Question

aA .265kg croquet ball makes an elastic head-on collision with a second ball intially at rest. The second ball moves off with half the original speed of the first ball. What is the mass of the second ball? What fraction of the original kinetic energy (thetaK/K)gets transferred to the second ball?

Answer 1

So far I know that te second mass is half of the first so,
\[M _{b2}=1/2M _{b1}\]
and that ball 2 is heavier than the first ball.

Answer 2

Also the ealstic colllsion formula:
\[1/2m _{a}v _{a}^{2}+1/2m _{b}v _{2}^{b}=1/2m _{a}v _{a}^{'}+1/2m _{b}v_{b}^{'}\]
where "'" stands for the final momentum in an elastic collsion.

Answer 3

The problem would simplify like so:
\[1/2m _{a}v _{a}^2=1/2m _{a}v a_{2}^{'}+1/2m _{b}(1/2v _{b}^2)\]

Answer 4

From here I'm stuck

Answer 5

Hmm..let's see,
Conservation of linear momentum states that:\[m1v1i+m2v2i=m1v1f+m2v2f\]

Based on the given condition (the second ball moves off with half the original speed of the first ball):\[v2f= 1/2 (v1i)\]

Now, plug in the available quantities and we get:\[0.265(v1i) + 0 = 0.265(v1f) + 1/2(m2)(v1i)\]

Finding m2 will give you :\[m2= [0.53(v1i-v1f)]/v1i\]

Now, we still have many unknowns. But remember the coefficient of restitution, e, for elastic collision. For perfectly elastic collision, e=1.0. Hence:\[e=[-(v1f-v2f)]/[v1i-v2i]=1\]

Plug in the values and solve for V1f, note that V2f=1/2(v1i), and v2i = 0 (at rest):\[1= [-(v1f-0.5v1i]/[v1i-0]\]

Solving for v1f: \[v1f = -1/2 (v1i)\]

Now go back to m2 (this is really what we are supposed to find) and substitute the value of v1f that we just obtained.\[m2 = [0.53 (v1i+0.5v1i)]/v1i= 0.795 kg\]

Answer 6

You should find exactly\[\frac{m _{2}}{m _{1}}=3\]

Answer 7

http://openstudy.com/users/mos1635#/updates/4f93a537e4b000ae9eca2ce5

Answer 8

I think it understand it now, but the ratio for kinetic energy doesn't appear to be correct when I enter it in.

Answer 9

try K2(final)/K1(initial)

Answer 10

It says: "The correct answer does not depend on the variables: K2,K1 ." when I try to enter it in.

Answer 11

no my frend
i ment to calculate K2(final)
and
to calculate K1(initial)
divide them and find the ratio
K2(final)/K1(initial)
sorry.......

Answer 12

K1=1/2 m1*V^2
K2' 1/2 3m1 *(V/2)^2
K2'/K1=.......=12:1 (i think. please check my calc...)

Answer 13

K2'/K1 cannot be greater than 1.
Forlumae by mos1635 are correct.
Numerical value is K2'/K1 = 3/4

Answer 14

thanks @Vincent-Lyon.Fr you are wright. again :)

Answer 15

You did the hard work ;-)

Answer 16

^ Thanks vincent.