Show that Momentum is conserved in a reference frame

[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!