91Ó°ÊÓ

We know that,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{K}{\mathit{A}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{K}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Where m is the mass of the object, v is the speed, K is the spring constant and A is the amplitude.

According to example 14.5, consider block has a speed of$v_{1x}$along x-axis,

$E_1=\frac{1}{2}(M+m)v_{{}_{1}X}^2=\frac{1}{2}K{A}_{{}_1}^2$

Where M is the mass of the block,

Just after the collision, the speed and mass of the block-putty system is$v_{2x}$and M+m respectively,

According to conservation of momentum

$\begin{array}{l}M{v}_{1x}=(M+m)v_{2x}\\ v_{2x}=\left(\frac{M}{M+m}\right)v_{1x}\end{array}$

Total energy of the block putty just after collision,

${\mathrm{E}}_2=\frac{1}{2}(M+\mathrm{m})v_{2x}^2$

$\begin{array}{r}{\mathrm{E}}_2=\frac{1}{2}(\mathrm{M}+\mathrm{m}){\left(\left(\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}\right){\mathrm{v}}_{1\mathrm{x}}\right)}^2\\ {\mathrm{E}}_2=\frac{1}{2}{\mathrm{Mv}}_{1×}^2\left(\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}\right)\\ {\mathrm{E}}_2=\frac{\mathrm{M}}{(\mathrm{M}+\mathrm{m}{)}^2}{\mathrm{E}}_1\\ \frac{1}{2}{\mathrm{KA}}_2^2=\left[\frac{\mathrm{M}}{(\mathrm{M}+\mathrm{m}{)}^2}\right]\frac{1}{2}K{A}_1^2\\ \frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}=\sqrt{\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}}\end{array}$

The amplitude after is collision is half of the original amplitude

Hence,

$\begin{array}{r}\frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}=\frac{1}{2}\\ \frac{1}{2}=\sqrt{\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}}\\ \mathrm{M}=\frac{1}{4}(\mathrm{M}+\mathrm{m})\\ 4\mathrm{M}=\mathrm{M}+\mathrm{m}\\ \mathrm{m}=3\mathrm{M}\end{array}$

The total mechanical energy after collision is given by,

$\begin{array}{r}{\mathrm{E}}_2=\frac{1}{2}K{\mathrm{A}}_2^2\\ \text{As}{\mathrm{A}}_2=\frac{1}{2}{\mathrm{A}}_1\\ {\mathrm{E}}_2=\frac{1}{2}\mathrm{K}{\left(\frac{1}{2}{\mathrm{A}}_1\right)}^2\\ {\mathrm{E}}_2=\frac{1}{4}\left(\frac{1}{2}{{\mathrm{KA}}_1}^2\right)=\frac{1}{4}{\mathrm{E}}_1\end{array}$

The total mechanical energy after the collision is one-fourth the total original mechanical energy this means that rest of the energy was converted into thermal energy.

$E_{Thermal}=\frac{3}{4}{E}_1$

Hence, the value of putty mass m=3M and the fraction of the original mechanical energy converted into thermal energy is $\frac{3}{4}$.