Version 001 circular and gravitation holland (2383) 1

Size: px

Start display at page:

Download "Version 001 circular and gravitation holland (2383) 1"

Maurice Pope
6 years ago
Views:

1 Version 00 circular and gravitation holland (383) This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. AP B 993 MC points Two objects of masses 4 kg and 35 kg are hung from the ends of a stick that is 70 cm long and has marks every 0 cm, as shown. ABCDEFG kg 35 kg If the mass of the stick is negligible, at whichofthepointsindicatedshouldacordbe attached if the stick is to remain horizontal when suspended from the cord?. F. A 3. B 4. C 5. E 6. G correct 7. D Let : l = 70 cm, m = 4 kg, and m = 35 kg. For static equilibrium, τ net = 0. Let x be the distance from the left end of the stick to the point of attachment of the cord: T = m gx m g(l x) = 0 (m +m )x = m l x = m l m +m = = 50 cm. (35 kg)(70 cm) 4 kg+35 kg Therefore the point should be point G. AP B 998 MC points Three forces act on an object. If the object is in translational equilibrium, which of the following must be true? I. The vector sum of the three forces must equal zero; II. The magnitude of the three forces must be equal; III. The three forces must be parallel.. I only correct. I, II and III 3. II only 4. II and III only 5. I and III only If an object is in translational equilibrium, the vector sum of all forces acting on it must equal zero. AP M 993 MC 35 A points A rod of negligible mass is pivoted at a point that is off-center, so that length l is different from length l. The figures show two cases in which masses are suspended from the ends of the rod. In each case the unknown mass m is balanced by a known mass M or M so that the rod remains horizontal. l l m M

2 Version 00 circular and gravitation holland (383) M l l What is the value of m in terms of the known masses?. m = M +M. m = M M 3. m = M M correct 4. m = M +M 5. m = M M Applying τ = 0 to balance the masses in both cases, ml = M l and M l = ml. m Which of the following expresses the condition required for the system to be in static equilibrium?. b m = a m. am = bm 3. a m = b m 4. am = bm correct 5. m = m In equilibrium, the total torque is zero, which gives am = bm. AP B 993 MC points Two spheres have equal densities and are subject only to their mutual gravitational attraction. Dividing, m = M M m m = M M m = M M. AP M 998 MC points Consider the wheel-and-axle system shown below. Which quantity must have the same magnitude for both spheres?. displacement from the center of mass. gravitational force correct 3. acceleration 4. velocity a b m m 5. kinetic energy Two spheres with the same density have different masses due to their relative sizes. Using Newton s third law, F = F. All of the other quantities(acceleration, velocity, kinetic energy, and displacement from the center of mass) have different magnitudes because the two spheres have different masses.

3 Version 00 circular and gravitation holland (383) 3 AP M 998 MC (part of ) 0.0 points A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the asteroid s radius and then falls straight down toward the surface of the asteroid. What forces, if any, act on the ball while it is on the way up?. Only a decreasing gravitational force that acts downward correct. Only a constant gravitational force that acts downward 3. Both a constant gravitational force that acts downward and a decreasing force that acts upward 4. Only an increasing gravitational force that acts downward 5. No forces act on the ball. There is no friction in the system, and the ball doesn t have any contact with other objects, so the only force acting on the ball is the attractive gravitational force, which acts downward. From F = G M m r ˆr, the force will decrease as the ball rises. 007 (part of ) 0.0 points The acceleration of the ball at the top of its path is. zero.. equal to one-fourth the acceleration at the surface of the asteroid. correct 3. equal to one-half the acceleration at the surface of the asteroid. 4. equal to the acceleration at the surface of the asteroid. 5. at its maximum value for the ball s flight. F = ma r, so a r and a (r) = 4 r 4 a. Weight of Spacecraft in Space 008 (part of ) 0.0 points The radius of Earth is about 6440 km. A 7770 N spacecraft travels away from Earth. What is the weight of the spacecraft at a height 6440 km above Earth s surface? Correct answer: 94.5 N. Let : r E = 6440 km, W = 7770 N, and h = 6440 km. By Newton s Universal Law of Gravitation W h W = r h r = W h = W r r h W r, so r rh = W r E (r E +h) = (7770 N) = 94.5 N. (6440 km) (6440 km+6440 km) 009 (part of ) 0.0 points What is the weight km above Earth s surface? Correct answer: N.

4 Version 00 circular and gravitation holland (383) 4 W = W r r h Let : h = km. = W r E (r E +h) = (7770 N) = N. (6440 km) (6440 km km) AP B 993 MC points Consider the following situations. A) An object moves in a straight line at constant speed. B) An object moves with uniform circular motion. C) An object travels as a projectile in a gravitational field with negligible air resistance. In which of the situations would the object be accelerated?. C only. A only 3. B and C only correct 4. B only 5. A and C only 6. None exhibits acceleration. 7. A and B only 8. All exhibit acceleration. A) The velocity of the object (its direction and magnitude) is unchanged, so it is not accelerated. B) The direction of the velocity constantly changes; the centripetal acceleration is directed toward the center of the motion. C) The projectile undergoes gratitational acceleration. AP B 993 MC points If Spacecraft X has twice the mass of Spacecraft Y, then what is true about X and Y? I) On Earth, X experiences twice the gravitational force that Y experiences; II) On the Moon, X has twice the weight of Y; III) When both are in the same circular orbit, X has twice the centripetal acceleration of Y.. III only. I only 3. I and II only correct 4. II and III only 5. I, II, and III I) gravitational force mass. II) weight mass. III) The centripetal acceleration is determined by a c = v r, so X and Y should have the same centripetal acceleration when they are in the same circular orbit. AP M 998 MC (part of ) 0.0 points A spring has a force constant of 593 N/m and an unstretched length of 6 cm. One end is attached to a post that is free to rotate in the center of a smooth table, as shown in the top view below. The other end is attached to a 4 kg disk moving in uniform circular motion on the table, which stretches the spring by 4 cm. Note: Friction is negligible.

5 Version 00 circular and gravitation holland (383) 5 0 cm 593 N/m 4 kg AP B 993 MC points The planet Krypton has a mass of kg and radius of m. Whatistheaccelerationofanobjectinfree fall near the surface of Krypton? The gravitational constant is N m /kg. WhatisthecentripetalforceF c onthedisk? Correct answer: 3.7 N. Let : r = 6 cm = 0.06 m, r = 4 cm = 0.04 m, m = 4 kg, and k = 593 N/m. The centripetal force is supplied only by the spring. Given the force constant and the extension of the spring, we can calculate the force as F c = k r = (593 N/m)(0.04 m) = 3.7 N. 03 (part of ) 0.0 points What is the work done on the disk by the spring during one full circle?. W = 0 J correct. W = J 3. W = 4.47 J 4. W =.43 J 5. W = J Since the force is always perpendicular to the movement of the disk, the work done by the spring is zero. Correct answer: m/s. Let : M = kg, R = m, and G = N m /kg. Near the surface of Krypton, the gravitation force on an object of mass m is F = G M m R, so the acceleration a of a free-fall object is a = g Krypton = F m = G M R = ( N m /kg ) kg (4 0 6 m) = m/s. AP B 998 MC points An object has a weight W when it is on the surface of a planet of radius R. What will be the gravitational force on the object after it has been moved to a distance of 4R from the center of the planet?. F = 4W. F = 4 W 3. F = W

6 Version 00 circular and gravitation holland (383) 6 4. F = 6 W correct 5. F = 6W On the surface of the planet, W = GM m R. When the object is moved to a distance 4R from the center of the planet, the gravitational force on it will be F = GM m (4R) = GM m 6R = 6 GM m R = 6 W. AP M 993 MC points A newly discovered planet has twice the mass oftheearth,buttheaccelerationduetogravity on the new planet s surface is exactly the sameasthe accelerationdue togravityonthe Earth s surface. What is the radius R p of the new planet in terms of the radius R of Earth?. R p = R. R p = 4 R 3. R p = R 4. R p = R 5. R p = R correct From Newton s second law and the law of universal gravitation, the gravitational force near the surface is F g = mg = G M m r g = GM r. M p = M e and g p = g e, so GM e R = GM p R p R = R p R p = R. = GM e R p Gravity on Planet X short points Planet X has a mass 4.6 times that of the Earth and a radius.5 times the radius of the Earth. What is the ratio of the acceleration due to gravity on the surface of Planet X to the acceleration due to gravity on the surface of the Earth? Correct answer: Let : M X = 4.6M E and R X =.5R E. The acceleration due to gravity is a = GM R M R, so g X g E = R E g X g E = M X M E R X ( RE.5R E = ) ( ) 4.6ME M E Gravity on Ceres 08 (part of ) 0.0 points TheasteroidCereshasamass kg and a radius of km. What is g on the surface? The value of the universal gravitational constant is N m /kg.

7 Version 00 circular and gravitation holland (383) 7 Correct answer: m/s. Let : M = kg, r = km, and G = N m /kg. The weight on the surface for an object of mass M is W = mg = G M m r g = G M r = ( N m /kg ) ( kg km (476.9 km) 000 m = m/s. ) 09 (part of ) 0.0 points How much would an 8.9 kg astronaut weigh on this asteroid? Correct answer: 5.86 N. The weight of the astronaut will be W = mg = (8.9 kg)( m/s ) = 5.86 N.

AP Physics QUIZ Gravitation

AP Physics QUIZ Gravitation Name: 1. If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of the force exerted by the satellite on the