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The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance 0.921 m apart on the same axle. From the angular displacement 27.4 degrees of the two bullet holes in the disks and the rotational speed 1050 rev/min of the disks, we can determine the speed v of the bullet. What is the speed of the bullet in units of m/s?

Okay to start the problem, I wrote out my known values:

d = o.921 m

r = 0.4605 m

initial angular velocity = 1050 rev/min

initial angular velocity = 1050 (rev/min) (1 min/ 60 sec) (2 x 3.14 rad/rev)

initial angular velocity = 109.9557429 rad/s

Theta = 27.4 degrees X ( 1 rev/ 360 degrees) = 0.0761111111

Those are my known values and their derivations. To solve the problem, I figured the speed of the bullet was reliant upon where the hole would be in the second disk. Thus I set up the equation:

theta = (1/2) alpha (t^2) + (initial angular velocity X t)

- Here I assumed alpha was 0, since I assumed the bullet was moving at a constant speed. Therefore:

theta = (initial angular velocity X t)

t = (0.0761111111/109.9557429)

t = 0.000692197688 seconds

With the time value found for the second disk, I figured that the bullet had to trave the diameter at that time. In other words, velocity is equal to diameter/time.

V = d/t

V = 0.921 m / 0.0006921976888 s

V = 1330.544749 m/s

This was my final answer, yet it is incorrect. Can anyone point out what I have done wrong and how to fix it? Sorry for the inconvienince.