91Ó°ÊÓ

The height of putty is$\mathrm{h}=30 \mathrm{cm}=0.3 \mathrm{m}$

The initial stretch of spring is $\mathrm{x}=0.0400 \mathrm{m}$

The mass of frame is: $\mathrm{m}=0.150\mathrm{kg}$

The mass of putty is: $\mathrm{M}=0.200 \mathrm{kg}$

The speed of frame is found by momentum conservation of frame and putty. The maximum distance travelled by frame from its initial equilibrium position is found by energy conservation.

The speed of putty before strike on frame is calculated as:

$v_p=\sqrt{2gh}$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{p}}=\sqrt{\left(2\right(9.8 \mathrm{m}/{\mathrm{s}}^2\left)\right(0.3 \mathrm{m})} \\ {\mathrm{v}}_{\mathrm{p}}=2.42 \mathrm{m}/\mathrm{s} \end{gathered}$$

Apply the momentum conservation for frame and putty.

$\left(\mathrm{m}+\mathrm{M}\right){\mathrm{v}}_{\mathrm{f}}={\mathrm{mv}}_{\mathrm{p}}$

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.150\mathrm{kg}+0.200\mathrm{kg}\right){\mathrm{v}}_{\mathrm{f}}=\left(0.200\mathrm{kg}\right)\left(2.42\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{f}}=1.38\mathrm{m}/\mathrm{s} \end{gathered}$$

The force constant of the spring is calculated as:

$kx=\left(m+M\right)g$

Here, k is the force constant and g is the gravitational acceleration.

$$\begin{gathered} \mathrm{k}\left(0.0400\mathrm{m}\right)=\left(0.150\mathrm{kg}+0.200\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{k}=85.75\mathrm{N}/\mathrm{m} \end{gathered}$$

The maximum distance moved by the frame from its equilibrium position is given as:

$$\begin{gathered} \frac{1}{2}k{\left(x+I\right)}^2=\frac{1}{2}\left(m+M\right){v_f}^2 \\ k{\left(x+I\right)}^2=\left(m+M\right){v_f}^2 \end{gathered}$$

Here, I is the maximum distance moved by frame.

Substitute all the values in the above equation.

$$\begin{gathered} \left(85.75N/m\right){\left(0.040m+I\right)}^2=\left(0.150kg+0.200kg\right){\left(1.38m/s\right)}^2 \\ 0.040m+I=0.088m \\ I=0.088m \end{gathered}$$

Therefore, the maximum distance moved by the frame from its equilibrium position is 0.048 m.