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For the Ballistic pendulum in the image below: enter image description here
Are we allowed to assume that after the bullet hit the log (mass M), then there's a conservation of energy? (Thus $\frac{1}{2}(M+m)V^2 = (M+m)gh$)

Why the wires don't do work?
They have components (Tension) both in the X and Y axis.

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The wires don't perform work because the direction of motion of the log is always perpendicular to the wires. If the wires don't stretch, then the motion of the log is circular with each wire as the radius of that circle. Therefore, the motion is perpendicular to the tension force. The formula for work $W = F\Delta x \cos \theta$ is zero for $\theta = 90^o$.

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  • $\begingroup$ You probably meant for $\theta=\frac{pi}{2}$.. But why the log is perpendicular? It has an angle with the wires.. $\endgroup$ – Dor Feb 12 '15 at 19:03
  • $\begingroup$ @Dor I fixed $\theta$ in my answers. The angle of the log is not important. The angle of the log's motion is important. $\endgroup$ – Mark H Feb 12 '15 at 19:05

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