Question

an open train car, with a mass of 2110 kg, coasts along a horizontal track at the speed 2.53 m/s. the car passes under a loading chute and, as it does so, gravel falls vertically into it for 3.01 s at the rate of 415 kg/s. ignore rolling friction. what is the cars speed ( v_{f} ) after the loading is completed? ( v_{f}= ) m/s

Explanation:

Step1: Calculate the mass of gravel

The mass of gravel \(m_{g}\) is given by the rate of gravel fall \(r\) times the time \(t\).
\(m_{g}=r\times t\)
Substituting \(r = 415\space kg/s\) and \(t=3.01\space s\), we get \(m_{g}=415\times3.01 = 1249.15\space kg\)

Step2: Apply conservation of momentum

The initial momentum \(p_{i}\) of the train - car is \(p_{i}=m_{c}v_{i}\), where \(m_{c}=2110\space kg\) and \(v_{i} = 2.53\space m/s\).
The final mass \(m_{f}=m_{c}+m_{g}\)
By conservation of momentum \(p_{i}=p_{f}\), i.e., \(m_{c}v_{i}=(m_{c}+m_{g})v_{f}\)
We can solve for \(v_{f}\):
\(v_{f}=\frac{m_{c}v_{i}}{m_{c}+m_{g}}\)
Substituting \(m_{c} = 2110\space kg\), \(v_{i}=2.53\space m/s\), and \(m_{g}=1249.15\space kg\)
\(v_{f}=\frac{2110\times2.53}{2110 + 1249.15}\)
\(v_{f}=\frac{5338.3}{3359.15}\)
\(v_{f}\approx1.59\space m/s\)

Answer:

\(1.59\)