A car of mass $1400$ $\mathrm{kg}$ moving south at $11$ $\mathrm{ms}^{−1}$ collides into another car of mass $1800$ $\mathrm{kg}$ moving east at $30$ $\mathrm{ms}^{−1}$. The cars are stuck together after the collision. Determine the velocity of the cars immediately after the collision.

I know the problem can be easily solved by conservation of momentum before and after collision. But I get confused in the direction in which the system will move after collision. I have solved many problem but I make a mistake every time. I have taken this very basic problem to understand the concept properly. Please throw some insight so that I can get a clear picture of after collision every time without making a mistake.


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  • $\begingroup$ You don't really need to know the direction......just use $m_1u_1+m_2v_2=(m1+m2)v'$...the direction can however be obtained by writing down momentum and KE equations, and solving them for v. $\endgroup$ – GRrocks Mar 17 '16 at 13:40

The way to approach this problem is to use the vector momenta and conserve them. If we orient our coordinate axis such that our $x$-coordinate is pointing North and our $y$-coordinate is pointing East, then the momentum of the first car is:

$$\vec{p}_{1}=m_{1}\vec{v}_{1} = 1400\begin{pmatrix}-11 \\ 0\end{pmatrix}\:\text{kg m s}^{-1}$$

Similarly, the momentum of the second car is:

$$\vec{p}_{2} = m_{2}\vec{v}_{2} = 1800\begin{pmatrix}0 \\ 30\end{pmatrix}\:\text{kg m s}^{-1}$$

Conservation of momentum means that:

$$\vec{p}_{\text{final}} = \vec{p}_{1} + \vec{p}_{2} = \begin{pmatrix}-15400 \\ 54000\end{pmatrix}\:\text{kg m s}^{-1}$$

But $\vec{p}_{\text{final}} = (m_{1} + m_{2})\vec{v}_{\text{final}}$ and so we can find the velocity of the cars after the collision:

$$\vec{v}_{\text{final}} = \frac{1}{1400+1800}\begin{pmatrix}-15400 \\ 54000\end{pmatrix}\:\text{m s}^{-1} = \frac{1}{16}\begin{pmatrix}-77 \\ 270\end{pmatrix}\:\text{m s}^{-1} \approx \begin{pmatrix}-4.8 \\ 16.9\end{pmatrix}\:\text{m s}^{-1}$$

So the joint cars are travelling east at approximately $16.9\:\text{m s}^{-1}$ and south at approximately $4.8\:\text{m s}^{-1}$.

  • $\begingroup$ I think the final velocity should be calculated only after finding the direction in which the system will move after collision. You haven't used vector here. $\endgroup$ – Hisenberg Mar 17 '16 at 13:50
  • $\begingroup$ @user109256 Finding the final velocity is equivalent to finding the direction in which the system will move after collision. What do you mean I haven't used vectors? $\endgroup$ – Thomas Russell Mar 17 '16 at 13:51
  • $\begingroup$ I mean you haven't visualized the direction in the system will move after collision. Considering cars as particles and then colliding and visualizing the final direction of the system on paper. $\endgroup$ – Hisenberg Mar 17 '16 at 13:59
  • 1
    $\begingroup$ @user109256 one doesn't need to draw arrows to correctly calculate vectors in Cartesian coordinates. $\endgroup$ – Ruslan Mar 17 '16 at 15:52

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