Question

Rewrite the equation m1v1+m2v2=m1v1'+m2v2
a. in the case of m1=m2=m (i.e., replace m1 and m2 with just m)
b. in the case of v1=v2=v
c. in the case of v1'=v2'=v'

Answer 1

I have no idea what this question is asking. Try to use the equation editor below to see if you can re-post the question more clearly.

Answer 2

\[m _{1}\times v_{1} + m _{2} \times v _{2} = m _{1} \times v _{1}\prime + m _{2}\times v _{2}\prime \]

Answer 3

Oh, okay. In that case, a should be really easy. Use the distributive property after you've made both m1 and m2 equal to m.

Answer 4

How do I make m1 and m2 equal to m? Do I just plug it in?

Answer 5

Yup. For example, if I have 5x, and x = 3, I have 5*3. If x = z, it's the same idea, but now I have 5*z.

Answer 6

So for a. would it just look like mv1+mv2=mv1'+mv2?

Answer 7

Almost. Now that you have m in both variables on both sides, can you combine them?

Answer 8

But if I did that wouldn't it turn out to be 0=0 because I would have to subtract mv1 from both sides and then mv2 from both sides which would equal 0.

Answer 9

Shouldn't. Because you have v1' on one side. Go ahead and show me how you would write it out. Even if you think it would end up as 0. I can point out where you're getting caught.

Answer 10

mv1+mv2=mv1'+mv2
-mv1
mv2=0+mv2
-mv2
0=0

Answer 11

Okay... so I think I see it. mv1' is different from mv1, so you can't subtract them.

Answer 12

Have you covered calculus, or derivatives at all in your course?

Answer 13

No, I'm in Algebra 2. This is in my Physics class.

Answer 14

Gotcha. Okay, well v1' and v1 are two different things. You may as well treat them as completely different variables; they are not interchangeable. If they were, you'd be totally correct and everything would cancel out.

Answer 15

So how do I simplify it?

Answer 16

Well, now that you know that you should treat them as different variables, and that you can use the distributive property, apply those two and you'll have an answer for a.

Which works like this: 5(x + 5) = 5x + 5*5

Answer 17

*that was an example of the distributive property. In this case, you'll probably want to use it in reverse...

Answer 18

m(v1+v2)=m(v1'+v2')?

Answer 19

Perfect. Now you can cancel out the m's.

Answer 20

So it's just v1+v2=v1'+v2'

Answer 21

That's as far as I would take it, unless you know the value of any of those variables.

Answer 22

It may not seem like a lot, but being able to eliminate a variable will save you loads of time. You want to get good and being able to recognize when you can do that. :)

Answer 23

*at being able to recognize

Answer 24

So, just to make sure I really know this,
b. v(m1+m2)=m1v1'+m2v2'
c. m1v1+m2v2=v(m1+m2)
Is that all right?

Answer 25

Looks like it. Double check what you "pulled out" in c, though. It shouldn't be just "v".

Answer 26

c. m1v1+m2v2=v'(m1+m2)?

Answer 27

There you go. :)

Answer 28

I would double check with at least one other person, though - always good to have more than one perspective.

Answer 29

Thanks a lot for all the help!

Answer 30

You're welcome!