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?

- anonymous

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- schrodinger

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- 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?

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## 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!

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