Let's get this over with

Classical Mechanics Level 2

Grain is loaded onto a train initially coasting at the speed $v_0 \text{ m/s}$ by dropping it vertically from a stationary silo at the rate $\alpha_m \text{ kg/s}$ . How long does it take for the train to pass the silo (in seconds)?

Details

The train is $l = 200$ m long, and has an empty weight of $m_0 = 500$ kg.
$v_0=45$ m/s
$\alpha_m = 20$ kg/s

The answer is 4.864.

1 solution

Pranshu Gaba
May 5, 2015

Let the train and its contents be the system.

Since there is no net force in the horizontal direction on the system, its (horizontal) momentum will remain conserved.

The initial momentum $p_0 = m_0 v_0$ .

At time $t$ , the mass of the system will be $m_0 + \alpha _m t$ , and let the velocity be $v$ . So, momentum at a time $t$ , is $p = (m_0 + \alpha _m t) v$ . Using conservation of momentum,

$p = p_0 \forall t \in [0, \infty)$

$m_0 v_0 = (m_0 + \alpha _m t) v$

$v = \frac{m_0 v_0 }{m_0 + \alpha _m t}$

Since $\int \mathrm{d}x = \int v ~ \mathrm{d}t$ ,

$\int _{0} ^{l} ~\mathrm{d}x =\displaystyle \int _{0} ^{t} \frac{m_0 v_0 }{m_0 + \alpha _m t} ~~ \mathrm{d}t$

On integrating, we get

$l = \frac{m_0 v_0} {\alpha _m} \ln\left(\frac{m_0 + \alpha _m t}{m_0} \right)$

After making $t$ the subject of the formula,

$t = \frac{m_0}{\alpha _ m} \left( e^{\frac{\alpha_m l}{m_0 v_0}} - 1\right)$

After substituting the given values, we get

$t \approx \boxed{4.864 \text{s}} ~~~~~~ _\square$

Exactly how i did that.

Kushal Patankar - 6 years, 1 month ago

What is this? I never got it in my school.

Hafizh Ahsan Permana - 6 years, 1 month ago

Can you tell me which step you didn't understand? I will try to explain it better.

Pranshu Gaba - 6 years, 1 month ago

You seem to have ignored the subtlety that there is in fact a time varying net force in the negative x direction that is strictly decreasing. This is due to the fact that the sand has 0 initial velocity in the x direction, then is rapidly accelerated to the instantaneous velocity of the train. In a time dt, the sand gains a forward momentum of $(\alpha dt)v(t)$ , and thus there is a backwards reaction force $F(t)=\alpha v(t)$ . The momentum equation is precisely $p_i - \int_0^t \alpha v(t) dt = (m_i +\alpha t )v(t)$ .

I was taking the train and the sand ALREADY in it to be my system, thus considering the newly entering sand to be not yet part of the system. My rationale was that your solution only accounted for the addition of mass to the system and its corresponding effect on momentum, but did not account for the force required to accelerate that mass from zero. If I am incorrect, which of course I could be, it would mean that it does not make a difference if the sand entered the train with its same velocity. My physical intuition says that a completely vertical entry of the sand presents more of a hindrance to the motion than one with equal velocity.

Would like to hear some input from the resident physics admin.

Mohamed Elsayed - 5 years, 2 months ago

Consider all the sand and the train to be the system. Then the time varying force you are referring to is in fact an internal force. The force applied by the train on the sand that accelerates it is equal and opposite to the force applied by the sand on the train that decelerates it.

In other words, the momentum of the train decreases by $\int_0^t \alpha v(t) dt$ and the momentum of the sand increases by $\int_0^t \alpha v(t) dt$ . However, since the sand and the train are both part of the system, the change in momentum due to the internal force between the sand and the train is $- \int_0^t \alpha v(t) dt + \int_0^t \alpha v(t) dt$ , that is zero.

The momentum of the system is simply $p = (m_{i} + \alpha t ) v(t)$

Pranshu Gaba - 5 years, 2 months ago

Let's get this over with

The answer is 4.864.

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