Physics – Errors in calculations help

Hi.
I would appreciate it if somebody could help me with the following problem, or point me in the direction of something i could read on the subject:
Determine, with an error range, the kinetic energy of the projectile as it hits the windscreen.
Length of Projectile – 10.0 (+_) 0.1cm
Diameter of projectile – 5.00 (+_) 0.05mm
Density of metal – 5.82 (+_) 0.05gcm-3
Impact velocity of projectile – 8.5 (+_) 0.3ms-1
Thanks!
Answer #1
The projectile should be considered a cylinder because it’s easier,then you can find it’s volume and from this the mass,then apply the kinetic energy formula.
LE: Drop a PM if isn’t working
Answer #2
cylinder’s volume = pi * radius^2 * height (length in this case) = 3.14 * 6.25 cm * 10 cm = 196.25 cm^3 = 0.000196 m^3 [SI]
mass = density * volume = 5820 Kg/m^3 * 0.000196 m^3 = 1.14 Kg [SI]
kinetic energy = 1/2 * mass * velocity^2 = 1/2 * 1.14 Kg * 8.5^2 m/s = 41.18 J [SI]
Others should verify this and correct me or assure you that it’s right.
Answer #3
Hi.
Really appreciate the replies!
That answer is about what i got (although I rounded the ansers off to different decimal places).
But how do i determine the error range?
Answer #4
Damn! I just saw something…The length is given in cm (10 +- 0.1) and the diameter in mm (5 +- 0.05) ?
Cause I’ve calculated with the diameter in cm.
cylinder’s volume = pi * radius^2 * height (length in this case) = 3.14 * 0.62 cm * 10 cm = 19.46 cm^3 = 0.000019 m^3 [SI] mass = density * volume = 5820 Kg/m^3 * 0.000019 m^3 = 0.11 Kg [SI] kinetic energy = 1/2 * mass * velocity^2 = 1/2 * 0.11 Kg * 8.5^2 m/s = 3.97 J [SI]
About the error range, I don’t know what’s expected of you.
Calculating the errors when working with volumes is a bit more complicated.
A more simplistic approach would be to consider a +- 1% tolerance, since 1% is given for the dimensions and density.
But, like I said, I really don’t know.