Collision (Task help for medicine entry exam)

[uberinferno]

Homework Statement

Two balls weighing 2.5 kg and 5 kg are subject to an elastic collision. Before the collision, the first and lighter ball was at rest, and the second one was moving at a speed of 3.5 m / s. Determine the kinetic energy of the lighter ball after the collision.
Solution: Ek=27,22J

Relevant Equations

Ek= (mv^2)/2
mv1+mv2=mv1prim+mv2prim

I am guessing you have to find the speed at which the lighter ball moves and then to use the formula for kinetic energy, which I've tried but I'm not getting it quite right.
We got
m1=5kg v1=3,5kg
m2=2.5kg v2=0
v1prim=? v2prim=? Ek=?
How do you get the speed of the lighter ball after collision if you do not have the speed of the heavier ball given after collision? (in the statement)

 

[PeroK]

According to the guidelines, you have to make an effort to solve this yourself. If you are not familiar with elastic collisions, then you need to study the material before attempting the problems.

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[uberinferno]

According to the guidelines, you have to make an effort to solve this yourself. If you are not familiar with elastic collisions, then you need to study the material before attempting the problems.

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

sorry, fixed it, read the guidelines
i've tried this, I'm stuck at this atm its why i am here

 

[PeroK]

sorry, fixed it, read the guidelines
i've tried this, I'm stuck at this atm its why i am here

You have given one key equation:

Relevant Equations::
mv1+mv2=mv1prim+mv2prim

Do you know what this equation is called? Is there a second equation? Hint: what does "elastic" mean?

 

[uberinferno]

You have given one key equation:
Do you know what this equation is called? Is there a second equation? Hint: what does "elastic" mean?

Law of constant impulse (i've learned all this from a book of my native language, sorry for mistranslations)
The other equation i know of is m1v1=m2v2
Elastic means that the kinetic energy of both systems is kept throughout the collision

 

[PeroK]

Law of constant impulse (i've learned all this from a book of my native language, sorry for mistranslations)

It's called conservation of momentum. One of the most important equation in all of physics.

The other equation i know of is m1v1=m2v2

That's not any equation that I know.

Elastic means that the kinetic energy of both systems is kept throughout the collision

There is only one system here (a system of two masses). Elastic means kinetic energy is conserved:

http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html

 

[uberinferno]

I used the bottom left formula and it helped. Thanks very much, my books did not contain any of these formulas, I've been bashing my head for a long time.

 

[PeroK]

View attachment 276498
I used the bottom left formula and it helped. Thanks very much, my books did not contain any of these formulas, I've been bashing my head for a long time.

It's important that you understand the steps to get those final equations. That's the real physics. If you can really understand what is going on there, it is very powerful.

Your book may be teaching physics in what may be called the "plug-and-chug" approach. This is where you are always given the numbers, you plug the numbers into the basic equations, then manipulate those equations with the specific numbers, rather than the general quantities $m_1, m_2, v_1, v_2$ etc.

Plug-and-chug is eventually a dead-end if you want to study real physics, but it may help get more students through simple exam questions without really understanding what they are doing.

That's my opinion anyway.

 

[PeroK]

View attachment 276498
I used the bottom left formula and it helped. Thanks very much, my books did not contain any of these formulas, I've been bashing my head for a long time.

Let me give you an example of what I mean by the "power" of those formulas. Let's look at the the three cases:

1) $m_1 < m_2$, hence $m_1 - m_2 < 0$

In this case, $v'_1 < 0$, which means the lighter mass rebounds off the larger mass, and changes its direction of motion..

2) $m_1 = m_2$

In this case, the first mass stops and remains at rest, and the second mass takes all the KE. That is to say: $$v'_2 = v_1, \ \ \text{and} \ \ v'_1 = 0$$

3) $m_1 > m_2$

In this case, $v'_1 > 0$, which means the large mass does not rebound off a small mass, but keeps moving in the same direction after the collision.

 

[uberinferno]

It's important that you understand the steps to get those final equations. That's the real physics. If you can really understand what is going on there, it is very powerful.

Your book may be teaching physics in what may be called the "plug-and-chug" approach. This is where you are always given the numbers, you plug the numbers into the basic equations, then manipulate those equations with the specific numbers, rather than the general quantities $m_1, m_2, v_1, v_2$ etc.

Plug-and-chug is eventually a dead-end if you want to study real physics, but it may help get more students through simple exam questions without really understanding what they are doing.

That's my opinion anyway.

i did understand how to get to the final formula, thing is my books are sh*t and i got nothing else to use, using the internet is kind off hard because the font confuses me and we've never done collisions in school, its like I'm trying to relearn everything
for eg. i had no idea there was that mv1^2+mv2^2=mv1prim^2+mv2prim^2 formula which you use in combination with mv1+mv2=mv1prim+mv2prim in order to get v1prim or v2prim (if 1 ball isn't moving), like that's all i needed in this case, bc my book did not contain it.
i understand how something is done when i just see it solved, its how i mostly learn but this helped too i guess i just needed the base formulas.
Thank you and I appreciate your effort of giving free help, u take ur time to be generous and help those in need and we students really need people like you.