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

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
$m = {m_0} + {m_c} = {m_0} + rt$