91Ó°ÊÓ

Given that a 1200-kg SUV is moving along a straight highway at 12.0 m/s.

Another car, with mass 1800 kg and speed 20.0 m/s, has its center of mass 40.0 m ahead of the center of mass of SUV.

Mass of the SUV ${\mathrm{m}}_{\mathrm{A}}=1200\mathrm{kg}$

Speed of SUV ${\mathrm{v}}_{\mathrm{A}}=12\mathrm{m}/\mathrm{s}$

Mass of lead car ${\mathrm{m}}_{\mathrm{B}}=1800\mathrm{kg}$$$" role="math" localid="1665053234412" width="9" height="19">

Speed of lead car ${\mathrm{v}}_{\mathrm{B}}=20\mathrm{m}/\mathrm{s}$

Let the origin be at center of mass of SUV.

Let x-axis lies along the line joining the both cars.

So ${\mathrm{x}}_{\mathrm{A}}=0$

And ${\mathrm{x}}_{\mathrm{B}}=40\mathrm{m}$

Now center of mass is$\mathbf{x}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{A}}{\mathbf{x}}_{\mathbf{A}}\mathbf{+}{\mathbf{m}}_{\mathbf{B}}{\mathbf{x}}_{\mathbf{B}}}{{\mathbf{m}}_{\mathbf{A}}\mathbf{+}{\mathbf{m}}_{\mathbf{B}}}$

Where${\mathbf{m}}_{\mathbf{i}}\mathbf{}\mathbf{\text{'}}\mathbf{s}$ are masses and${\mathbf{x}}_{\mathbf{i}}\mathbf{}\mathbf{\text{'}}\mathbf{s}$ are positions

Now center of mass is$\mathrm{x}=\frac{{\mathrm{m}}_{\mathrm{A}}{\mathrm{x}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{x}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}$

So

$$\begin{gathered} \mathrm{x}=\frac{\left(12000\mathrm{kg}\right)\left(0\right)+\left(18000\mathrm{kg}\right)\left(40\mathrm{m}\right)}{1200\mathrm{kg}+1800\mathrm{kg}} \\ =\frac{0+72000}{3000} \\ =24\mathrm{m} \end{gathered}$$

Hence, the center of mass is between the two cars, 24 m to the right of SUV and lead car.

Momentum of system is $\mathrm{P}={\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}$

So

$$\begin{gathered} \mathrm{P}=\left(1200\mathrm{kg}\right)\left(12\mathrm{m}/\mathrm{s}\right)+\left(1800\mathrm{kg}\right)\left(20\mathrm{m}/\mathrm{s}\right) \\ =5.04×{10}^4\mathrm{kgm}/\mathrm{s} \end{gathered}$$

Hence, momentum of the system is $5.04×{10}^4\mathrm{kg}\mathrm{m}/\mathrm{s}$.

Speed of center of mass is$\mathrm{v}=\frac{{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}$

$$\begin{gathered} \mathrm{v}=\frac{\left(1200\mathrm{kg}\right)\left(12\mathrm{m}/\mathrm{s}\right)+\left(1800\mathrm{kg}\right)\left(20\mathrm{m}/\mathrm{s}\right)}{1200\mathrm{kg}+1800\mathrm{kg}} \\ =16.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence speed of centre of mass is 16.8 m/s.

Momentum is $\mathrm{P}=\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{v}$

So $$\begin{gathered} \mathrm{P}=\left(1200\mathrm{kg}+1800\mathrm{kg}\right)\left(16.8\mathrm{m}/\mathrm{s}\right) \\ =5.04××{10}^4\mathrm{kgm}/\mathrm{s} \end{gathered}$$

Hence, momentum of the system is $5.04××{10}^4\mathrm{kgm}/\mathrm{s}$.

Which is same as in part (b).