A 3.0 g coin moving to the right at 26.0 cm/s makes an elastic head-on colli

[lettertwelve]

Homework Statement

A 3.0 g coin moving to the right at 26.0 cm/s makes an elastic head-on collision with a 12.0 g coin that is initially at rest. After the collision, the 3.0 g coin moves to the left at 12.5 cm/s.
(a) Find the final velocity of the other coin.
_____cm/s to the right
(b) Find the amount of kinetic energy transferred to the 12.0 g coin.
_____Joules

Homework Equations

for part a, i know you have to do conversions to get to M and Kg.
for part b, .5mv^2

The Attempt at a Solution

i know that KE of the 2nd coin = (m of second coin)(.5)(Velocity of second coin ^2)

but I'm STUCK on part A!

 

[dwintz02]

Use conservation of momentum:

p(initial)=p(final)

p=m*v

 

[lettertwelve]

i found part A, and it's 9.6cm to the right.
but now i am stuck on part B.
no matter what i do, it's not working!

...i tried this:
KE of second coin=(Mass of second coin).5(Velocity of 2nd coin^2)
so it'd look like:
(.12)(.5)(9.6^2)
BUT since it's 9.6 CM/s, to convert it to m/s it would be .096
(.12)(.5)(.096^2)
which is: .00055296


why is it wrong!?

 

[kudoushinichi88]

What's the answer given in the book? Btw, 12.0 g is 0.012 kg.

 

[cristo]

I'll unmark this as solved since you still seem to be asking questions on it.

 

[haha]

ok i think this is how u do it
first u need to find kinetic energy before contact and find the kinetic energy of the coin A
and u subtract the original kinetic to the kinetic of coin A u will get the kinetic of coin B
(assumed that the moving coin is A and the resting coin is B)