# Momentum Problems...

## Recommended Posts

I have a few momentum based problems for basic physics that I'm drawing a complete blank on...

1. A fully dressed person is at rest in the middle of a pond on perfectly frictionless ice and must get to shore. How can this be accomplished?

(it wants the answer in terms of conservation of momentum)

2. The previous problem answered in terms of Newtons 3rd law.

and

3. A railroad diesel engine weighs four times as much as a freight car. If the diesel engine coasts at 5 km/h into a freight car that is initially at rest, how fast do the two coast after they couple together?

I've gotten this far: mass of diesel engine = 4 x mass of freight car. Momentum of diesel engine = 20 kgm/s, Momentum of freight car = 0 kgm/s making total Momentum before 20 kgm/s. So I set up the equation 20 kgm/s = v x (4mf + mf) (mf = mass of freight car).

I don't know what I'm supposed to do from there.

##### Share on other sites

1&2) do you mean the person is on a piece of ice, and the piece of ice is on the pond? if so, you may use the momentum of the ice as your source of momentum...

3) your equation is largely correct, but 20 should be changed into 20mf, now you should be able to do that.

##### Share on other sites

For #1. "Fully clothed" is a hint, though there are solutions that are available otherwise.

3. A railroad diesel engine weighs four times as much as a freight car. If the diesel engine coasts at 5 km/h into a freight car that is initially at rest, how fast do the two coast after they couple together?

I've gotten this far: mass of diesel engine = 4 x mass of freight car. Momentum of diesel engine = 20 kgm/s, Momentum of freight car = 0 kgm/s making total Momentum before 20 kgm/s. So I set up the equation 20 kgm/s = v x (4mf + mf) (mf = mass of freight car).

I don't know what I'm supposed to do from there.

You've equated the initial and final momentum, which is correct. Do that with the variables, rather than the numbers. (dttom has already pointed out that you don't actually know the value of the momentum)

##### Share on other sites

Oh ok I see why it should be 20mf and not 20 kgm/s for the total momentum. So then I get 20mf/5mf = V which makes 4mf = V. Is that all I can do? I thought it wanted an actual speed...but since I dont know mf (mass of the freight car) how can i find it?

i know the speed has got to be less than 5 km/h because now their coupled mass is greater, reducing the velocity.

i think the other problem means a person on a frozen pond that is completely frictionless, although it doesn't specify very clearly.

##### Share on other sites
Oh ok I see why it should be 20mf and not 20 kgm/s for the total momentum. So then I get 20mf/5mf = V which makes 4mf = V. Is that all I can do? I thought it wanted an actual speed...but since I dont know mf (mass of the freight car) how can i find it?

i know the speed has got to be less than 5 km/h because now their coupled mass is greater, reducing the velocity.

i think the other problem means a person on a frozen pond that is completely frictionless, although it doesn't specify very clearly.

Careful of your units. It's 20 km/hr * mf = 5 mf, and mf cancels

##### Share on other sites
For #1. "Fully clothed" is a hint, though there are solutions that are available otherwise.

Myself, I would use breathing.

##### Share on other sites

oh yeah it is 20 mf*km/hr so then the mf cancels and I'm left with velocity. I got all turned around on that problem, thanks for the help.

the other problem I'm just assuming creating any momentum in the right direction would get them to the shore...That's all I really want to say I don't know any technical way to word it lol.

##### Share on other sites

the other problem I'm just assuming creating any momentum in the right direction would get them to the shore...That's all I really want to say I don't know any technical way to word it lol.

You can use action/reaction to discuss this.