for physics students or past students

[amar]

When something falls on a moving train, it slows the train down due to the conservation of momentum. However, when that same thing later falls out of the train, it does not affect the trains velocity. This is apparently due to the object moving at the same velocity as the train. I don't get this. 
 
Do we need to know how to do circuit analysis with the circuit element of grounding? I did these in Unit 1. I cannot find any in my Units 3 & 4 book or any other book I have seen. The ones that do, are actually voltage divider questions, not ones like analyzing resistance, voltage current, etc. Can someone find me a question to practice this, as I'm terrible at it!

[appianway]

When something falls off the moving train, the velocity of the train should NOT be reduced.

Let's also presume that no force acts on the object to make it drop (perhaps there's a trapdoor or something?). As dP = F/dT, and F = 0, dP also equals 0, hence the momentum of the object stays the same. The momentum of the object is initially mv (with m as its mass), and the momentum of the train is v(m + t), where t is the mass of the train. Due to the conservation of momentum, the initial velocity (which equals mv + mt) is the final velocity, and subtracting the momentum of the object (mv), we reach the momentum of the train, which stays at mt.

[amar]

When something falls off the moving train, the velocity of the train should NOT be reduced.

Let's also presume that no force acts on the object to make it drop (perhaps there's a trapdoor or something?). As dP = F/dT, and F = 0, dP also equals 0, hence the momentum of the object stays the same. The momentum of the object is initially mv (with m as its mass), and the momentum of the train is v(m + t), where t is the mass of the train. <b>Due to the conservation of momentum, the initial velocity (which equals mv + mt) is the final velocity, and subtracting the momentum of the object (mv), we reach the momentum of the train, which stays at mt.</b>

I did not understand the bolded part of your explanation appianway could you please re-explain it. i'm bad at understanding in terms of symbols

but since the mass in changing, and there is conserv of momentum,

should the velocity change?

[appianway]

OK. Think about it like this.

The initial momentum of the system is the momentum is the mass of the train and the mass of the object multiplied by the velocity. We can also consider the two individually, and such the momentum of the object is its mass multiplied by the velocity. However, this momentum does not change at the moment when it leaves the train, and therefore the final momentum of the object is its mass multiplied by its velocity. Due to the conservation of momentum, the initial momentum is the final momentum (which is the sum of the final momentum of the object and train). We can then minus the momentum of the object to find the momentum of the train, which is the same as the inital momentum as we subtract mv from v(m+t) = mv + mt - mv = mt

[mark_alec]

Train has mass: M
Object has mass: m
System has mass: M+m

Initially
Train has velocity: v
Object has velocity: v
Momentum: v(M+m)

Finally
Object has velocity: v
Momentum is still: v(M+m)
Therefore, train has final velocity: v.