Two dimensional Collisions

PROBLEM:

A 1000 kg car traveling north-east at speed 20 m/s collides with a 1800 kg car traveling west at speed 16 m/s. After the collision the two cars stick together and skid for a distance. Calculate the speed and direction of the wreckage immediately after the collision. Is it an elastic or inelastic collision?

SOLVE:

Figure 1 depicts the described collision, where direction west – east is $x$ -axis, south – north is $y$ -axis. South-west – north-east direction forms an angle of $45^{\circ}$ with $x$ -axis. This is a two-dimensional (2D) event as it has components in both $x$ and $y$ directions.

Figure 1

Let $m_{1}$ and $m_{2}$ and $\vec{v}_{1}$ and $\vec{v}_{2}$ be the masses and velocities of car 1 and car 2, then $m_{1} =$ 1000 kg, $m_{2} =$ 1800 kg, speed $v_{1} = |\vec{v}_{1}| =$ 20 m/s and speed $v_{2} = |\vec{v}_{2}| =$ 16 m/s.

If we only consider the interaction between car 1 and car 2, and ignore all surrounding factors, such as the friction between the cars and the road surface, then this is an isolated system. The total linear momentum of an isolated system must be conserved, therefore, if $\vec{p}$ and $\vec{p'}$ are total linear momenta of the system before and after the collision, respectively, then

$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} \textbf{\scriptsize{(1)}} \qquad & $\vec{p}$ &= & $\vec{p'}$ \\ or&& & \\ \textbf{\scriptsize{(2)}} &$m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2}$ & = &$(m_{1} + m_{2})\vec{v'}$ \end{tabular} }$

where $\vec{v'}$ is the velocity of the wreckage immediately after the collision (whose $x$ and $y$ components are $\vec{v'}_{x}$ and $\vec{v'}_{y}$ , respectively). Note that
$\renewcommand{\arraystretch}{1.5}\textsf{ $\vec{p}$ = $m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2}$, and }$
$\vec{p'}$ = $(m_{1} + m_{2})\vec{v'}$

As this is a 2D collision, the total linear momentum of the system must be conserved in both $x$ and $y$ directions separately, thus,

$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} \textbf{\scriptsize{(3)}} \qquad & $m_{1} v_{1x} + m_{2} v_{2x} = (m_{1} + m_{2})v'_{x}$ \\ and& & & \\ \textbf{\scriptsize{(4)}} &$m_{1} v_{1y} + m_{2} v_{2y} = (m_{1} + m_{2})v'_{y}$ \end{tabular} }$

From Figure 1, it can be seen that $\vec{v}_{1x}$ and $\vec{v}_{1y}$ (the $x$ and $y$ components of velocity $\vec{v}_{1}$ ) have values of

$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} & $v_{1x} = |\vec{v}_{1x}| $ = &20$\cos{45^{\circ}} = $ &10$\sqrt{2}$ m/s \\ &$v_{1y} = |\vec{v}_{1y}| $ = &20$\sin{45^{\circ}} = $ &10$\sqrt{2}$ m/s \end{tabular} }$

Similarly, values of $x$ and $y$ components of velocity $\vec{v}_{2}$ can be calculated

$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} & $v_{2x}$ & = & $-16$ m/s \\ & $v_{2y}$ & = & $0$ m/s \end{tabular} }$

Re-arranging equations (3) and (4), $v'_{x}$ and $v'_{y}$ can be obtained as follows

$\renewcommand{\arraystretch}{2} \textsf{ \begin{tabular}{r c l} $v'_{x} $ &= &$\frac{m_{1} v_{1x} + m_{2} v_{2x}} {m_{1} + m_{2}}$ \\ &= &$\frac{(1000 \times 10 \sqrt{2}) + (1800 \times (-16)}{1000 + 1800}$ \\ & = &$-$5.2349 m/s \end{tabular} }$

$\renewcommand{\arraystretch}{2} \textsf{ \begin{tabular}{r c l} $v'_{y} $ &= &$\frac{m_{1} v_{1y} + m_{2} v_{2y}} {m_{1} + m_{2}}$ \\ &= &$\frac{(1000 \times 10 \sqrt{2}) + (1800 \times (0)}{1000 + 1800}$ \\ & = &5.0508 m/s \end{tabular} }$

As a result, the speed of the wreckage right after the collision is

$v' = \sqrt{v'^{2}_{x} + v'^{2}_{y}} =$ 7.2743 m/s

Direction of the wreckage can be determined by the angle $\theta$ , formed by $\vec{v'}$ with the $x$ -axis.

$\renewcommand{\arraystretch}{2} \textsf{ \begin{tabular}{l r c l} & $\tan \theta $ & = &$\frac{v'_{y}}{v'_{x}}$ \\ & & = & -0.9649 \\ \implies & $\theta$ & = &$(-0.7675 + k\pi)$ rad \\ & & = & $-43.9745^{\circ} + k 180^{\circ}$ \end{tabular} }$

where $k = 0, \pm1, \pm2, \quad \dots$

Based on the values of $v'_{x}$ and $v'_{y}$ , it can be deduced that $k$ = 1, hence $\theta =$ -43.9745 $^{\circ}$ + 180 $^\circ =$ 136.0255 $^{\circ}$ or 46.0255 $^{\circ}$ to the west of the north axis.

To determine whether a collision is elastic or inelastic, we consider the total kinetic energy of the system. A collision is elastic if there is no loss in the system’s kinetic energy after the collision.

Let $K$ and $K'$ be the kinetic energy of the system before and after the collision, respectively, then

Kinetic energy of the system before the collision:

$\renewcommand{\arraystretch}{2} \textsf{ \begin{tabular}{r c l} $K$ &= &$\frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$ \\ &= &$\frac{1000 \times 20^{2}}{2} + \frac{1800 \times 16^{2}}{2}$ \\ & = &430400 J \end{tabular} }$

Kinetic energy of the system after collision:

$\renewcommand{\arraystretch}{2} \textsf{ \begin{tabular}{r c l} $K'$ &= &$\frac{1}{2}(m_{1} + m_{2})v'^{2} $ \\ &= &$ \frac{(1000 + 1800) \times 7.2743^{2}}{2} $ \\ & = & 74082 J \end{tabular} }$

As $K$ ≠ $K'$ , this is an inelastic collision.

Note that a collision in which the involved objects stick together and move with a common final velocity is termed perfectly inelastic collision. In this type of collision, the maximum amount of energy is lost from the system’s kinetic energy. Also note that, the total energy of the system is conserved, as kinetic energy of the system is only transformed into internal energy (cars’ deformation) and heat after the collision.