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**1.A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.**

What fraction of the original kinetic energy is converted into internal energy in the collision?

What fraction of the original kinetic energy is converted into internal energy in the collision?

Here is my attempt:

Ki = (1/2)mv^2

kf = (m+M) L^2

Now I'm a bit confused what they mean by fraction, so I initial thought I had to divide Kf/Ki , which is:

2(m + M)L^2 / (mv^2)

but WebAssign marked it wrong. I have one more try and I want to try the other way, Ki/Kf but I don't want to get it wrong.

Can anyone help me and tell me if I'm doing this right or not. Thanks

Kevin