Freight Car Velocity: Newton's 1st Law Explained

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In the discussion about a railway freight car carrying sand, the question centers on whether the car's velocity changes as sand leaks out. According to Newton's 1st Law, the car's velocity should remain constant in the absence of external forces. However, participants clarify that as the sand exits the car, it carries momentum away, which does not affect the car's velocity directly. Newton's 2nd Law indicates that while the mass of the car decreases, the momentum of the sand leaving does not result in an increase in the car's velocity. Ultimately, the car's velocity remains unchanged despite the loss of mass from the sand.
mini_mia
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Hey guys!

I was wondering if I could get some help with this question:

A railway freight car that is carrying sand is traveling at a uniform velocity on a straight level track. A leak develops in the flat bottom, and the sand begins to pour out of a hole onto the ground. Assuming there aren't any external forces on the car (including friction), does the velocity of the car decrease, increase or remain the same?

According to Newton's 1st law, it says that the velocity of an object will remain constant unless it's acted upon by an external or unbalanced force, which there isn't any in this question. However, I am a bit confused about the sand falling out. Do this affect the velocity of the car? Thanks if you can help me!
 
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What does Newton's second law say about this situation? Remember that Newton's second law can be written as a relation between the net force and the rate of change of momentum.
 
Okay, so Newton's 2nd Law in terms of momentum is Δp = F Δt , which is also
F Δt = m Δv.

So in isolating velocity from the formula, as the mass decreases, the velocity would increase?
 
That is incorrect. Yes the mass of the sand is leaving the car, but the sand itself has a velocity and therefore takes it momentum with it.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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