# Inelastic collision problem [closed]

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.

• 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. Mar 17, 2016 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 $4.8\:\text{m s}^{-1}$.

• 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. Mar 17, 2016 at 13:50
• @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? Mar 17, 2016 at 13:51
• 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. Mar 17, 2016 at 13:59
• @user109256 one doesn't need to draw arrows to correctly calculate vectors in Cartesian coordinates. Mar 17, 2016 at 15:52