Question

An empty freight car A of mA = 2.75 x 104kg coasts along a horizontal track at 3.20 m/s until it couples to a stationary car B of mB = 5.00 x 104kg. There is negligible friction with the track, and the brakes are off. What is the speed of the two cars as they move along the track after the interaction?

Answer 1

We can find the speed of the two cars after the interaction from the Law of Conservation of Momentum:

$$m_Av_A+m_Bv_B=(m_A+m_B)v,$$

here, $m_A=2.75\cdot 10^4\ kg$ is the mass of the freight car A, $v_A=3.20\ \dfrac{m}{s}$ is the speed of the freight car A, $m_B=5.0\cdot 10^4\ kg$ is the mass of the stationary car B, $v_B=0 \ \dfrac{m}{s}$ is the speed of the car B and $v$ is the speed of the two cars after the interaction

Then, from this equation we can find $v$:

$$v=\dfrac{m_Av_A}{m_A+m_B}=\dfrac{2.75\cdot 10^4\ kg\cdot 3.20\ \dfrac{m}{s}}{2.75\cdot 10^4\ kg+5.0\cdot 10^4\ kg}=1.13\ \dfrac{m}{s}.$$

Answer:

$v=1.13\ \dfrac{m}{s}.$