A 30g bullet is fired horizontally into a 12kg block which is suspended on a long cord. What speed of bullet would cause the center of gravity of the block to rise 8cm?(adsbygoogle = window.adsbygoogle || []).push({});

energy approach

change of kinetic energy for bullet = change of potential for bullet and block

[tex]\frac{1}{2}mv^2 = mgh[/tex]

[tex]\frac{1}{2}(0.03)v^2 = (12.03)(9.8)(0.08)[/tex]

[tex]v = 25.08 \frac{m}{s}[/tex] <-------------------------------

momentum approach

change in kinetic energy (block and bullet) = change in potential (block and bullet)

[tex]\frac{1}{2}mv^2 = mgh[/tex]

[tex]\frac{1}{2}(12.03)v^2 = (12.03)(9.8)(0.08)[/tex]

[tex]v = 1.252[/tex]

momentum of bullet = momentum of bullet and block

[tex]mv = mv[/tex]

[tex](0.03)v = (12.03)(1.252)[/tex]

[tex]v = 502 \frac{m}{s}[/tex] <-------------------------------

Half the class thinks the answer is 25.08, the other half thinks it's 502. The answer in the back of the book is 502.

Why would the answer be 502? All of the energy in the system comes from the bullet, thus, the bullet's initial kinetic energyshouldequal the final potential energy of the block and the bullet. Where would all the energy go if this system is 100% efficient.

Look at the energy difference, assuming the answer is 502m/s.

Here is the energy of the bullet initially.

[tex]E = \frac{1}{2}(0.03)(502)^2[/tex]

[tex]E = 3780 J[/tex]

Here is the final energy of the block and bullet

[tex]E = (12.03)(9.8)(0.08)[/tex]

[tex]E = 9.432 J[/tex]

Where did the energy go???

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# Homework Help: Momentum and energy disagreement

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