Two hard steel balls collide in a perfectly elastic collision. The first ball has a mass of 1kg and is traveling at 5m/s east. The second ball has a mass of 2kg and is initially traveling at 7 m/s west along the same path so they collide head on. After the collision the second ball is traveling at 1m/s east. What is the velocity of the 1st ball after the collision? **How do I solve and show the work for this? :/ Thanks!!

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Two hard steel balls collide in a perfectly elastic collision. The first ball has a mass of 1kg and is traveling at 5m/s east. The second ball has a mass of 2kg and is initially traveling at 7 m/s west along the same path so they collide head on. After the collision the second ball is traveling at 1m/s east. What is the velocity of the 1st ball after the collision? **How do I solve and show the work for this? :/ Thanks!!

Physics
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here we have to apply the law of conservation of total momentum and the law of conservation of kinetic energy, since the collision is elastic
the situation, before collision is: |dw:1433610623171:dw|
ok!

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so, total momentum is: \[\Large {m_1}{v_1} - {m_2}{v_2} = 1 \times 5 - 2 \times 7 = ...\]
ok! so we get 5-14? = -9 ?
minus sign, means that the vector of the resultant momentum is oriented towards west
ohh okay! so -9, meaning 9 m/s to the west is the velocity of the 1st ball after collision?
no, -9 is the magnitude and sign of the resultant momentum
ohh okay! so it's like this? 5-14=-9 =resultant momentum ?
yes! now total kinetic energy is: \[\large KE = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 = \frac{1}{2} \times 1 \times 25 + \frac{1}{2} \times 2 \times 49 = ...\]
ok! so we get this? 61.5 ?
ok! correct!
now let's consider the situation after the collision
ohhh so 61.5 is just the KE right? so we solve for velocity now? :/
|dw:1433611229406:dw|
no, we have to apply the conservation of total momentum first
oh ok! how do we do that?
I call with u the speed of the first ball after collision
ok!
so we can write: conservation of total momentum: \[\Large \begin{gathered} {m_1}u + {m_2}v{'_2} = - 9 \hfill \\ 1 \times u + 7 \times 1 = - 9 \hfill \\ \end{gathered} \]
what is u?
-16?
in my reasoning i have considered positive velocity if the velocity is towards east, and negative velocity if that velocity is towards west
oh! so u=-16?
16 m/s to the west?
correct! that means the first ball, after the collision is goinf towards west
going*
ahh so that is our solution? :O
yes!
yay!! thank you!:D
:)

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