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$$
\begin{array}{l}{\text { Three odd-shaped blocks of chocolate have the following }} \\ {\text { masses and center-of-mass coordinates: }(1) 0.300 \mathrm{kg},(0.200 \mathrm{m},} \\ {0.300 \mathrm{m}) ;(2) 0.400 \mathrm{kg},(0.100 \mathrm{m},-0.400 \mathrm{m}) ;(3) 0.200 \mathrm{kg},} \\ {(-0.300 \mathrm{m}, 0.600 \mathrm{m}) . \text { Find the coordinates of the center of mass }} \\ {\text { of the system of three chocolate blocks. }}\end{array}
$$

$$
x_{c m}=\frac{m_{A} x_{A}+m_{B} x_{B}+M_{C} x_{C}}{m_{A}+m_{B}+m_{C}}
$$

$$
y_{c m}=\frac{M_{A} y_{A}+m_{B} y_{B}+M_{C} y_{C}}{M_{A}+m_{B}+M _C}
$$

$$
x_{\mathrm{cm}}=\frac{(0.3)(0.2)+(0.4)(0.1)+(0.2)(-0.3)}{0.3+0.4+0.2}=0.0444 \mathrm{m}
$$

$$
y_{c m}=\frac{0.3(0.3)+(0.4)(-0.4)+0.2(0.6)}{0.3+0.4+0.2}=0.0556 \mathrm{m}
$$

$$
\begin{array}{l}{\text { A } 1200-\text { kg station wagon is moving along a straight highway }} \\ {\text { at } 12.0 \mathrm{m} / \mathrm{s} \text { . Another car, with mass } 1800 \mathrm{kg} \text { and speed } 20.0 \mathrm{m} / \mathrm{s} \text { , }} \\ {\text { has its center of mass } 40 \mathrm{m} \text { anead of the center of mass of the sta- }} \\ {\text { tion wagon (Fig. E8.54). (a) Find the position of the center of mass of }} \\ {\text { the system consisting of the two automobiles. (b) Find the magnitude }}\end{array}
$$

$$
\begin{array}{l}{\text { of the total momentum of the system from the given data. (c) Find the }} \\ {\text { speed of the center of mass of the system. (d) Find the total momen- }} \\ {\text { tum of the system, using the speed of the center of mass. Compare }} \\ {\text { your result with that of part (b). }}\end{array}
$$

$$
m_{A}=1200 \mathrm{kg}
$$

$$
m_{B}=1800 \mathrm{Kg}
$$

$$
M=M_{A}+M_{B}=3000 \mathrm{kg}
$$

$$
V_{A}=12 m / s
$$

$$
V_{B}=20 \mathrm{m/s}
$$

$$
x_{c m}=\frac{m_{A} x_{A}+m_{B} x_{B}}{m_{A}+M_ B}=\frac{0+180 0(40)}{3000}=24 m
$$

$$
C .M=24 m \quad \text { to the Right } \quad (40) \ ??  \quad 16 m
$$

$$
P_{x}=M_{A} v_{A x}+m_{B} v_{B x}=(1200)(12)+(1800)(20)=5.04 \times 10^{4} kg.m/s
$$

$$
V_{cmx}=\frac{m_A v_{Ax}+m_B v_{Bx}}{M_A+M_B}= \frac{1200(12)+(1800)(20)}{3000}=16.8 \mathrm{m/s}
$$

$$
P_{x}=V_{C m}(M_A+M_ B)=16.8 * 3000=5.04*10^{4} \mathrm{kg}. \mathrm{m} / \mathrm{s}
$$

$$
\begin{array}{l}{\text { A system consists of two particles. At } t=0 \text { one par- }} \\ {\text { ticle is at the origin; the other, which has a mass of } 0.50 \mathrm{kg} \text { , is on }} \\ {\text { the } y \text { -axis at } y=6.0 \mathrm{m} \text { . At } t=0 \text { the center of mass of the system }} \\ {\text { is on the } y \text { -axis at } y=2.4 \mathrm{m} \text { . The velocity of the center of mass is }} \\ {\text { given by }\left(0.75 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} \hat{\imath} \text { (a) Find the total mass of the system. }} \\ {\text { (b) Find the acceleration of the center of mass at any time } t .} \\ {\text { (c) Find the net external force acting on the system at } t=3.0 \mathrm{s}}\end{array}
$$

$$
y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}}{m_{1}+m_{2}}
$$

$$
m_{1}+m_{2}=\frac{m_{1} y_{1}+m_{2} y_{2}}{y_{cm}}=\frac{0+0.5(6)}{2.4}=1.25 \mathrm{kg}
$$

$$
\text {total mass}(M)=1.25=m_{1}+m_{2} \rightarrow  m_{1}=1.25 - 0.5=0.75 \mathrm{kg}
$$

$$
\vec{a}_{c m}=\frac{\overrightarrow{d v}_{c m}}{d t}=(1.5 t) \hat{i}
$$

$$
\sum \vec{F}_{e x t}=M \vec{a}_{c m}=1.25(1.5 t) \hat{\imath}
$$

$$
@ \  t=3 s \longrightarrow 1.25(1.5)(3) \hat i=5.6 \hat i \quad \mathrm{N}
$$

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