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Orbital Acceleration - General Physics - Solved Exam, Exams of Physics

Bharat Ratna Dr. B. R. Ambedkar UniversityPhysics

This is the Solved Exam of General Physics which includes Orbital Acceleration, Component of Ball’s Velocity, Horizontal Component, Smallest Acceleration, Order of Magnitude, Gravitational Potential Energy, Kinetic Energy etc. Key important points are: Orbital Acceleration, Component of Ball’s Velocity, Horizontal Component, Smallest Acceleration, Order of Magnitude, Gravitational Potential Energy, Kinetic Energy, Spring Constant

Typology: Exams

2012/2013

Uploaded on 03/07/2013

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Physics 202A FINAL EXAM Chapters 1 - 15 Fall 2012
1
Name:_______________________________ PC _____ _____ _____
Solve the following problems in the space provided. Use the back of the page if needed. Each problem is
worth 10 points. You must show your work in a logical fashion starting with the correctly applied and
clearly stated physical principles. The equations you need are on the equation sheet. Your score will be
maximized if your work is easy to follow because partial credit will be awarded.
1. In Game 1 of the World Series, Pablo Sandoval hit three homeruns. On his third homer, the ball left
the bat at 50m/s at an angle of 33˚. Find (a)the horizontal component of the ball’s velocity, (b)the vertical
component of the ball’s velocity, (c)the time the ball was in the air and (d)the distance the ball travelled.
You may ignore air resistance.
Given: xo = 0, yo = 0, ax = 0, y = 0, ay = -9.8m/s2, vo = 50m/s, and θ = 33˚.
Find: vox = ?, voy = ?, t = ?, and x = ?
(a)Using the vector components,
vox =vocos
θ
=50 cos 33˚⇒
vox =42 m
s
.
(b)and
voy =vosin
θ
=50 sin 33˚⇒
voy =27 m
s
(c)Using the kinematic equation,
y=yo+voyt+1
2ayt2⇒
0=0+voyt+1
2ayt2⇒t=−2voy
ay
=2(27)
9.8
⇒
t=5.5s
.
(d)Using the kinematic equation,
x=231m
.
2. Below is a table of data on four moons of Jupiter discovered by Galileo. Rank these moons in order of
the magnitude of their orbital acceleration. That is, rank first the one with the highest acceleration and
rank last the one with the smallest acceleration. Explain your reasoning.
These moons are in circular orbit around Jupiter so the
experience centripetal acceleration which is given by,
ac=v2
r
.
Io:
ac=v2
r
=(17.3)2
422
=0.709
Callisto:
ac=(8.20)2
1883
=0.036
Ganymede:
ac=v2
r
=(10.9)2
1070
=0.111
Europa:
ac=v2
r
=(13.7)2
671
=0.280
The ranking is: Io >Europa> Ganymede> Callisto
Moon
Orbital
Radius
(x106m)
Orbital
Speed
(x103m/s)
Io
422
17.3
Callisto
1883
8.20
Ganymede
1070
10.9
Europa
671
13.7
x
y
v
θ
pf3
pf4
pf5

Partial preview of the text

Download Orbital Acceleration - General Physics - Solved Exam and more Exams Physics in PDF only on Docsity!

Name:_______________________________ PC _____ _____ _____ Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 1 0 points. You must show your work in a logical fashion starting with the correctly applied and clearly stated physical principles. The equations you need are on the equation sheet. Your score will be maximized if your work is easy to follow because partial credit will be awarded.

  1. In Game 1 of the World Series, Pablo Sandoval hit three homeruns. On his third homer, the ball left the bat at 50m/s at an angle of 33˚. Find (a)the horizontal component of the ball’s velocity, (b)the vertical component of the ball’s velocity, (c)the time the ball was in the air and (d)the distance the ball travelled. You may ignore air resistance. Given: xo = 0, yo = 0, ax = 0, y = 0, ay = - 9.8m/s^2 , vo = 50m/s, and θ = 33˚. Find: vox = ?, voy = ?, t = ?, and x =? (a)Using the vector components, vox = vo cos θ = 50 cos 33 ˚⇒ vox =^42 m s. (b)and voy = vo sin θ = 50 sin 33 ˚⇒ voy = 27 m s (c)Using the kinematic equation, y = yo + voyt + 12 ayt^2 ⇒ 0 = 0 + voyt + 12 ayt^2 ⇒ t = − 2 voy ay

⇒ t^ =^ 5.5 s^. (d)Using the kinematic equation, x = xo + voxt + 12 axt^2 = 0 + ( 42 )(5.5) + 0 ⇒ x^ =^231 m^.

  1. Below is a table of data on four moons of Jupiter discovered by Galileo. Rank these moons in order of the magnitude of their orbital acceleration. That is, rank first the one with the highest acceleration and rank last the one with the smallest acceleration. Explain your reasoning. These moons are in circular orbit around Jupiter so the experience centripetal acceleration which is given by, ac = v 2 r

Io: ac = v 2 r

2 422

Callisto: ac =

(8.20)^2

Ganymede: ac = v^2 r

(10.9)^2

Europa: ac = v^2 r

(13.7)^2

The ranking is: Io >Europa> Ganymede> Callisto Moon Orbital Radius (x10^6 m) Orbital Speed (x10^3 m/s) Io 422 17. Callisto 1883 8. Ganymede 1070 10. Europa 671 13. x y v θ

Physics 202A FINAL EXAM Chapters 1 - 15 Fall 2012

  1. At the right are six different crates at the top of six different ramps. The mass of each crate as well as the height and length of each ramp are given. Rank them from greatest to least based upon (a)their gravitational potential energy at the top, (b)their kinetic energy at the bottom, and (c)the speed of the crate at the bottom. You must explain your reasoning for full credit. (a)Gravitational potential energy is given by mgh, so F > A=B > C > D > E. (b)The Law of Conservation of Energy requires that the potential energy at the top equal the kinetic energy at the bottom, so the ranking is the same - F > A=B > C > D > E. (c)Using the definition of kinetic energy and remembering energy is conserved, Uo + Ko = U + K ⇒ mgh + 0 = 0 + 12 mv^2 ⇒ v = 2 gy. So, ranking by the height of the ramp, D > A=B > F > C=E.
  2. Two blocks of masses 100g and 300g are placed on a horizontal, frictionless surface. A light spring of spring constant 8.00N/m is attached to one of them. The blocks as pushed together compressing the spring 5.00cm and held in place by a cord. When the cord is cut, the small block heads off at a speed of 0.387m/s. Find the speed of the large block. Given: m 1 = 0.100kg, m 2 = 0.300kg, k = 8.00N/m, x = 0.0500m, and v 1 = 0.387m/s. Find: v 2 =? Applying the Law of Conservation of Momentum, 0 = Mv 2 − mv 1 ⇒ v 2 = (^) Mm v 1. Plugging in the numbers, v 2 = 100300 (0.387) ⇒ v 1 =^ 0.129^ m s. This problem can also be solved by using Conservation of Energy. Physics Ranking Tasks 68 Mechanics Sliding Masses on Incline—Kinetic Energy 64 Rank, in order from greatest to least, the final kinetic energies of the sliding masses the instant before they reach the bottom of the incline. All surfaces are frictionless. All masses start from rest. A B C D E F 10 m 5 m 5 m 10 m 5 m 7.5 m 0.5 m 2 m 0.5 m 1 m 1 m 0.75 m 10 kg 15 kg 1 kg 1 kg 5 kg 10 kg Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least Or, all masses will have the same kinetic energies. _________ Please carefully explain your reasoning. How sure were you of your ranking? (circle one) Basically Guessed Sure Very Sure 1 2 3 4 5 6 7 8 9 10 (^64) P. Marquard, D. Maloney before po =^0 after p = Mv 2 − mv 1 v 1 v 2
  1. At the right are the first six standing waves on identical guitar strings. Rank them from greatest to least based upon (a)velocity, (b)wavelength, and (c)frequency. You must explain your reasoning for full credit. (a)The speed of waves depends upon the properties of the string. Since the strings are identical, the speeds are all equal. A = B = C = D = E = F (b)The distance between nodes is half a wavelength so, A > B > C > D > E > F (c)Using the wave speed, λ f = v , and the fact that all the speeds are equal, the frequencies are inversely related to the wavelengths. So, F > E > D > C > B > A
  2. A 10.0kg rock rests on the bottom of a lake. You can lift the rock off the bottom with a force of 6 0.0N. Find (a)the magnitude of the buoyant force, (b)the volume of the rock, and (c)the density of the rock. Given: m = 10.0kg and FL = 60.0N. Find: FB = ?, V =? and ρ =? (a)Apply the Second Law, Σ F = ma ⇒ Flift + FB − mg = 0 ⇒ FB = mg − Flift. Plugging in the numbers, FB = ( 10 )(9.8) − 60 ⇒ FB = 38.0 N. (b)Using Archimedes Principle, FB = ρ Vg ⇒ V =

FB

ρ g

Plugging in the values, V =

⇒ V^ =^ 3.88 x^10 − (^3) m 3 . (c)The definition of density is, ρ = m V

3.88 x 10 −^3 ⇒ ρ = 2580 kg m^3.

A B C D E F

Flift

FB

Fg

x

  1. You happen to know that your glass of iced tea has 0.200kg of ice remaining when it reaches a temperature of 0˚C. You are so fascinated by conversation you are having that you don’t actually drink any tea but do notice that the last of the ice melts in 20.0 minutes. (a)Find the energy required to melt the ice. (b)Explain where this energy came from. Given: m = 0.200kg and L = 3.34x10^5 J/kg. Find: Q =? (a)Since the temperature isn’t changing, this is just a phase change. So,

Q = mL = (0.2)(3.34 x 105 ) ⇒ Q = mL = 6.68 x 10

4

J.

(b)This energy comes from the surroundings – mostly the air around the glass.

  1. Consider a cubic meter of air in this room at atmospheric pressure (101.3kPa) and 25.0˚C. Find (a)the number of molecules, (b)the average kinetic energy of the molecules, and (c)the total kinetic energy of all the molecules. Given: P = 1.013x10^5 Pa, V = 1.00m^3 , and T = 25.0˚C = 298K. Find: N = ?, Kav = ?, and K =? (a)Using the Ideal Gas Law, PV^ =^ NkT^ ⇒^ N^ =^

PV

kT

(1.013 x 105 )( 1 ) (1.38 x 10 − 23 )( 298 ) ⇒ N = 2.46 x (^1025). (b)The average kinetic energy is related to the temperature, Kav = 12 mvrms 2 = 32 kT = 32 (1.38 x 10 − 23 )( 298 ) ⇒ Kav =^ 6.17 x^10

− 21 J

(c)The total kinetic energy is just the average times the number of molecules, K = NKav = (2.46 x 1025 )(6.17 x 10 −^21 ) ⇒ K = 1.52 x 10 5 J.