anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
the situation, before collision is: |dw:1433610623171:dw|
anonymous
  • anonymous
ok!

Looking for something else?

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

More answers

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

Looking for something else?

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