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You need to to use conservation of momentum.
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.
what's the definition of momentum?
mass x velocity
Yes. So p1 = ? and p2 = ?
0 and 6ms?
\[ p_1 = 0 \ kg.m/s \] yes. But \[ p_2 = m_2.v_2 = ... what? \]
what's m2 = ? what's v2 = ?
\[ v_2 = 6 m/s \] \[ m_2 = 100 kg \] 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' ?
is it 6 m/s...that's all i need to know..
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 \[ p_1' + p_2' = m_1v_1' + m_2v_2' = (m_1 + m_2)v \]
Now, set that equal to the total momentum before the collision and you can solve for v.
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.
Work the algebra here. Set that last expression for the total momentum equal to the total momentum before the collision.
it has to be one of them, its a multiple choice question.
can you please just tell me the answer i'll ask my teacher tomorrow
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.
**correction: p1 + p2 = 600, because m2.v2 = (100)(6) = 600. Hence you have that \[ (m_1 + m_2)v = 600 \]
10 and 6
Now, what's the value of \[ m_1 + m_2 \]?
well one is 6...so the other has to be 10!
they both weight 100 kg. Hence \[ m_1 + m_2 = ... what? \] and hence what is the solution to \[ (m_1 + m_2)v = 600 \ kg.m/s \]
if they weigh 100, it must be 6
\[ m_1 = m_2 = 100 \ kg \] hence \[ m_1 + m_2 = 200 \ kg \] therefore \[ (200 \ kg) v = 600 \ kg.m/s \] hence \[ v = ...what? \]
v = 3 m/s, yes. Makes intuitive sense as well, as that's half the original velocity.
god i'm stupid. thank you!
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.
just did. thank you!