Elastic Collision with 280kg Ball: Mass & KE Transfer Calculation

[rie_06]

280kg ball has elastic collision with 2nd ball which is initially at rest. Then 2nd ball moves with .5 velocity of 1st ball - what is mass of the 2nd ball? what percentage of KE gets transferred to 2nd ball

Homework Equations

I believe i use m1v1 + m2v2 = m1v1 +m2v2 (before & after), conservation of momemtum

then i think i would use 1/2mv2 + 1/2mv2 = 1/2mv2 + 1/2mv2 (energy before = energy after) conservation of energy
Maybe my algebra is just rusty, but I'm very confused.

The Attempt at a Solution

but i am having trouble just solving this the first equation. so if the m's cancel out i have v1 = v1 + v2 (the initial velocity of m2 = 0)
and then v2 = .5v1 so i have v1 = v1 + .5v1 is any of this correct? I'm confused

 

[neutrino]

If the m's cancel, the why would they ask you to find the mass of the second ball? And what's the initial velocity of the first ball?

Provide the question as it is in the book.

 

[radou]

First of all, you equation 'm1v1 + m2v2 = m1v1 +m2v2' implies 0 = 0, so it would be better to use v1' and v2' for the velocities after the collision.

 

[rie_06]

i agree - that's where my confusion is!
The question in the book does not state a velocity for the first ball.

 

[rie_06]

thanks for suggesting using v1' and v2' I did use them in my equation but forgot to mention them in the post
so m1v1 = m1v1' + m2v2' i then get v1 = v1' + .5v1 then .5 = v1'
does that make sense? should i continue with the next equation from here - or do i need to find the mass of the 2nd ball before continuing - i am still confused.

 

[radou]

thanks for suggesting using v1' and v2' I did use them in my equation but forgot to mention them in the post
so m1v1 = m1v1' + m2v2' i then get v1 = v1' + .5v1 then .5 = v1'
does that make sense?

The masses are not equal, at least we have no right to assume they are, so they can not cancel out.

 

[rie_06]

ok - let me try it without cancelling the mass
m1v1 = m1v1' + m2v2'
.280(v1) = .280(v1') + m2(.5v1)
is that how i should proceed?

 

[radou]

ok - let me try it without cancelling the mass
m1v1 = m1v1' + m2v2'
.280(v1) = .280(v1') + m2(.5v1)
is that how i should proceed?

Looks correct now, except it should be 0.5v1', unless I'm missing something.

 

[rie_06]

you are correct - let me work it some more and see what i come up with - thank you very much!

 

[rie_06]

i meant you are correct on the 0.5v1'
Thank you

 

[rie_06]

in solving .280(v1) = .280(v1)' + m2(0.5v1)
to get m2 by itself - i divide both sides by 0.5v1
.14 = .14 + m2
m2 = 0
doesn't seem right? what am i doing wrong?