Answer to Question #144539 in Physics for Saad

Question

A car with mass 1800 kg traveling south at a speed of 35 m/s collides at an intersection with a 2500 kg van traveling west at a speed of 40m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected.

Answer 1

Let's write down the momentum conservation law:

$$m_1\mathbf{v}_1 + m_2\mathbf{v}_2 = (m_1+m_2)\mathbf{v}'$$

where $m_1 = 1800kg$, $v_1 = 35m/s$, $m_2 = 2500kg$, and $v_2 = 40m/s$. $\mathbf{v}'$ is their velocity after the collision. Its magnitude is:

$$v' = \dfrac{|m_1\mathbf{v}_1 + m_2\mathbf{v}_2|}{m_1 + m_2}$$

Since $m_1\mathbf{v}_1$ and $m_2\mathbf{v}_2$ form right triangle, the magnitude of their sum will be:

$$|m_1\mathbf{v}_1 + m_2\mathbf{v}_2| = \sqrt{(m_1v_1)^2 + (m_2v_2)^2}$$

Thus, obtain:

$$v' = \dfrac{\sqrt{(m_1v_1)^2 + (m_2v_2)^2}}{m_1 + m_2}\\ v' = \dfrac{\sqrt{(1800\cdot 35)^2 + (2500\cdot 40)^2}}{1800 + 2500} \approx 27.49\space m/s$$

The direction is shown in the figure.