Question

In the railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.20 m/s and collides with an initially stationary, fully loaded boxcar of mass 5.20m. The two cars couple together on collision.A. What is the speed of the two cars after the collision?B. Suppose instead that the two cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one was moving at 1.20 m/s?

Answer 1

Given data

*The given mass of the empty freight car is m_1 = m

*The given initial speed of the empty freight car is v_1 = 1.20 m/s

*The mass of the fully loaded boxcar is m_2 = 5.20 m

*The initial speed of the fully loaded boxcar is v_2 = 0 m/s

(A)

The formula for the speed of the two cars after the collision is given by the conservation of momentum as

`\begin{gathered} P_i=P_f \\ m_1v_1+m_2v_2=(m_1+m_2)v_{} \\ v=(m_1v_1+m_2v_2)/((m_1+m_2)) \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} v=((m)(1.20)+(5.20m)(0))/((m+5.20m)) \\ =0.19\text{ m/s} \end{gathered}`

Hence, the speed of the two cars after the collision is v = 0.19 m/s

(B)

As from the given data, the two cars are at rest after the collision. It means the final speed of the two cars equals to zero (v_f = 0 m/s).

The formula for the speed of the loaded box car before the collision is given by the conservation of momentum as

`\begin{gathered} p_i=p_f \\ m_1v_1+m_2v_l=(m_1+m_2)v_f \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} m(1.20)+(5.20m)(v_l)=(m+5.20m)(0)_{} \\ v_l=-(1.20m)/(5.20m) \\ =-0.23\text{ m/s} \end{gathered}`