# Momentum problems?

1. Dec 30, 2015

### David112234

1. The problem statement, all variables and given/known data
Problem 8.68
A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 5.10 m/s . Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.

PartA
An object with a mass of 28.0 kg is thrown sideways out of the car with a speed of 2.40 m/s relative to the car's initial velocity.

PartB
An object with a mass of 28.0 kg is thrown backward out of the car with a velocity of 5.10 m/s relative to the initial motion of the car.

PartC
An object with a mass of 28.0 kg is thrown into the car with a velocity of 5.80 m/s relative to the ground and opposite in direction to the initial velocity of the car.

2. Relevant equations
conservation of momentum?
M1V1+M2V2=M1V1+M2V2 (after)

3. The attempt at a solution

I am unsure about how to approach these problems, should I use newtons second law? Or should I use conservation of momentum? I am use to using conservation of momentum when two objects collide, but not the opposite

Here is my attempt for Part A
M1=170
V1= 5.10
M2=28
V2=2.40

170(5.10) +28(5.10)=(170-28)V1 + 28(2.40)

I put 170-28 because after the object is thrown out of the cart the carts total mass should be less.

I would like to know if this approach is correct before I submit my answers

2. Dec 30, 2015

### jambaugh

You'll want to pay attention to conservation of momentum. Energy won't be conserved because some external (or rather internal) energy is used to eject the cargo in that case and energy is presumably stored or converted to heat when the cargo is thrown into the car since it doesn't bounce out again (inelastic collision).

So you want to pick a specific inertia frame and calculate total momenta before and after cargo-car interactions for each case. You should use the final velocity variable v and solve for it in the conservation equations.

3. Dec 30, 2015

### CWatters

Regarding the left hand side of that equation. The problem says...

4. Dec 30, 2015

### David112234

Looks like I missed that detail, would this be correct then?

142(5.10) +28(5.10)=142V1 + 28(2.40)

5. Dec 30, 2015

### CWatters

Almost. Check out what Jam said about frames of reference. The problem says..

So the final velocity of the object is not 2.4m/s relative to the ground (which is the reference you used for the other velocities).

6. Dec 30, 2015

### David112234

How would that work, I know if it was thrown directly behind the cart it would be 2.4 -5.10 relative to the ground, if it was in front it would be 2.4+5.10, now if it is trown perpendicular to the cart its x component would be 5.10 and its y component would be 2.4.
So would I have to do vector addition to find the objects velocity relative to the ground?

7. Dec 30, 2015

### haruspex

You would, if you needed to know that overall velocity, but you don't.
Treat components along and perpendicular to the track separately.

8. Dec 30, 2015

### David112234

alright, so I it and found the object to be traveling 5.636 relative to the ground
then I do conservation of momentum with each of the components

170 (5.10) = 142V + 28 (5.10) the x component of the object
170 (5.10) = 142V + 28 (2.40) the y component of the object

Vx = 5.1
Vy = 5.632

This does not seem to make sense to me ,after trowing the object out the window the train would not move much in the y direction, it might go faster in the x since its mass decreased

Well i requested the answer for the assignment and it was 5.1, which was the x velocity I got, but what about the numbers for the Y velocity I got?

Last edited: Dec 30, 2015
9. Dec 30, 2015

### haruspex

You are mixing x and y components there. The cart has no velocity perpendicular to the track, either before or after. You cannot apply conservation of momentum in that direction because there is an external force (keeping the cart on the track).
As you calculated, the momentum it lost in the x direction was in direct proportion to the mass it lost, so no change to velocity.