- anonymous

In an isolated system, two roller skaters, each with a mass of 100 kg, collide. Skater 1 was initially at rest, while Skater 2 was moving at 6 m/s. They move off together. What is their speed?

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- JamesJ

You need to to use conservation of momentum.

- JamesJ

Let p1 be the momentum of the first skater before the collision on p1' afterwards. Likewise p2 and p2' for skater 2.
Then by conservation of momentum
\[ p_1 + p_2 = p_1' + p_2' \]
Now, use that the definition of momentum to find the joint velocity afterwards.

- anonymous

so 6m/s?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- JamesJ

what's the definition of momentum?

- anonymous

mass x velocity

- JamesJ

Yes. So p1 = ? and p2 = ?

- anonymous

0 and 6ms?

- JamesJ

\[ p_1 = 0 \ kg.m/s \] yes.
But
\[ p_2 = m_2.v_2 = ... what? \]

- anonymous

10m/s?

- JamesJ

what's m2 = ?
what's v2 = ?

- anonymous

3 m/s?

- JamesJ

\[ v_2 = 6 m/s \]
\[ m_2 = 100 kg \]
Hence
\[ p_2 = m_2.v_2 = (100 kg)(6 m/s) = 6000 kg.m/s \]
Therefore the total momentum before the collision is
\[ p_1 + p_2 = 0 + 6000 = 6000 \]
Now what about afterwards? What do you know about v1' and v2' ?

- anonymous

is it 6 m/s...that's all i need to know..

- JamesJ

No. That's wrong. After the collision, they both move off together so their velocities are the same. Call it v:
\[ v = v_1' = v_2' \]
Hence the total momentum after the collision is
\[ p_1' + p_2' = m_1v_1' + m_2v_2' = (m_1 + m_2)v \]

- anonymous

0!

- JamesJ

Now, set that equal to the total momentum before the collision and you can solve for v.

- anonymous

0 m/s!

- JamesJ

No. That can't possibly be right. If they are both stationary, v = 0, then their total momentum is zero. But we know it must be equal to 6000.

- anonymous

so 10??

- JamesJ

Work the algebra here. Set that last expression for the total momentum equal to the total momentum before the collision.

- anonymous

it has to be one of them, its a multiple choice question.

- anonymous

can you please just tell me the answer i'll ask my teacher tomorrow

- JamesJ

You're just guessing the answers. I'm trying to show you how to find it.
By Conservation of Momentum, momentum before and after the collision are equal:
\[ p_1 + p_2 = p_1' + p_2' \]
We know that
\[ p_1 + p_2 = 6,000 \ kg.m/s \]
We also know that
\[ p_1' + p_2' = (m_1 + m_2)v \]
where v is their shared velocity after the collision. Hence we have that
\[ (m_1 + m_2)v = 6000 \]
Now solve for v.

- JamesJ

**correction: p1 + p2 = 600, because m2.v2 = (100)(6) = 600. Hence you have that
\[ (m_1 + m_2)v = 600 \]

- anonymous

10 and 6

- JamesJ

Now, what's the value of
\[ m_1 + m_2 \]?

- anonymous

well one is 6...so the other has to be 10!

- JamesJ

they both weight 100 kg. Hence
\[ m_1 + m_2 = ... what? \]
and hence what is the solution to
\[ (m_1 + m_2)v = 600 \ kg.m/s \]

- anonymous

if they weigh 100, it must be 6

- JamesJ

\[ m_1 = m_2 = 100 \ kg \]
hence
\[ m_1 + m_2 = 200 \ kg \]
therefore
\[ (200 \ kg) v = 600 \ kg.m/s \]
hence
\[ v = ...what? \]

- anonymous

3!!!!!!!

- JamesJ

v = 3 m/s, yes. Makes intuitive sense as well, as that's half the original velocity.

- anonymous

god i'm stupid. thank you!

- JamesJ

Now, do yourself a favor, and write this solution out again on a blank piece of paper. When you can do that without looking at this or anything else, then you understand the problem.

- anonymous

just did. thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.