Question

A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 10 m/s. The three vehicles become entangled and slide as one body. What are their speed and direction just after the collision?

Short answer

Speed of the resulting body is 4.19 m/s and the direction is $17.35°$above x-axis.

Step-by-step solution

Step 1. Given information is Mass of the truck = 2100 kgMasses of 2 cars are 1200 kg and 1500 kg.

We need to find out the speed of the entangled vehicles and their direction just after the collision.

Step 2. Applying Law of conservation of momentum for both axes.

Let 2100 kg truck as object 1

1200 kg car as object 2

1500 kg car as object 3

Using Law of conservation of mass,

For x-axis,

$$\begin{gathered} m_1{v}_1+m_3{v}_3=\left(m_1+m_2+m_3\right)v_{final.x} \\ 2100kg\times 2m/s+1500kg\times 10m/s=4800kg.v_{final.x} \\ {v}_{final.x}=\frac{19200kgm/s}{4800kg} \\ {v}_{final.x}=4m/s \end{gathered}$$

For y-axis,

$$\begin{gathered} m_2{v}_2=(m_1+m_2+m_3)v_{final.y} \\ 1200kg\times 5m/s=4800\times v_{final.y} \\ {v}_{final.y}=\frac{6000kgm/s}{4800kg} \\ {v}_{final.y}=1.25m/s \end{gathered}$$

Speed of the resulting body is,

$$\begin{gathered} v_{final}=\sqrt{{\left(v_{final.x}\right)}^2+{\left(v_{final.y}\right)}^2} \\ =\sqrt{{\left(4\right)}^2+{\left(1.25\right)}^2} \\ =4.19m/s \end{gathered}$$

and angle is

$$\begin{gathered} \tan \theta =\frac{v_{final.y}}{v_{final.x}} \\ \tan \theta =\frac{1.25m/s}{4m/s} \\ \theta =17.35° \end{gathered}$$