A 6.0kg metal ball moving at 4.0m/s hits a 6.0 kg ball of putty at rest and sticks to it. The two go on at 2.0 m/s.(adsbygoogle = window.adsbygoogle || []).push({});

a) How much energy does the metal ball lose in the collision?

b) How much energy does the putty ball gain in the collision?

c) What happened to the rest of the energy?

For a):

[tex]E_{klose} = E_{k} - E_{k'}[/tex]

[tex]E_{k} = 1/2mv^2

= 1/2(6.0kg)(4.0m/s)^2 - 1/2(6.0kg)(2.0m/s)^2

= 36J

[/tex]

Another method I tried was:

[tex]

W = F \Delta d

=m( \Delta v/ \Delta t)*(v_{f}+v_{i})/2* \Delta t

=6.0kg(-2m/s)*(3m/s)

=-36J

[/tex]

b) is simple:

[tex]

E_{k} = 1/2mv^2

= 1/2(6.0kg)(2.0m/s)^2

= 12J

[/tex]

The only problem I have with this question is the last part, C. The thing is, the momentum was conserved completely, so the energy could not be lost from friction or sound. Here is an example of trying to find the velocity of the ball and putty combined if no energy was lost to heat/sound/friction/etc:

[tex]1/2(6.0kg)(v)^2 + 1/2(6.0kg)(v)^2 = 1/2(6.0kg)(4.0m/s)^2[/tex]

[tex]2v^2 = 48J*2/6.0kg[/tex]

[tex]v = \sqrt{8}[/tex]

Which is impossible because of

[tex]mv = m^{'}v^{'}[/tex]

[tex]6.0kg*4.0m/s = 12.0kg*\sqrt{8}m/s[/tex]

Which is clearly unequal.

I asked this question to my teacher and he said that the conservation of energy does not always work. That is a ridiculous answer :uhh:

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# Metal ball sticks to putty. Where does the energy go?

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