Question

Two vehicles are traveling when they enter an intersection and crash and stick together. Both have a mass of 1,650 kg and both are traveling at 15 m/s. If one is headed North and the other is headed East, after the collision they end up traveling NE together at what speed (in m/s)? Please input your answer as a positive number with two decimal places.

Answer 1

10.61 m/s

Step-by-step explanation:

To find the final speed, we will use the conservation of momentum in each direction, so we can write the following equations:

`\begin{gathered} p_(ix)=p_(fx) \\ m_1v_(1x)+m_2v_(2x)=(m_1+m_2)v_(fx) \\ \text{and} \\ p_(iy)=p_(fy) \\ m_1v_(1y)+m_2v_(2y)=(m_1+m_2)v_(fy) \end{gathered}`

Where m1 and m2 are the mass of the vehicles, v1 and v2 are their respective velocities in each direction, and Vfx and Vfy are the final velocities in each direction.

If one of the vehicles is headed north, its horizontal velocity Vx = 0 m/s. In the same way if the other is headed east, its vertical velocity Vy = 0m/s

Therefore, we can replace the values and solve the first equation as:

`\begin{gathered} 1650(0)+1650(15)=(1650+1650)v_(fx) \\ 24750=3300v_(fx) \\ (24750)/(3300)=v_(fx) \\ 7.5m/s=v_(fx) \end{gathered}`

In the same way, we can solve the second equation as:

`\begin{gathered} 1650(15)+1650(0)=(1650+1650)v_(fy) \\ 24750=3300v_(fy) \\ (24750)/(3300)=v_(fy) \\ 7.5m/s=v_(fy) \end{gathered}`

Now, we know that the vertical and horizontal speed was 7.5 m/s. So, we can calculate the final speed using the Pythagorean theorem as:

`\begin{gathered} v=\sqrt[]{(v_(fx))^2+(v_(fy))^2} \\ v=\sqrt[]{(7.5)^2+(7.5)^2} \\ v=\sqrt[]{56.25+56.25} \\ v=\sqrt[]{112.5}=10.61\text{ m/s} \end{gathered}`

So, the speed after the collision was 10.61 m/s