Asked Nov 21, 2019

To test the speed of a bullet, you create a pendulum by attaching a 4.70 kg wooden block to the bottom of a 1.20 m long, 0.800 kg rod. The top of the rod is attached to a frictionless axle and is free to rotate about that point.

You fire a 10 g bullet into the block, where it sticks, and the pendulum swings out to an angle of 25.0°. What was the speed of the bullet?

(Treat the wooden block as a particle.)


Expert Answer

Step 1

Moment of inertia of rod + bullet + wood is,


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m.12 +m2 +m2 I = (0.800kg) (1.20m) -+(4.70kg)(1.20m)+(001kg)(1.20m) 3 7.17kg-m2

Step 2

Using conservation of angular momentum,


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ту- Io I@ m2l

Step 3

Using Work-Energy...


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Ia2mgl(1-cos)+ (m2 + m, ) g| )g(1-cose) 2m+(m2m =gl(1-cose) I 2(0.800 kg)+(4.70 kg +0.01kg) 7.17kg-m (9.8m/s) (1.20m) (1- cos 25°) =0.98 rad/s


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