In the railroad car collision described (400,000 kg car moving at 8.0 m/s couples to a stationary 1,200,000 kg car and they move together), what is the speed of both cars after they connect? A) 1.0 m/s B) 2.0 m/s C) 4.0 m/s D) 8.0 m/s

What kind of collision is this?

Because the two railroad cars connect and move together afterward, this is a perfectly inelastic collision. In that case, kinetic energy is not conserved, but momentum is conserved.

Write the momentum conservation equation

Total momentum before equals total momentum after:

$$m_1v_1 + m_2v_2 = (m_1+m_2)v_f$$

Here,

• $m_1=400{,}000\,\text{kg}$, $v_1=8.0\,\text{m/s}$
• $m_2=1{,}200{,}000\,\text{kg}$, $v_2=0\,\text{m/s}$ (stationary)

Solve for the shared final speed

$$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2} =\frac{(400{,}000)(8.0)+(1{,}200{,}000)(0)}{400{,}000+1{,}200{,}000}$$

$$v_f=\frac{3{,}200{,}000}{1{,}600{,}000}=2.0\,\text{m/s}$$

Match to the answer choices

$2.0\,\text{m/s}$ corresponds to B.