1)In the absence of air resistance, a projectile that lands at the elevation fro
ID: 1434404 • Letter: 1
Question
1)In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45? angle. Suppose a projectile of mass m is launched with speed v0 into a headwind that exerts a constant, horizontal retarding force F? wind =?Fwindi^.
By what percentage is the maximum range of a 0.900 kg ball reduced if Fwind = 0.100 N ?
2)A 400 g steel block rotates on a steel table while attached to a 1.20 m -long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.81 Nperpendicular to the tube. The maximum tension the tube can withstand without breaking is 50.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60.
If the block starts from rest, how many revolutions does it make before the tube breaks?
Figure 1of 1 Air 1.2 m Tube PivotExplanation / Answer
1) In the absence of air resistance,
• range x = V²sin(2) / g, so for = 45º
x = V² / g
• flight time t = 2·V·sin/g, so for = 45º
t = 2*V*2/2 / g = (V/g)*2
The acceleration in the x-direction due to the wind is
a = F / m = 0.100N / 0.900kg = 0.111m/s²
so the reduction in range is
s = ½at² = ½ * 1.5m/s² * [(V/g)*2]² = 0.111m/s² * V² / g²
and finally
s / x = (0.111m/s²*V² / g²) / (V²/g) = 0.111m/s² / g
s / x = 0.111 / 9.8 = 0.0113 = 1.13%
2) The linear (tangential) acceleration is
a_t = F / m = 4.81N / 0.400kg =12.02 m/s²
and after time "t" the velocity is
v = a_t * t =12.02m/s² * t
and so the centripetal acceleration is
a_c = v²/r = (12.02m/s² * t)² / 1.20m = 120.40m/s * t²
The tension is
T = m*a_c, so
50.0 N = 0.400kg * 120.40m/s * t²
t = 1.01 s
= v / r = 12.02m/s² * 1.01s / 1.2m =10.11 rad/s
so the average angular velocity avg =5.055 rad/s
= avg*t = 5.055rad/s * 1.01s = 5.10 rads = 0.811 revolutions.
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