Answer to Question #172925 in Physics for Margarette

Question

A car with a mass of 2500 kg and moving at 3 m/s northeast collides with a 3000 kg car moving northwest at 2.5 m/s . Find their final velocities. Assume a perfectly inelastic collision.

Answer 1

Let's choose the axes as shown in the figure below.

In a perfectly inelastic collision the momentum conservation law still holds. Thus, the total momentum before the collision in the projection on the x-axis is:

$$p_{x} = -m_1v_1$$

is equal to the total momentum after the collision in the projection on the x-axis:

$$p_{x}' = (m_1 + m_2)v_x'$$

where $m_1 = 2500kg, m_2 = 3000kg$ are the masses of the cars, $v_1 = 3m/s, v_2 = 2.5m/s$ are their speeds before the collision, and $v_x'$ is their speed after the collsion in x-direction. Thus, obtain:

$$-m_1v_1 = (m_1 + m_2)v_x'\\ v_x' = -\dfrac{m_1v_1}{m_1 + m_2} = -\dfrac{2500\cdot 3}{2500+3000} \approx -1.36m/s$$

Similarly, the momentum conservation law in the projection on y-axis gives:

$$m_2v_2 = (m_1 + m_2)v_y'$$

where $v_y'$ is car's speed after the collsion in y-direction. Thus, obtain:

$$m_2v_2 = (m_1 + m_2)v_y'\\ v_y' = -\dfrac{m_2v_2}{m_1 + m_2} = -\dfrac{3000\cdot 2.5}{2500+3000} \approx 1.36m/s$$

Thus, the final velocity is:

$$\mathbf{v}' = (-1.36, 1.36)\space m/s$$

Answer. $\mathbf{v}' = (-1.36, 1.36)\space m/s$.