Respuesta :
Answer:
Approximately [tex]6.7\; {\rm m\cdot s^{-1}}[/tex] at approximately [tex]63^{\circ}[/tex] west from north ([tex]{\rm N63^{\circ}W}[/tex].)
Explanation:
The velocity of both vehicles can be described with a two-dimensional vector:
[tex]\begin{aligned}\begin{bmatrix}(\text{north-south velocity}) \\ (\text{west-east velocity})\end{bmatrix}\end{aligned}[/tex].
(Note that the two directions are perpendicular to one another.)
For example, since the cookie vehicle is travelling north at [tex]5\; {\rm m\cdot s^{-1}}[/tex], its velocity vector will be:
[tex]\begin{aligned}v_{a} &= \begin{bmatrix}5 \\ 0\end{bmatrix}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Likewise, the velocity vector of the milk vehicle travelling west at [tex]15\; {\rm m\cdot s^{-1}}[/tex] will be:
[tex]\begin{aligned}v_{a} &= \begin{bmatrix}0 \\ 15\end{bmatrix}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].
The momentum vector of the [tex]m_{a} = 6000\; {\rm kg}[/tex] cookie vehicle will be:
[tex]\begin{aligned}p_{a} &= m_{a} \, v_{a} \\ &= (6000\; {\rm kg})\, \begin{bmatrix}5 \\ 0\end{bmatrix}\; {\rm m\cdot s^{-1}} \\ &= \begin{bmatrix}30000 \\ 0\end{bmatrix}\; {\rm kg\cdot m\cdot s^{-1}}\end{aligned}[/tex].
The momentum vector of the [tex]m_{a} = 4000\; {\rm kg}[/tex] milk vehicle will be:
[tex]\begin{aligned}p_{a} &= m_{a}\, v_{a} \\ &= (4000\; {\rm kg})\, \begin{bmatrix}0\\ 15\end{bmatrix}\; {\rm m\cdot s^{-1}} \\ &= \begin{bmatrix}0\\ 60000\end{bmatrix}\; {\rm kg\cdot m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the total momentum of the two vehicles before the collision will be:
[tex]\begin{aligned}p_{a} + p_{b} &= \begin{bmatrix}30000\\ 0\end{bmatrix}\; {\rm kg\cdot m\cdot s^{-1}} + \begin{bmatrix}0\\ 60000\end{bmatrix}\; {\rm kg\cdot m\cdot s^{-1}} \\ &= \begin{bmatrix}30000\\ 60000\end{bmatrix}\; {\rm kg\cdot m\cdot s^{-1}} \end{aligned}[/tex].
Let [tex]v[/tex] denote the velocity vector of the two vehicles right after they collide. With a total mass of [tex](m_{a} + m_{b}) = (6000\; {\rm kg} + 4000\; {\rm kg}) = 10000\; {\rm kg}[/tex], the total momentum of the two vehicles right after the collision will be: [tex]p = (m_{a} + m_{b})\, v[/tex].
Momentum is conserved. Hence, right after collision, the total momentum of the two vehicles will stay the same. Thus,
[tex]\begin{aligned}(m_{a} + m_{b})\, v = p = p_{a} + p_{b}\end{aligned}[/tex].
[tex]\begin{aligned}v &= \frac{p}{m_{a} + m_{b}} \\ &= \frac{p_{a} + p_{b}}{m_{a} + m_{b}} \\ &= \frac{\begin{bmatrix}30000 \\ 60000\end{bmatrix}\; {\rm kg \cdot m\cdot s^{-1}}}{10000\; {\rm kg}} \\ &= \begin{bmatrix}3 \\ 6\end{bmatrix}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Since the two directions (north-south and west-east) are perpendicular to each other, the Pythagorean Theorem can be applied to find the magnitude of this velocity:
[tex]\begin{aligned}\| v \| &= \left(\sqrt{3^{2} + 6^{2}}\right)\; {\rm m\cdot s^{-1}} \\ &\approx 6.7\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The angle between this velocity and the direction of north can be found as:
[tex]\begin{aligned}\arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right) &= \arctan \left(\frac{6}{3}\right) \approx 63^{\circ}\end{aligned}[/tex].
