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## Homework Statement

A 11.0 g rifle bullet is fired with a speed of 360 m/s into a ballistic pendulum with mass 9.00 kg, suspended from a cord 70.0 cm long.

A.) Compute the initial kinetic energy of the bullet

B.) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.

C.) Compute the vertical height through which the pendulum rises.

## Homework Equations

## The Attempt at a Solution

A.) I just applied 1/2mv^2 to get the initial KE of the bullet to be 713 J (it's correct)

B.) Since the kinetic and potential energy of the pendulum right when the bullet fires is zero, I claim that:

[tex]713 = \frac{1m_{b}v^{2}_{b}}{2} + \frac{1m_{p}v^{2}_{p}}{2}[/tex]

[tex]713 = \frac{(.011)v^{2}_{b}}{2} + \frac{(9)v^{2}_{p}}{2}[/tex]

Now I am assuming that the velocity of the bullet and pendulum are the same since they are now attached so:

[tex]713 = \frac{1}{2}(.011+9)v^{2}[/tex]

[tex]1426 = (.011+9)v^{2}[/tex]

[tex]1426 = (9.011)v^{2}[/tex]

[tex]\frac{1426}{9.011} = v^{2}[/tex]

[tex]\sqrt{\frac{1426}{9.011}} = v[/tex]

[tex]v = 12.6 m/s[/tex]

??

C.)

[tex]\frac{1}{2}(9.011)(12.6)^{2} = (9.011)(9.8)h[/tex]

[tex]h = 8.1m[/tex]

There is no way this is right, the shaft of the pendulum is only 70cm long!