Question

In the railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.10 m/s and collides with an initially stationary, fully loaded boxcar of mass 4.60m. 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 1.10 m/s?

Answer 1

Given,

The mass of the empty freight car is m

The initial velocity of the empty freight car is u=1.10 m/s

The mass of the fully-loaded freight car is 4.60m

The initial velocity of the fully-loaded car is zero.

From the law of conservation of momentum, the total momentum of the system remains constant. That is, the total momentum of the cars before the collision is equal to their total momentum after the collision.

A.

Given the cars will stick together after the collision.

From the law of conservation momentum,

`\begin{gathered} mu=(m+4.60m)v \\ \Rightarrow mu=m(1+4.60)v \\ \Rightarrow u=5.60v \end{gathered}`

Where v is the speed of the two cars after the collision.

On substituting the known values,

`\begin{gathered} 1.10=5.60v \\ \Rightarrow v=(1.10)/(5.60) \\ =0.2\text{ m/s} \end{gathered}`

Thus the speed of the two cars after the collision is 0.2 m/s

B.

If the two cars are at rest after the collision it means the momentum of the two cars after the collision is zero.

Therefore, from the law of conservation of momentum,

`mu+4.60m* u_0=0`

Where u₀ is the velocity of the loaded car before the collision.

On rearranging the above equation,

`\begin{gathered} u+4.60u_0=0 \\ \Rightarrow u_0=(-u)/(4.60) \end{gathered}`

On substituting the known value,

`\begin{gathered} u_0=(-1.10)/(4.60) \\ =-0.24\text{ m/s} \end{gathered}`

The negative sign indicates that the loaded car was initially moving in the direction opposite to the direction of motion of the empty car.

Thus the initial speed of the loaded car is 0.24 m/s.