\,law\,of\,conservation\,of\,\\
mechanical\,energy\\
Loss\,in\,kinetic\,energy\, = Gain\,in\,potential\\
energy\,of\,cylinder\,of\,spring\\
\frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = \frac{1}{2}k{x^2}\\
\frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{{m{R^2}}}{2}} \right){\left( {\frac{v}{R}} \right)^2} = \frac{1}{2}k{x^2}
\end{array}$
$\begin{array}{l}
\,\,\,\,\,\,\,(v = R\omega \,and\,for\,solid\,cylinder,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = \frac{1}{2}m{R^2})\\
\frac{1}{2}m{v^2} = \frac{1}{4}m{v^2} = \frac{1}{2}k{x^2}\\
\frac{3}{4}m{v^2} = \frac{1}{2}k{x^2}\,or\,{x^2} = \frac{3}{2}\,\frac{{m{v^2}}}{k}\\
Here,\,m = 3\,kg,\,v = 4\,m\,{s^{ – 1}},\,k = 200\,N\,{m^{ – 1}}\\
Subsitituting\,the\,given\,values,\,we\,get\\
{x^2} = \frac{{3 \times 3 \times 4 \times 4}}{{2 \times 200}} \Rightarrow {x^2} = \frac{{36}}{{100}}\,or\,x = 0.6\,m
\end{array}$
