Question

When an empty freight train mass \[{m_0}\] starts, loading of coal in the train begins at a constant rate \[r\] from a stationary hopper. If the track is horizontal and engine pull \[F\] is constant, deduce expression for speed of the train at a function of time \[t\] . Neglect all resistive forces.

Answer 1

Hint: The mass of the train would be increasing in time due to the coal. We can use Newton's second law to calculate the acceleration of the train.
Formula used: In this solution we will be using the following formulae;
 \[F = ma\] where \[F\] is the force acting on a body, \[m\] is the mass of the body, and \[a\] is the acceleration of the body.
 \[a = \dfrac{{dv}}{{dt}}\] , where \[v\] is the instantaneous velocity of an accelerating body, and \[t\] is time at which the body has such velocity, \[\dfrac{{dv}}{{dt}}\] signifies instantaneous rate of change of velocity with time.

Complete Step-by-Step solution:
Initially, the freight train was empty with an initial mass of \[{m_0}\] . Now, we are told that freight trains are being loaded with coal at a constant rate of \[r\] (i.e. rate of loading of the mass of coal). Hence, mass after a particular time \[t\] would be
 \[m = {m_0} + rt\]
Now, the force said to act on the train is \[F\] and is constant. Hence from newton’s second law, we may write that
 \[F = ma\] where \[F\] is the force acting on a body, \[m\] is the mass of the body, and \[a\] is the acceleration of the body
Hence,
 \[a = \dfrac{F}{m} = \dfrac{F}{{{m_0} + rt}}\]
But
 \[a = \dfrac{{dv}}{{dt}}\] , where \[v\] is the instantaneous velocity of an accelerating body, and \[t\] is time at which the body has such velocity, \[\dfrac{{dv}}{{dt}}\] signifies instantaneous rate of change of velocity with time.
Hence, we have
 \[\dfrac{{dv}}{{dt}} = \dfrac{F}{{{m_0} + rt}}\]
Hence, the velocity would be
 \[v = \int_0^v {dv} = \int_0^t {\dfrac{F}{{{m_0} + rt}}dt} \]
Hence, by integrating the above we get
 \[v = \dfrac{F}{r}\ln \left( {\dfrac{{{m_0} + rt}}{{{m_0}}}} \right)\]
Which is the velocity as a function of time.

Note: For clarity, we get the equation \[m = {m_0} + rt\] through the following reasoning. We are given that the coal was loaded at a constant rate of \[r\] . This Implies that the rate of change of mass of coal is \[r\] as in
 \[r = \dfrac{m}{t}\] , hence, the mass after a time \[t\] is
 \[{m_c} = rt\] . This would be added to the mass of the empty freight train, hence total mass is
 \[m = {m_0} + {m_c} = {m_0} + rt\]