Two vehicles are approaching an intersection. One is a 2500-kg pickup traveling at 14.0m/s from east to west (the-x-direction), and the other is a 1500-kg sedan going from south tonorth (the +y-direction) at 23.0m/s. (a) Find the x- and y-componentsof the net momentum of this system. (b) What are the magnitudeand direction of the net momentum?

The Mass of the pickup,$m_A=2500\mathrm{kg}$

The velocity of pick up,$v_A=v_{\mathrm{Ax}}=-14\mathrm{m}/\mathrm{s}$

The Mass of the sedan$m_B=1500\mathrm{kg}$

The velocity of sedan$v_B=v_{By}=23.0\mathrm{m}/\mathrm{s}$

As we know the momentum of an object is

$\overrightarrow{\mathbf{p}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{v}}$

If${\mathit{p}}_{\mathbf{x}}$ is x component of momentum and${\mathit{p}}_{\mathbf{y}}$ is y component of momentum

Then,

$$\begin{gathered} {\mathit{p}}_{\mathbf{x}}\mathbf{+}{\mathit{p}}_{\mathbf{y}}\mathbf{=}\mathit{m}{\overrightarrow{\mathbf{v}}}_{\mathbf{x}}\mathbf{+}\mathit{m}{\overrightarrow{\mathbf{v}}}_{\mathbf{y}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{p}\mathbf{=}\sqrt{{\mathbf{p}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{p}}_{\mathbf{y}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{p_x}{p_y}\right) \end{gathered}$$

Where${\mathit{v}}_{\mathbf{x}}$ is x component of velocity and${\mathit{v}}_{\mathbf{y}}$ is y component of velocity.

The x component of the net momentum of the system can be given as,

$$\begin{gathered} p_{Ax}+p_{Bx}=m_A{v}_{Ax}+m_B{v}_{Bx} \\ {p}_{\mathit{x}}=2500\mathrm{kg}\left(-14\mathrm{m}/\mathrm{s}\right)+1500\mathrm{kg}\left(0\right) \\ \mathit{}{p}_{\mathit{x}}=-35000\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The x component of the net momentum of the system is -35000kg.m/s .

The y component of the net momentum of the system can be given as.

$$\begin{gathered} p_{Ay}+p_{By}=m_A{v}_{Ay}+m_B{v}_{By} \\ {p}_{\mathit{y}}=2500\mathrm{kg}\left(0\right)+1500\mathrm{kg}\left(23\mathrm{m}/\mathrm{s}\right) \\ \mathit{}{p}_{\mathit{x}}=34500\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The y component of the net momentum of the system is 34500kg.m/s

$$\begin{gathered} p_x=-35000kg.\mathrm{m}/\mathrm{s} \\ {\mathrm{p}}_{\mathrm{y}}=345000\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The magnitude of the momentum

$$\begin{gathered} p=\sqrt{p_x^2+p_y^2} \\ =\sqrt{{\left(-35000\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2+{\left(34500\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2} \\ =49145\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The Magnitude of net momentum is 49145kg.m/s

The direction of net momentum

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{p_x}{p_y}\right) \\ ={\tan }^{-1}\left(\frac{34500\mathrm{kg}.\mathrm{m}/\mathrm{s}}{-35000\mathrm{kg}.\mathrm{m}/\mathrm{s}}\right) \\ =44.9° \end{gathered}$$

Since the x-component is negative and the y-component is positive

Therefore, the direction of net momentum is$44.9°$ North-of west.