Module 8 : Specialized Traffic Studies
Lecture 42 : Accident Studies
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Angular collision

Angular collision occurs when two vehicles coming at right angles collies with each other and bifurcates in different direction. The direction of the vehicles after collision in this case depends on the initial speeds of the two vehicles and their weights. One general case is that two vehicles coming from south and west direction after colliding move in its resultant direction as shown in Figure 1.
Figure 1: Angular collision of two vehicles resulting in movement in resultant direction
\begin{figure}\centerline{\epsfig{file=qfAcciAngularCollision.eps,width=8cm}}\end{figure}
The mass of the car 1 is $ m_1$ kg and the car 2 is $ m_2$ kg and the initial velocity is $ v_1$ m/s and $ v_2$ m/s respectively. So as the momentum is the product of mass and velocity. The momentum of the car 1 and car 2 is $ m_1v_1$ kgm/s and $ m_2v_2$ kgm/s respectively. By the law of conservation of momentum the final momentum should be equal to the initial momentum. But as the car are approaching each other at an angle the final momentum should not be just mere summation of both the momentum but the resultant of the two, Resultant momentum = $ \sqrt {(m_1v_1)^2+(m_2v_2)^2}$ kg m/s. The angle at which they are bifurcated after collision is given by $ \tan^{-1}(h/b)$ where h is the hypotenuse and b is the base. Therefore, the cars are inclined at an angle. Inclined at an angle = $ \tan^{-1}(m_2v_2/ m_1v_1)$. Now, since the mass of the two vehicles are same the final velocity will proportionally be changed. The general schematic diagrams of collision are shown in Figs. 2 to  4.
Figure 2: After collision movement of car 1 north of west and car 2 in east of north
\begin{figure}\centerline{\epsfig{file=qfAcciMovementAfterCollision1.eps,width=8cm}} % (Source: [1])} \end{figure}
Figure 3: After collision movement of car 1 and car 2 in north of east
\begin{figure}\centerline{\epsfig{file=qfAcciMovementAfterCollision2.eps,width=8cm}}\end{figure}
Figure 4: After collision movement of car 1 north of east and car 2 in south of east
\begin{figure}\centerline{\epsfig{file=qfAcciMovementAfterCollision3.eps,width=8cm}} % (Source: [7])} \end{figure}

Numerical example

Vehicle A is approaching from west and vehicle B from south. After collision A skids $ 60^0$ north of east and B skids $ 30^0$ south of east as shown in Figure 4. Skid distance before collision for A is 18 m and B is 26 m. The skid distances after collision are 30m and 15 m respectively. Weight of A and B are 4500 and 6000 respectively. Skid resistance of pavement is 0.55 m. Determine the pre-collision speed.

Solution

Let: initial speed is $ v_{A1}$ and $ v_{B1}$, speed after skidding before collision is $ v_{A2}$ and $ v_{B2}$, speed of both the vehicles A and B after collision is $ v_{A3}$ and $ v_{B3}$, final speed is $ v_{A4}$ and $ v_{B4}$ is 0, initial skid distance for A and B is $ s_{A1}$ and $ s_{B1}$, final skid distance for A and B is $ s_{A2}$ and $ s_{B2}$, and weight of vehicle A is $ W_A$ and Weight of vehicle B is $ W_B$.
  1. After collision:  Loss in kinetic energy of each cars= Work done against skid resistance (can be obtained from equation. [*])
    $\displaystyle \frac{W_Av_{A3}^2}{2g} = W_A~f~s_{A2}$      

    As $ v_{A4}$ = 0, it is not considered in the above equation
    $\displaystyle v_{A3}$ $\displaystyle =$ $\displaystyle \sqrt {2gfs_{A2}}$  
    $\displaystyle v_{A3}$ $\displaystyle =$ $\displaystyle 18~m/s$  

    Similarly, we calculate $ v_{B3}$ using the similar formula and using $ s_{B2}$
    $\displaystyle v_{B3} = 12.7~m/s$      

  2. At collision: Momentum before impact is momentum after impact (resolving along west-east direction and using equation. [*])
    $\displaystyle \frac{W_A}{g} \times v_{A2} + 0$ $\displaystyle =$ $\displaystyle \frac{W_B}{g} \cos B v_{B3} + \frac{W_A}{g} \cos A v_{A3}$  
    $\displaystyle v_ {A2}$ $\displaystyle =$ $\displaystyle \frac{W_B}{W_A} \cos B v_{B3} + \cos A v_{A3}$  
      $\displaystyle =$ $\displaystyle \frac{6}{4.5} \cos 30 \times 12.7 + \cos 60 \times 18$  
    $\displaystyle v_ {A2}$ $\displaystyle =$ $\displaystyle 23.66~m/s.$  

    Resolving the moments along south- north direction
    $\displaystyle \frac{W_B}{g} \times v_{B2} + 0$ $\displaystyle =$ $\displaystyle \frac{W_A}{g} \sin A v_{A3} - \frac{W_B}{g} \sin B v_{B3}$  
    $\displaystyle v_{B2}$ $\displaystyle =$ $\displaystyle \frac{W_A}{W_B} \sin A v_{A3} - \sin B v_{B3}$  
      $\displaystyle =$ $\displaystyle \frac{4.5}{6} \times \sin 60 \times 12.7 - \sin 30 \times 18$  
    $\displaystyle v_{B2}$ $\displaystyle =$ $\displaystyle 5.34~m/s$  

  3. Before collision:  Loss in kinetic energy of each cars= Work done against skid resistance (can be obtained from equation. [*])
    $\displaystyle \frac{W_A(v_{A1}^2 - v_{A2}^2)}{2g}$ $\displaystyle =$ $\displaystyle W_A.f.s_{A2}$  
    $\displaystyle v_{A1}$ $\displaystyle =$ $\displaystyle \sqrt {2gfs_{A1} + v_{A2}^2}$  
      $\displaystyle =$ $\displaystyle 27.45 m/s = 99~km/hr$  

    Similarly, using the same equation and using $ s_{B2}$
    $\displaystyle v_{B1}$ $\displaystyle =$ $\displaystyle \sqrt{2gfs_{B1} + v_{B2}^2}$  
      $\displaystyle =$ $\displaystyle 17.57 m/s = 63.26~km/hr$  

    Answer: The pre-collision speed of the vehicle A (approaching from west) is $ v_{A1}$ = 99 km/hr and vehicle B (approaching from south) is $ v_{B1}$ = 63.26 km/hr.