Question

For the arrangement shown in figure, the spring is initially compressed by $3cm$. when the spring is released the block collides with the wall and rebounds to compress the spring again. The maximum compression in the spring after collision (coefficient of restitution is $0.7$ ) is ______$cm$. (Take $\sqrt{5}=2.24,k=10^{4}N/m$ )

Answer 1

Answer: 2.24

From Energy conservation,
$\frac{1}{2}m{v}_0{}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{x}_0{}^2$
$\Rightarrow v_0=\sqrt{\frac{k}{m}\left(x_0{}^2-x^2\right)}$
.... [velocity of block just before collision]
Here $x_0=0.03m,x=0.01m$,
$k=10^{4}N/m,m=1kg$
$v_0=\sqrt{\frac{10^4}{1}\left[(0.03{)}^2-(0.01{)}^2\right]}$
$=\sqrt{10^4\times 10^{-4}[9-1]}$
$=2\sqrt{2}m/s$
After collision, $v=e{v}_0$
$=(0.7)2\sqrt{2}$
$=\frac{1}{\sqrt{2}}\times 2\sqrt{2}$
$=2m/s$
$\cdots$. [ $\because$ Take $0.7\approx \frac{1}{\sqrt{2}}$ ]
maximum compression in the spring ;
$\frac{1}{2}k{x}_m^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$
$\Rightarrow x_m=\sqrt{x^2+\frac{m}{k}{v}^2}$
$=\sqrt{(0.01{)}^2+\frac{1(2{)}^2}{10^4}}$
$=\frac{\sqrt{1+4}}{10^2}=\frac{2.24}{10^2}m$
$\therefore x_m=2.24cm$