AP Physics C - Problem Drill 18: Gravitation and Circular Motion

Transcription

1 AP Physics C - Problem Drill 18: Gravitation and Circular Motion Question No. 1 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 1. Two objects some distance apart gravitationally attract each other. If the distance between them is suddenly tripled, how will that affect the gravitational attraction? Question 01 (A) 3 times as great (B) 1/3 as great (C) 1/9 as great (D) no change will occur (E) 9 times as great Since the distance is on the bottom of the fraction in the universal gravitation law, increasing the distance would lessen the force between the objects. There is a d squared in the bottom of the fraction in the universal gravitation law. Consider that affect on the final answer. C. Correct! Since the universal gravitation law is an inverse square law, tripling the distance will make the force 1/9 of its previous value. The universal gravitation law incorporates a distance term. This change in distance will affect the force somehow. Examine the universal gravitational law. Notice that there is a d squared in the bottom of the formula. Examine the universal gravitational law. m m F= G d 1 g The masses and gravitational constant remain unchanged. The distance triples. Making the distance larger will definitely lessen the force. However, the distance is squared. This emphasized that change even more. Squaring a tripled distance will reduce the force by a factor of 9. The force will be 1/9 of its previous value. The correct answer is (C).

2 Question No. of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as. Calculate the gravitational force of attraction between the earth and moon, when the moon is at distance of 3.8 x 10 5 km, the mass of the moon is kg, the mass of the earth is kg. Question 0 (A) 8.00 x N (B).1 x 10 8 N (C).1 x 10 6 N (D).1 x N (E) 8.00 x 10 8 N Gravitational force follows an inverse square law. Check that your use of exponents. C. Correct! Use the Universal Law of Gravitation and keep track of the exponents. Check that your use of exponents. Gravitational force follows an inverse square law. Check your use of exponent. Known: Distance moon to earth, d = 3.8 x 10 5 km Mass of earth, m e = kg Mass of moon, mm = kg Unknown: Force, F g =? N Define: m /kg Use the universal law of gravitation: mm e m G g F = where Universal Gravitational Constant, G = 6.67 x N d Output: kg kg kg F = N m /kg = N m /kg g 5 11 ( km) km 6 F =.1 10 N g Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (C).

3 Question No. 3 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as 3. Harry is attracted to Sally (in a gravitational sense!). Assume that Harry and Sally have a mass of 100kg each. If they stand 4 m apart, what is the gravitational force attracting them? Question 03 (A) 4. x N (B) 4. x 10-8 N (C) 65 N (D) 1.7 x 10-7 N (E) 4. x 10 3 N Don t forget that two masses are need for the gravitation formula. In this case, both masses are 100 kg. B. Correct! Use Newton s law of universal gravitation. Be sure to input the correct constant G. Also be sure to square the distance. Don t forget the universal gravitational constant, G, in the equation. Don t forget that the distance is squared in the denominator of the gravitation formula. Check that you are using the correct exponents in the calculation. Known: Mass of Harry and Sally, m1 =m =100 kg Distance apart, d = 4 m Unknown: F g =? N Define : Use Newton s law of universal gravitation: m /kg mm 1 F G g d = where Universal Gravitational Constant, G = 6.67 x N Output: Substitute the known values and carefully calculate: -11 N m ( )( 100kg)(100kg) kg -8 F = = N g (4m) Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. Notice how the units cancel to leave Newton as the final label. Also notice how the final answer is very small. This force would be negligible in almost all cases. This is due to the relatively small masses involved. The correct answer is (B).

4 Question No. 4 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 4. Where is the difference in weight the greatest? Question 04 (A) You would weigh the same at all points inside and on the earth. (B) Greatest difference between weight at North Pole and South Pole. (C) Greatest difference is between weight at the equator and at the poles. (D) Greatest difference is between weight at center of the earth, and weight on its surface. (E) Greatest difference is between weight at the surface and weight half way to the center of the earth. Remember weight depends on gravitational force. Weight is essentially the same at both poles. There is a slight difference in weight at the poles and equator due to flattening of the earth at the poles and the spin of the earth, but it is small. D. Correct! At the center of the earth you would be weightless. Weight does decrease the closer you get to the earth s center but consider where the weight is least. There is a slight difference in weight at the poles and equator due to flattening of the earth at the poles and the spin of the earth but it is small. Weight is essentially the same at either pole. Weight does decrease the closer you get to the earth s center, at the center (assuming that the earth s density is uniform) an object would be weightless, since the mass of the surrounding material would be equal the gravitational forces would cancel each other out giving a net gravitational force of zero, which means no weight. The correct answer is (D).

5 Question No. 5 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 5. The picture shows a ball on a string being swung around clockwise which arrow shows the direction at which the ball would fly off at if the string broke? Question 05 (A) A, anticlockwise around the circle. (B) B, perpendicular to the direction it was travelling when the string broke. (C) C at a tangent to the circle, in the direction it was travelling when the string broke. (D) D continuing around the circle in the original direction of travel. (E) E inwards along the radius of the circle of travel. Consider which way the ball is travelling just before the string breaks. E D A C B Consider which way the ball is travelling just before the string breaks, what forces will be acting after the string breaks. C. Correct! The ball will continue to move in the direction that it was travelling at the instant the string broke, that is at a tangent to the circle, the tangent forms an angle of 90 to the radius of the circle. There is no longer a centripetal force acting to keep the ball moving in a circle. This is the direction of the centripetal force that is present when the string is intact; when the string breaks the force is no longer present. When the string breaks there is no longer a centripetal force acting to keep the ball moving in a circle. The ball will continue to move in the direction that it was travelling at the instant the string broke, that is at a tangent to the circle, remember the tangent forms an angle of 90 to the radius of the circle. The correct answer is (C).

6 Question No. 6 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 6. Which of these is a correct statement about centrifugal force? Question 06 (A) Centrifugal force acts to keep an object moving in a circle. (B) Centrifugal force is a center seeking force. (C) Centrifugal force is a real force that results from an object moving in a circle. (D) Centrifugal force is the apparent force that seems to be acting to pull an object away from the center when it moves in a circle. (E) When you swing a ball on a string in a circle you apply a centrifugal force to the string, which is transferred to the ball. This is not a centrifugal force. This is Centripetal force not Centrifugal force. Centrifugal force is an apparent force. D. Correct! If such a force existed the ball from the previous questions would fly outward from the circle not at a tangent. When you swing a ball on a string in a circle you apply an inward force on the string, which is transferred to the ball, the ball exerts an equal but opposite force back on the string. BUT the force you apply is not a centrifugal force. Consider the previous question where a ball was swung around in a circle, when you do this it feels like a force is pulling on your hand, it could be interpreted that there is a force acting on the ball to pull it out from the circle. This is what we refer to as a Centrifugal force (a center fleeing force) In fact you are applying an inward force to the string, which is applied to the ball, and we know from Newton s third law the ball applies an equal but opposite force to the string that you feel on your hand. If the Centrifugal force existed we would expect that if string broke the ball would go flying out wards in the direction the force, but we know that it actually goes at a tangent to the circle. The correct answer is (D).

7 Question No. 7 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 7. In a cyclotron, an electromagnet exerts a force of 7.5 x N on a beam of protons. Each proton has a mass of 1.67x10-7 kg. The electromagnet causes the protons to travel in a circular path of radius 1. m. What is the velocity of the proton beam? Question 07 (A).3 x 10 7 m/s (B) 3. x 10-0 m/s (C) 5.5 x 10 7 m/s (D) 5.4 x m/s (E) 4.3 x 10-8 m/s A. Correct! Use the formula: F c = mv /r, rearrange to give v=sqrt(f c r/m). Be sure of your substitutions. You accidentally inverted the radius and mass. This answer seems very slow for protons in a cyclotron. Use the centripetal force formula. You re looking for the velocity. Algebraically rearrange to solve for this velocity. This would be faster than the speed of light. Don t forget to take the square root since you are looking for v, not v squared. Use the centripetal force formula. You re looking for the velocity. Algebraically rearrange to solve for this velocity. Known: Force on proton, F c = 7.5 x N Mass proton, m p = 1.67x10-7 kg Radius of circular path = 1. m Unknown: Velocity v =? m/s Define: Use the formula for centripetal force and rearrange to find velocity: mv F= c r v Remember that N equals a kg m/s. This gives m/s as your final answer units. = F r c m Output: Substitute values: -13 ( kg m/s )(1.m) 7 v = =.3 10 m/s kg Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (A).

8 Question No. 8 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 8. Imagine that a car is traveling in a circle on a track that is banked at a 5 angle. If the radius of the circle is 50 m, how fast could the car travel without drifting? Ignore friction and air resistance. Question 08 (A) 15 m/s (B) 3.8 x 10-5 m/s (C) 110 m/s (D) 8 m/s (E) None of the above A. Correct! The centripetal force is equal to mass times the acceleration of gravity times the tangent of the angle. Equate this to mass times velocity squared divided by radius. The masses cancel out. Substitute all the given information and solve for velocity. The g used in this problem is the acceleration from gravity, not the universal gravitational constant. This answer choice is extremely slow for any car race. The tangent of the angle must be used, not just the angle itself. Remember that the centripetal force in this case is a component of the normal force. Don t forget to take the square root since you are looking for v, not v squared. You might think that the mass is necessary for the problem. However, the mass cancels out when the two expressions for centripetal force are equated. Known: Radius of curvature, r = 50 m Banking angle, θ = 5 F c Unknown: Velocity, v=? m/s Define: The centripetal force is the force horizontal to the road, to find it we must use trig and combine with the formula for centripetal force and velocity. opp F F c c tan θ = = = adj W mg Normal force 5 5 Weight velocity Rearrange the force equation to find mv F = mg tan θ = c r v = r g tan θ v = r g tan θ v = 50 m 9.8 m/s tan 5 Output: v = 490 m /s v = 15 m/s Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (A).

9 Question No. 9 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 9. Two satellites are orbiting the earth, satellite 1 orbits is at 300 km above the Earth, satellite orbits is at 600 km above the Earth, what can be said about their relative velocities? The earth s radius is 6.4 x 10 6 m. Question 09 (A) There is insufficient information. (B) The velocity of satellite 1 will be twice that of satellite. (C) The velocity of satellite 1 will half that of satellite. (D) The velocity of satellite 1 will be 4 times that of satellite. (E) None of the above You can calculate the relative velocities; use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. Remember that the satellites are orbiting about the center of the earth; draw a picture to help find the radius of orbits. Remember that the satellites are orbiting about the center of the earth, draw a picture to help find the radius of orbits, and then use the fact that the centrifugal force is the same as the gravitational pull. Remember that the satellites are orbiting about the center of the earth, draw a picture to help find the radius of orbits, and then use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. E. Correct! Remember that the satellites are orbiting about the center of the earth, draw a picture to help find the radius of orbits, and then use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. You can calculate the relative velocities; use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. F G = F c Consider any satellite of mass m, and an earth mass of M m M m v G = r r M v = G r The velocity of the satellites does not depend on its mass. Remember that the satellites are orbiting about the center of the earth so the radius of orbits must include the radius of the earth, do not forget to convert from km to m for satellites. r t = r e + r o r 1 = 6.4 x 10 6 m x 10 6 m =6.7 x 10 6 m r = 6.4 x 10 6 m x 10 6 m =7.0 x 10 6 m If we compare velocity of satellite 1 to velocity of satellite v 1 M M r 6 7.0x 10 m v r r r x 10 m = G G = = = 1.0 The correct answer is (E).

10 Question No. 10 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 10. If the Hubble space telescope orbits 600 km above the earth, how fast is it orbiting above the earth? The radius of the earth is 6.4 x 10 6 m. The mass of the earth is 6 x 10 4 kg. Question 10 (A).4 x 10 3 m/s (B).9 x 10 9 m/s (C).6 x 10 4 m/s (D) 7.6 x 10 3 m/s (E) Cannot calculate without mass of telescope. Equate the centripetal force to the gravitational force. The mass of the satellite cancels out. This number is way too large, greater than the speed of light. Be sure to use the universal gravitational constant for G. This value isn t the acceleration from gravity 9.8 m/s/s that was used so often before. Don t just use the satellite altitude for the radius. The middle of the satellite orbit is the center of the earth. Add the satellite altitude to the radius of the earth for a more accurate radius. D. Correct! Equate the centripetal force to the gravitational force. The mass of the satellite cancels out. Rearrange for velocity, and then substitute the known values. For the radius, be sure to add the altitude of the satellite to the radius of the earth. This will give the distance of the satellite from the center of the earth. You might think that the mass of the telescope is necessary, but it isn t. Like in so many equations, the mass cancels out. Equate the centripetal force to the gravitational force. Known: Orbit radius, r o = 600 km Earth s radius, r e = 6.4 x 10 6 m Earth s mass M = 6 x 10 4 kg Unknown: Velocity of the satellite, v =? m/s Define: First, change the telescopes altitude into meters. Next add this distance to the radius of the earth. This will give the total distance from the center of the earth to the telescope. r t = r e + r o The centripetal force on the satellite equals the gravitational pull on the telescope. F G = F c m M m v G = r r M v = G r Remember: G = 6.67 x N m /kg Output: r o = 600 km = 6 x 10 5 m r t = 6.4 x 10 6 m + 6 x 10 5 m = 7 x 10 6 m x 10 kg v = 6.67 x 10 N m /kg 6 7 x 10 m v = 7.6 x 10 3 m/s Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (D). 4