91Ó°ÊÓ

The change in the momentum is given as$\frac{\mathbf{m}}{\mathbf{L}\mathbf{}}{\mathbf{v}}^{\mathbf{2}}\mathbf{t}$

a) consider a longitudinal wave that travels through the slinky by comprising a part of the slinky with a length of (vt), where (v) is the speed of the pulse "the speed of the propagation of the wave". So, the mass of this part of the slinky is $\left(\mathrm{μ\pm ¹³Ù}\right)$, where (g) is the mass per unit length of the slinky. Hence, we can write the change in the longitudinal momentum as follows

Change in momentum = (The mass of the compressed part) x The change in velocity

=$\left(\mathrm{μ\pm ¹³Ù}\right)×\mathrm{V}$

Notice that, $\mathrm{μ}=\frac{\mathrm{m}}{\mathrm{L}}$where (m) is the mass of the slinky and (L) is the length of the slinky, and change in the momentum of the pulse between the first point of the slinky with zero velocity and the final point of the slinky with velocity (v).

Thus, Change in momentum =$\frac{\mathrm{m}}{\mathrm{L}}{\mathrm{v}}^2\mathrm{t}$

In order to find the impulse for the wave pulse in the slinky we make use of the principle that states

that the impulse equals the change in the momentum, so, we need first to find the force acting on the compressed part when it reaches the end of the slinky, that force can be determined using Hook's law

In the case, k= k' and $∆\mathrm{X}$= L, so, the magnitude of F becomes

$\mathrm{F}=\mathrm{k}\text{'}\mathrm{L}$

$\mathrm{Impulse}=\mathrm{Ft}=\mathrm{k}\text{'}\mathrm{Lt}$

Equating the Impulse and the Change in momentum

$\mathrm{k}\text{'}\mathrm{Lt}=\frac{\mathrm{m}}{\mathrm{L}}{\mathrm{v}}^2\mathrm{t}$

$$\begin{gathered} {\mathrm{v}}^2=\frac{\mathrm{k}\text{'}{\mathrm{L}}^2}{\mathrm{m}} \\ \mathrm{v}=\mathrm{L}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \\ \mathrm{v}=\left(2\mathrm{m}\right)\sqrt{\frac{1.5\mathrm{N}/\mathrm{m}}{0.25\mathrm{kg}}} \\ \mathrm{v}=4.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocity is $4.9\mathrm{m}/\mathrm{s}$