In Class Test Monday, March 26, 2018 PHYS 251 University Physics NO NOTES OR REF
ID: 2031357 • Letter: I
Question
In Class Test Monday, March 26, 2018 PHYS 251 University Physics NO NOTES OR REFERENCE TO BOOKS, ALL PHONES PUT AWAY Part I: A roller coaster starts to move from rest, at a starting height of 100 meters. Assuming a frictionless situation, calculate the speed at points A and B. The heights at points A and B are 25 meters and 50 meters, respectively. At point B, the track is angled at 30 degrees above the horizontal. The radius of curvature of the track at point A is 10 meters. 1a) calculate the speeds of the roller coaster at points A andbB 1b) calculate the force the track exerts on the roller coaster at point A eneray when it reaches point B What will be its speed at point ioExplanation / Answer
1a)
from energy conservation
initial mechanical energy Ei = m*g*h
total mechanical energy at A , EA = m*g*hA + (1/2)*m*VA^2
total energy is conserved
EA = E
m*g*hA + (1/2)*m*vA^2 = m*g*h
9.8*25 + (1/2)*vA^2 = 9.8*100
speed at point A , VA = 38.3 m/s
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total mechanical energy at B , EB = m*g*hB + (1/2)*m*VB^2
total energy is conserved
EB = E
m*g*hB + (1/2)*m*vB^2 = m*g*h
9.8*50 + (1/2)*vB^2 = 9.8*100
speed at point B , VB = 31.3 m/s
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1b)
Force F = mg + m*vA^2/r
Force F = m*(g + vA^2/r) = m*(9.8 + 38.3^2/10) = m*156.5 N
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1c)
initial speed vB = 31.3 m/s
along vertical
initial velocity voy = vB*sin30
acceleration ay = -g = -9.8 m/s^2
initial position y0 = 50 m
final position y = ?
final velocity at maximum height vy = 0
from equation of motion
vy^2 - voy^2 = 2*ay*(y-y0)
0 - (31.3*sin30)^2 = -2*9.8*(y-50)
y = 62.5 m <<--------ANSWER
after striking ground
final position y = 0
y-y0 = voy*t + (1/2)*ay*t^2
0-50 = 31.3*sin30*t - (1/2)*9.8*t^2
time taken to strike ground t = 5.17 s
along horizontal
initial velocity vox = vB*cos30
acceleration ax = 0
displacement = x
x = vo*t + (1/2)*ax*t^2
x = 31.3*cos30*5.17 = 140.1 m
It hits the ground at 140.1 m <<<---ANSWER
after hitting the ground
along vertical
vy = voy + ay*t = 31.3*sin30*5.17 - 9.8*5.17 = 30.24 m/s
along vertical
vx = vox + ax*t = 31.3*cos30 = 27.1 m/s
speed v = sqrt(v^2+vy^2) = 40.6 m/s <<-----ANSWER
velocity v = 27.1 i - 30.24 j
direction = tan^-1(-30.24/27.1) = 48.13 degrees below horizontal
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part 2
E - Eloss = EB
Eloss = 0.25*E
m*g*h - 0.25*m*g*h = m*g*hB + (1/2)*m*vB^2
0.75*g*h = g*hB + (1/2)*vB^2
0.75*9.8*100 = 9.8*50 + (1/2)*vB^2
vB = 22.13 m/s