Show that Momentum is conserved in a reference frame

[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.

 

[Tonia]

Does it hqave something to do with multiplying by the 1.43?

 

[Tonia]

2000kg times 20 m/s = 40,000
1500kg times 0m/s = 0
I think I'm doing this wrong.

 

[SammyS]

11.4285714

2000kg times 20 m/s = 40,000
1500kg times 0m/s = 0
I think I'm doing this wrong.

Before the collision:
Total momentum = (momentum of 2000 kg car) + (momentum of 1500 kg car)

After collision:
Total momentum = momentum of the two cars together.

 

[Tonia]

total momentum of car before collision = (mom. of 2,000 kg car) 40,000 + (mom. of 1500kg car) 0
total momentum after collision = 40,000 + 0 = 40,000

 

[Tonia]

Or is it 20,000 + (-15,000) = 5,000?? before collision

 

[Tonia]

The momentum after collision would be the 5,005 that I got from multiplying the 1.43 m/s times the (M+m)??

 

[SammyS]

Or is it 20,000 + (-15,000) = 5,000?? before collision

Yes.

The momentum after collision would be the 5,005 that I got from multiplying the 1.43 m/s times the (M+m)??

Right.

Considering significant digits. those answers are the same.

However, if you use the value of 1.4285714 m/s (from 11.4285714 before rounding), you get very very close to the same thing.

 

[Tonia]

thank you for your help!

 

[SammyS]

thank you for your help!

You're welcome !

 

[Tonia]

One more question: How would you get the same answer using the 1.4285714 number?

 

[SammyS]

One more question: How would you get the same answer using the 1.4285714 number?

Try it.

Multiply 1.4285714 (m/s) times 3500 (kg) .

 

[Tonia]

oh that's right, I forgot. 1.43 m/s times (M+m) = 1.43 m/s times 3500kg = 5, 005. Nevermind. Thanks!