Question 14SP.4: For the 200-kg space vehicle of Sample Prob. 14.1, it is kno......

STRATEGY: Because there are no external forces, use the conservation of linear momentum. Although it is not immediately apparent, you will also need to use the conservation of angular momentum to solve this problem.

MODELING and ANALYSIS: Choose the space vehicle as your system.After the explosion, the system is composed of three parts: A, B, and C. Fig. 1 shows the momenta of the system before and after the explosion. From the conservation of linear momentum, you have

\mathbf{L}_{1}=\mathbf{L}_{2}\colon\qquad\qquad\quad m\mathbf{v}_{0}=m_{A}\mathbf{v}_{A}+m_{B}\mathbf{v}_{B}+m_{C}\mathbf{v}_{C}             (1)

From conservation of angular momentum about point O, you have

(\mathbf{H}_{o})_{1}=(\mathbf{H}_{o})_{2}\colon\qquad0=\mathbf{r}_{A}\times m_{A}\mathbf{v}_{A}+\mathbf{r}_{B}\times m_{B}\mathbf{v}_{B}+\mathbf{r}_{C}\times m_{C}\mathbf{v}_{C}             (2)

Recall from Sample Prob. 14.1 that v_{0}= (150 m/s)i and

m_{A}=100\,\,\mathrm{kg}\qquad m_{B}=60\,\,\mathrm{kg}\qquad m_{C}=40\,\,\mathrm{kg}

\mathbf{r}_{A} = (555 m)i – (180 m)j + (240 m)k

\mathbf{r}_{B} = (255 m)i – (120 m)k

\mathbf{r}_{C} = (105 m)i + (450 m)j – (420 m)k

Then, using the information given in the statement of this problem, rewrite Eqs. (1) and (2) as

200(150i) = 100(270i − 120j + 160k) + 60\left[(\nu_{B})_{x}\mathbf{i}+(\nu_{B})_{z}\mathbf{k}\right]+ 40 \left[(\nu_{C})_{x}\mathbf{i}+(\nu_{C})_{y}\mathbf{j}+(\nu_{C})_{z}\mathbf{k}\right]             (1′)

0=100\left|\begin{array}{ccc}i & j & k \\555 & -180 & 240 \\270 & -120 & 160 \end{array}\right|+60\left|\begin{array}{ccc}i & j & k \\255 & 0 & -120 \\\left(v_B\right)_x & 0 & \left(v_B\right)_z \end{array}\right|+40\left|\begin{array}{ccc}i & j & k \\105 & 450 & -420 \\\left(v_C\right)_x & \left(v_C\right)_y & \left(v_C\right)_z\end{array}\right|        (2′)

Equate the coefficient of j in Eq. (1′) and the coefficients of i and k in Eq. (2′). After reductions, you obtain the following three scalar equations:

(\nu_{C})_{y}-300=0

450(\nu_{C})_{z}+420(\nu_{C})_{y}=0

105(\nu_{C})_{y}-450(\nu_{C})_{x}-45\,000=0

which yield, respectively,

(\nu_{C})_{y}=300\qquad\ (\nu_{C})_{z}=-280\qquad(\nu_{C})_{x}=-30

The velocity of part C is thus

\mathbf{v}_{C} = − (30 m/s)i + (300 m/s)j – (280 m/s)k ◂

Equating the coefficients of the i and k terms on each side of Eq. (1′) and solving for the unknown components of the velocity of B gives

(\nu_{B})_{x}=70\;\mathrm{m/s}\qquad(\nu_{B})_{z}=-80\;\mathrm{m/s}

So

\nu_{A}=\sqrt{(270\;\mathrm{m}/\mathrm{s})^{2}+(-120\;\mathrm{m}/\mathrm{s})^{2}+(160\;\mathrm{m}/\mathrm{s})^{2}}=336.0\;\mathrm{m}/\mathrm{s}

\nu_{B}={\sqrt{(70\;{\mathrm{m/s}})^{2}+(0)^{2}+(-80\;{\mathrm{m/s}})^{2}}}=106.3\;{\mathrm{m/s}}

\nu_{C}={\sqrt{(-30\,\mathrm{m}/s)^{2}+(300)^{2}+(-280\,\mathrm{m}/s)^{2}}}=411.5\,\mathrm{m}/s

The initial kinetic energy is

T_{1}={\frac{1}{2}}m\nu_{0}^{2}={\frac{1}{2}}(200\;{\mathrm{kg}})(150\;\mathrm{m/s})^{2}=2250\;{\mathrm{kJ}}

The final kinetic energy is

T_{2}={\textstyle{\frac{1}{2}}}m_{A}\nu_{A}^{2}+{\textstyle{\frac{1}{2}}}m_{A}\nu_{A}^{2}+{\textstyle{\frac{1}{2}}}m_{A}\nu_{A}^{2}

          ={\frac{1}{2}}(100\;{ k g})(336.0\;\mathrm{m/s})^{2}+{\frac{1}{2}}(60\;{ k g})(106.3\;\mathrm{m/s})^{2}+{\frac{1}{2}}(40\;{ k g})(411.5\;\mathrm{m/s})^{2}

          = 9370 kJ

So

\Delta T=T_{2}-T_{1}=9370\,{\mathrm{kJ}}-2250\,{\mathrm{kJ}}\qquad\qquad\Delta T=7120\,{\mathrm{kJ}}  ◂

REFLECT and THINK: The negative signs for (v_{C})_{x}~and~(v_{C})_{z} indicate that the velocity is not directed as shown in Fig. 1. We also notice that the directions of the components of v_{C} are opposite to those of v_{A}. Given the lack of external forces, it seems reasonable to expect a more symmetric spread of velocities in all directions. You should also notice that the explosion added a lot of energy to the system.