91Ó°ÊÓ

In an isolated system of two or more bodies the total momentum of the system remains conserved for any process.

Let the +x direction is the same as Rebecca’s initial velocity as shown in diagram.

The conservation momentum of both thex andy components are as follows:

Now,x component is solved as follows:

$$\begin{gathered} m_R{v}_{R0}=m_R{v}_{R1}\cos {53.1}^{∘}−m_D{v}_{D1}\cos \mathrm{β} \\ \left(45\mathrm{kg}\right)×13\mathrm{m}/\mathrm{s}=\left(\right(45\mathrm{kg})×8\mathrm{m}/\mathrm{s}×0.6)−\left(65\mathrm{kg}\right)v_{D1}\cos \mathrm{β} \\ \left(65\mathrm{kg}\right)v_{D1}\cos \mathrm{β}=369\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{D}1}\mathrm{³¦´Ç²õβ}=5.68\mathrm{m}/\mathrm{s}......\left(1\right) \end{gathered}$$

For, y component is solved as follows:

$$\begin{gathered} 0=m_R{v}_{R1}\sin (53.1{)}^{∘}−m_D{v}_{D1}\sin \mathrm{β} \\ 0=(4\mathrm{kg})×(8\mathrm{m}/\mathrm{s})×0.8−(65\mathrm{kg}\left)v_{D1}\sin \mathrm{β} \\ \right(65\mathrm{kg})v_{D1}\sin \mathrm{β}=288\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{D}1}\mathrm{²õ¾\pm ²Ôβ}=4.43\mathrm{m}/\mathrm{s}......\left(2\right) \end{gathered}$$

Divide equation (2) by equation (1) gives:

$$\begin{gathered} \frac{v_{D1}\sin \mathrm{β}}{v_{D1}\cos \mathrm{β}}=\frac{4.43\mathrm{m}/\mathrm{s}}{5.68\mathrm{m}/\mathrm{s}} \\ \tan \mathrm{β}=0.78 \\ \mathrm{β}={\tan }^{−1}\left(0.78\right) \\ \mathrm{β}={38}^{∘} \end{gathered}$$

Substitute into equation (1) gives,

$$\begin{gathered} v_{D1}=\frac{5.68m/s}{\cos 38°} \\ =7.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, Daniel velocity is 7.2 m/s making an angle of$38°$ from the initial direction.

The initial kinetic energy is solved when only Rebecca is moving as follows:

$$\begin{gathered} {\mathrm{K}}_0=\frac{1}{2}{\mathrm{m}}_{\mathrm{R}}{\mathrm{v}}_{\mathrm{R}0}^2 \\ =\frac{1}{2}\left(45\mathrm{kg}\right){\left(13\mathrm{m}/\mathrm{s}\right)}^2 \\ =3800\mathrm{J} \end{gathered}$$

The final kinetic energy is solved when only Rebecca and Daniel are moving as follows:

$$\begin{gathered} {\mathrm{K}}_1=\frac{1}{2}{\mathrm{m}}_{\mathrm{R}}{\mathrm{v}}_{\mathrm{R}1}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{D}}{\mathrm{v}}_{\mathrm{D}1}^2 \\ =\frac{1}{2}\left(45\mathrm{kg}\right){\left(8\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}\left(65\mathrm{kg}\right){\left(7.2\mathrm{m}/\mathrm{s}\right)}^2 \\ =3125\mathrm{J} \end{gathered}$$

The change in kinetic energy is solved as follows:

$$\begin{gathered} ∆K=K_1-K_0 \\ =\left(3125-3800\right) \\ =-675\mathrm{J} \end{gathered}$$

Thus, the kinetic energy is not conserved, and this is an inelastic collision.

Hence, the change in kinetic energy of the skaters is -675 J.