Question

A 6000 kg truck full of Girl Scout cookies traveling north at 5 m/s collides with a

4000 kg tanker truck full of milk traveling west at 15 m/s. The two remain locked
together after the collision. What is their velocity after the collision?

Answer 1

Approximately
`6.7\; {\rm m\cdot s^(-1)}` at approximately
`63^(\circ)` west from north (
`{\rm N63^(\circ)W}`.)

Step-by-step explanation:

The velocity of both vehicles can be described with a two-dimensional vector:

`\begin{aligned}\begin{bmatrix}(\text{north-south velocity}) \\ (\text{west-east velocity})\end{bmatrix}\end{aligned}`.

(Note that the two directions are perpendicular to one another.)

For example, since the cookie vehicle is travelling north at
`5\; {\rm m\cdot s^(-1)}`, its velocity vector will be:

`\begin{aligned}v_(a) &= \begin{bmatrix}5 \\ 0\end{bmatrix}\; {\rm m\cdot s^(-1)}\end{aligned}`.

Likewise, the velocity vector of the milk vehicle travelling west at
`15\; {\rm m\cdot s^(-1)}` will be:

`\begin{aligned}v_(a) &= \begin{bmatrix}0 \\ 15\end{bmatrix}\; {\rm m\cdot s^(-1)}\end{aligned}`.

When an object of mass
`m` travels at a velocity of
`v`, the momentum
`p` of that object will be
`p = m\, v`.

The momentum vector of the
`m_(a) = 6000\; {\rm kg}` cookie vehicle will be:

`\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}`.

The momentum vector of the
`m_(a) = 4000\; {\rm kg}` milk vehicle will be:

`\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}`.

Hence, the total momentum of the two vehicles before the collision will be:

`\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}`.

Let
`v` denote the velocity vector of the two vehicles right after they collide. With a total mass of
`(m_(a) + m_(b)) = (6000\; {\rm kg} + 4000\; {\rm kg}) = 10000\; {\rm kg}`, the total momentum of the two vehicles right after the collision will be:
`p = (m_(a) + m_(b))\, v`.

Momentum is conserved. Hence, right after collision, the total momentum of the two vehicles will stay the same. Thus,

`\begin{aligned}(m_(a) + m_(b))\, v = p = p_(a) + p_(b)\end{aligned}`.

`\begin{aligned}v &= (p)/(m_(a) + m_(b)) \\ &= (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}`.

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:

`\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}`.

The angle between this velocity and the direction of north can be found as:

`\begin{aligned}\arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right) &= \arctan \left((6)/(3)\right) \approx 63^(\circ)\end{aligned}`.