Show that Momentum is conserved in a reference frame

[Tonia]

Homework Statement

https://answers.yahoo.com/question/index?qid=20150607101609AAs8C9t
Show that momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

Homework Equations

Mass M for 2000 kg car, Mass m for 1500 kg car. To the observer on the ground: mv + 0 = (M + m) v` v` = Mv/M + m
The two cars are one mass after the collision (M + m), with velocity v`:
v` = Mv/M + m
After the collision, M + m is moving at a speed of v` - v
M (V -v) - mv = (M + m) (v` - v)
MV - Mv -mv + Mv + mv = (M + m) v` v` = Mv/(M + m)

The Attempt at a Solution

[/B]
To the observer on the ground: mv + 0 = (M + m) v` v` = Mv/M + m
M (V -v) - mv = (M + m) (v` - v)
MV - Mv -mv + Mv + mv = (M + m) v` v` = Mv/(M + m)

 

[unscientific]

Homework Statement

https://answers.yahoo.com/question/index?qid=20150607101609AAs8C9t
Show that momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

Homework Equations

Mass M for 2000 kg car, Mass m for 1500 kg car. To the observer on the ground: mv + 0 = (M + m) v` v` = Mv/M + m
The two cars are one mass after the collision (M + m), with velocity v`:
v` = Mv/M + m
After the collision, M + m is moving at a speed of v` - v
M (V -v) - mv = (M + m) (v` - v)
MV - Mv -mv + Mv + mv = (M + m) v` v` = Mv/(M + m)

The Attempt at a Solution

[/B]
To the observer on the ground: mv + 0 = (M + m) v` v` = Mv/M + m
M (V -v) - mv = (M + m) (v` - v)
MV - Mv -mv + Mv + mv = (M + m) v` v` = Mv/(M + m)

Let's assume that the 2000kg car is moving towards the right initially. In the reference frame 10m/s to the right, how fast is each car moving before collision? After collision, they move off as one with a certain speed. What is that speed?

 

[Noctisdark]

Hi there, to prove that momentum is conserved, show that Mv₁ + mv₂ = Mv'₁ + mv'₂, in the moving frame of reference whose direction is the velocity of the car M we don't have to worry about dealing with vectors, let it all be scalar where (+) is the direction of M, what are the new speeds of cars M and m ? good luck !

 

[Tonia]

Note: I wrote 2.00 kg car when I meant a 2000 kg car.
Before the collision, the larger car is moving at 20.0 m/s, and the car at rest is moving at 0.0 m/s. The speed they move (together) is: Mv/M+m = (2000 kg times 20.0 m/s) divided by (2000 kg + 1500 kg
= 40000 divided by 3500
= 11.4285714
= 11.43 m/s of cars together after collision.

 

[SammyS]

Note: I wrote 2.00 kg car when I meant a 2000 kg car.
Before the collision, the larger car is moving at 20.0 m/s, and the car at rest is moving at 0.0 m/s. The speed they move (together) is: Mv/(M+m) = (2000 kg times 20.0 m/s) divided by (2000 kg + 1500 kg
= 40000 divided by 3500
= 11.4285714
= 11.43 m/s of cars together after collision.

Be careful to use enough parentheses to say what you mean: Mv/(M+m) rather than Mv/M+m .

What you have so far is the result in a reference frame stationary with respect to Earth.

Take that result and calculate the momentum of each vehicle in the moving reference frame, before & after the collision. Is momentum conserved ?

 

[Tonia]

I am not sure how to.

 

[SammyS]

I am not sure how to.

The 2000 kg car travels at 20 m/s and the reference frame travels in the same direction at 10 m/s.

What is the velocity of the 2000 kg car with respect to the reference frame?

 

[Tonia]

10 m/s?? Because 20 m/s - 10 m/s = 10 m/s?

 

[SammyS]

10 m/s?? Because 20 m/s - 10 m/s = 10 m/s?

Right.

What is the velocity of the 1500 kg car with respect to the moving reference frame?

 

[Tonia]

0 m/s?

 

[SammyS]

0 m/s?

No.

If the 1500 kg car had the same velocity as the moving reference frame, the velocity of the 1500 kg car with respect to the moving reference frame would be 0 .

So I ask again:
What is the velocity of the 1500 kg car with respect to the moving reference frame?

 

[Tonia]

10m/s?

 

[SammyS]

10m/s?

In what direction?

If the 2000kg car and the reference frame move to the right with respect to Earth, that make their velocities both positive.

That also makes the 1500 kg car move to the left with respect to the moving reference frame. Correct?​

 

[Tonia]

10 m/s west(left)

 

[SammyS]

10 m/s west(left)

Correct.

So, we would consider that to be -10 m/s .

What is the velocity of the two cars after the collision (in the moving frame) ?

 

[Tonia]

11.43 m/s

 

[SammyS]

11.43 m/s

No. That's relative to Earth.

What is that relative to the moving reference frame?

 

[Tonia]

10 m/s? I'm sorry I'm just having difficulty understanding it.

 

[SammyS]

10 m/s? I'm sorry I'm just having difficulty understanding it.

After the collision, the two cars have a velocity (to the right, which is positive) of 11.43m/s relative to Earth.

The moving reference frame has a velocity of 10 m/s (to the right also).

What is the velocity of the two cars after the collision (relative to the moving reference frame) ?

 

[Tonia]

11.43 m/s minus 10 m/s?

 

[SammyS]

11.43 m/s minus 10 m/s?

Correct.

Is that about 1.43 m/s ?

It's positive, so to the right.

Now you have these three velocities in the moving reference frame.
Can you find the momentum of each car before the collision and the two cars together after the collision -- all in the moving reference frame?

 

[Tonia]

My calculator says 1.43
So I have 10m/s, the velocity of the 2000kg car with respect to the reference frame,
10 m/s , the veloc. of the 1500 kg car with respect to the reference frame, and
1.43 m/s, the velocity of the two cars after collision with respect to the refer. frame.
I am not sure which equation to use to find the momentum. I assume I need to use a momentum equation, but I'm not sure which one, there are so many in my textbook.

 

[SammyS]

My calculator says 1.43
So I have 10m/s, the velocity of the 2000kg car with respect to the reference frame,
10 m/s , the veloc. of the 1500 kg car with respect to the reference frame, and
1.43 m/s, the velocity of the two cars after collision with respect to the refer. frame.
I am not sure which equation to use to find the momentum. I assume I need to use a momentum equation, but I'm not sure which one, there are so many in my textbook.

I thought we established -10 m/s , for the velocity of the 1500 kg car with respect to the moving reference frame,

Really? You don't know how to calculate momentum?

 

[Tonia]

I'm studying on my own. I have a physics book. Maybe I should wait until I take physics in class.

 

[Tonia]

I forgot to put the negative sign on it.

 

[Tonia]

Is the momentun mass times velocity?

 

[SammyS]

Is the momentun mass times velocity?

Yes it is!

 

[Tonia]

So 10m/s velocity times 2000kg = 20,000
-10 m/s velocity times 1500 kg = -15,000
and 1.43 m/s velocity times (M +m) = 1.43 m/s times (2,000kg + 1500 kg) = 1.43 m/s times 3500 = 5,005

 

[SammyS]

So 10m/s velocity times 2000kg = 20,000
-10 m/s velocity times 1500 kg = -15,000
and 1.43 m/s velocity times (M +m) = 1.43 m/s times (2,000kg + 1500 kg) = 1.43 m/s times 3500 = 5,005

Is momentum conserved -- or nearly so ?

 

[Tonia]

I think so

 

[SammyS]

I think so

How can you check it?

 

[Tonia]

Does checking it have something to do with this: v` = Mv/M + m ??

 

[SammyS]

Does checking it have something to do with this: v` = Mv/M + m ??

Don't over-complicate this. Don't waste all of those momenta you have just calculated.

Again to be clear, do this in the moving reference frame.

Add up the momentum prior to the collision.

How does that compare to the momentum after the collision?

 

[Tonia]

Momentum before collision(with respect to the reference frame):
2000kg times 10m/s = 20,000
1500kg times -10m/s = -15,000
both are the same as the momentum after the collision, so momentum is conserved?

 

[Tonia]

Oops I forgot that I already calculated that, now I'm confused.