Exam 1 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01
Fall Term 2006
Exam 1 Solutions Part I Concept Questions: (20 points). Circle your answer. There are five questions and each question is worth 4 points. 1) (4 points) A pendulum bob swings down and is moving fast at the lowest point in its swing. T is the tension in the string, T ! W = mr" 2 W is the gravitational force exerted on the pendulum bob. Which free-body diagram below best represents the forces exerted on the pendulum bob at the lowest point? The lengths of the arrows represent the relative magnitudes of the forces.
Solution: d). The bob is undergoing circular motion. It is accelerating towards the center. Newton’s Second Law gives T ! W = mr" 2 hence T = W + mr! 2 so T > W .
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2) (4 points) Which of the following expressions gives a reasonable formula for the period of the moon orbiting around the earth? Let G be the universal gravitational constant, re,m be the average distance between the earth and the moon, mm be the mass of the moon, and me be the mass of the earth. a) T ! G me mm / r 2 . e,m
b) T ! G me mm / re,m . 3 c) T ! re,m / Gmm .
3 d) T ! re,m / Gme .
2 e) T ! re,m / Gme .
2 f) T ! re,m / Gmm .
Solution: (d). The answer cannot depend on mass of the moon by the Equivalence Principle (the moon’s mass cancels from each side of Newton’s second Law). The period must be some product of the given
T ! re,m aG b me c dim[G] = dim[force ! m 2 ! kg -2 ] = dim[kg ! m ! s-2 ! m 2 ! kg -2 ] = dim[s-2 ! m 3 ! kg -1 ] dim[me ] = kg 3 e,m
dim[(r
/ Gme ) ] = (m 3 /(s-2 ! m 3 ! kg --1 )(kg))1/2 = s 1/ 2
Using the above results:
dim[T ] ! dim[re,m aG b me c ] = (m a )(s-2b " m 3b " kg -b )(kg c ) = s Thus the powers must satisfy: a + 3b = 0, ! 2b = 1, ! b + c = 0 . Solving we have that b = !1 / 2 , a = !3b = 3 / 2 , and c = b = !1 / 2 .
So 3 T ! re,m3/ 2G "1/ 2 me "1/ 2 = re,m / Gme .
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3) (4 points) A large truck collides head-on with a small car. During the collision a) the truck exerts a greater force on the car than the car exerts on the truck. b) the car exerts a greater force on the truck than the truck exerts on the car. c) the truck exerts the same force on the car as the car exerts on the truck. d) the truck exerts a force on the car but the car does not exert a force on the truck. Solution: c). Newton’s Third Law. 4) (4 points) A child throws a ball in the air. The path of the ball is a parabola as shown in the picture. Ignore air resistance. When the ball is at the highest point in its trajectory, which of the following statements is true?
a) The magnitudes of the velocity and acceleration are zero. b) The magnitude of the velocity is at a minimum but not equal to zero and the acceleration is not equal to zero. c) The magnitude of the velocity is equal to zero, and the magnitude of the acceleration is not equal to zero. d) The magnitude of the velocity is at a minimum but not equal to zero and the magnitude of the acceleration is zero. e) Neither the magnitudes of acceleration nor velocity are at their minimum values. Solution: b). At the top of the flight the vertical component of the velocity is zero. During the flight the horizontal component of the velocity does not change. Therefore at the top, the magnitude of the velocity is at a minimum. The acceleration is always downwards and non-zero.
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5) (4 points) An object that is free to slide on a table is attached to a spring. The other end of the spring is attached firmly to the wall. In the graphs shown, x is the x-component of the displacement of the object (away from the wall) with respect to the equilibrium position of the object, and Fx is the x-component of the force that the spring exerts on the object.
Which graph could represent the x-component of the force, Fx , as function of x ?
(a)
(b)
(c)
(d)
Solution: b). With this choice of coordinates the force is always a restoring force. When x > 0 , the spring is stretched so the force on the object points in the negative ˆi , hence the x-component of the force is negative, F < 0 . When the spring is x compressed x < 0 , the force on the object points in the positive ˆi , hence the xcomponent of the force is positive, Fx > 0 .
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Part II Analytic Problems: (80 points). Answers without work shown will not be given any credit. Good luck! Problem 1: Circular Motion (20 points) A car of mass M traveling at a constant speed v enters a curve of radius r that is banked at an angle ! towards the center of the curve. The coefficient of static friction between the tires and the road is µ .
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a) (8 points) Suppose µ tan ! < 1 . What is the maximum speed vmax with which the car can enter the curve so that it does not slide off the road? (Answer in terms of the quantities M , ! , g , r , and µ , although you may not need them all.)
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Problem 3: Walking the Plank (20 points) A uniform plank of length l and mass m is pivoted at one end and supported by a vertical rope at the other end. (Express your answers in terms of the quantities m , g , and l , although you may not need them all.)
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A person of mass 4m walks across the plank and stops so that the center of mass of the person is a distance 3 l / 4 from the pivot point. . (Express your answers in terms of the quantities m , g , and l , although you may not need them all.)
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Problem 4: Towing a Sled (20 points) A mother tows her daughter on a sled on level ice. The friction between the sled and the ice is negligible, and the tow rope makes an angle of ! to the horizontal. The combined mass of the sled and the child is M . The sled has an acceleration in the horizontal direction of magnitude a . As we will learn to justify in a few weeks, the child and sled can be treated in this problem as if they comprised a single particle. Express your answers in terms of the given variables M , ! , g ,and a .
a) (6 points) Calculate the tension, T , in the rope.
b) (4 points) What is the magnitude of the normal force, N , exerted by the ice on the sled?
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The mother then tows her daughter up a hill that is inclined by an angle ! with respect to the horizonal. The friction between the sled and the ice is negligible, and the tow rope makes an angle of ! to the hill. The sled has the same magnitude of acceleration, a . Express your answers in terms of the given variables M , ! , ! , g ,and a .
c) (6 points) Calculate the new tension, T ! , in the rope.
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d) (4 points) What is the magnitude of the new normal force, N ! , exerted by the ice on the sled?
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Fall Term 2006
Exam 1 Solutions Part I Concept Questions: (20 points). Circle your answer. There are five questions and each question is worth 4 points. 1) (4 points) A pendulum bob swings down and is moving fast at the lowest point in its swing. T is the tension in the string, T ! W = mr" 2 W is the gravitational force exerted on the pendulum bob. Which free-body diagram below best represents the forces exerted on the pendulum bob at the lowest point? The lengths of the arrows represent the relative magnitudes of the forces.
Solution: d). The bob is undergoing circular motion. It is accelerating towards the center. Newton’s Second Law gives T ! W = mr" 2 hence T = W + mr! 2 so T > W .
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2) (4 points) Which of the following expressions gives a reasonable formula for the period of the moon orbiting around the earth? Let G be the universal gravitational constant, re,m be the average distance between the earth and the moon, mm be the mass of the moon, and me be the mass of the earth. a) T ! G me mm / r 2 . e,m
b) T ! G me mm / re,m . 3 c) T ! re,m / Gmm .
3 d) T ! re,m / Gme .
2 e) T ! re,m / Gme .
2 f) T ! re,m / Gmm .
Solution: (d). The answer cannot depend on mass of the moon by the Equivalence Principle (the moon’s mass cancels from each side of Newton’s second Law). The period must be some product of the given
T ! re,m aG b me c dim[G] = dim[force ! m 2 ! kg -2 ] = dim[kg ! m ! s-2 ! m 2 ! kg -2 ] = dim[s-2 ! m 3 ! kg -1 ] dim[me ] = kg 3 e,m
dim[(r
/ Gme ) ] = (m 3 /(s-2 ! m 3 ! kg --1 )(kg))1/2 = s 1/ 2
Using the above results:
dim[T ] ! dim[re,m aG b me c ] = (m a )(s-2b " m 3b " kg -b )(kg c ) = s Thus the powers must satisfy: a + 3b = 0, ! 2b = 1, ! b + c = 0 . Solving we have that b = !1 / 2 , a = !3b = 3 / 2 , and c = b = !1 / 2 .
So 3 T ! re,m3/ 2G "1/ 2 me "1/ 2 = re,m / Gme .
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3) (4 points) A large truck collides head-on with a small car. During the collision a) the truck exerts a greater force on the car than the car exerts on the truck. b) the car exerts a greater force on the truck than the truck exerts on the car. c) the truck exerts the same force on the car as the car exerts on the truck. d) the truck exerts a force on the car but the car does not exert a force on the truck. Solution: c). Newton’s Third Law. 4) (4 points) A child throws a ball in the air. The path of the ball is a parabola as shown in the picture. Ignore air resistance. When the ball is at the highest point in its trajectory, which of the following statements is true?
a) The magnitudes of the velocity and acceleration are zero. b) The magnitude of the velocity is at a minimum but not equal to zero and the acceleration is not equal to zero. c) The magnitude of the velocity is equal to zero, and the magnitude of the acceleration is not equal to zero. d) The magnitude of the velocity is at a minimum but not equal to zero and the magnitude of the acceleration is zero. e) Neither the magnitudes of acceleration nor velocity are at their minimum values. Solution: b). At the top of the flight the vertical component of the velocity is zero. During the flight the horizontal component of the velocity does not change. Therefore at the top, the magnitude of the velocity is at a minimum. The acceleration is always downwards and non-zero.
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5) (4 points) An object that is free to slide on a table is attached to a spring. The other end of the spring is attached firmly to the wall. In the graphs shown, x is the x-component of the displacement of the object (away from the wall) with respect to the equilibrium position of the object, and Fx is the x-component of the force that the spring exerts on the object.
Which graph could represent the x-component of the force, Fx , as function of x ?
(a)
(b)
(c)
(d)
Solution: b). With this choice of coordinates the force is always a restoring force. When x > 0 , the spring is stretched so the force on the object points in the negative ˆi , hence the x-component of the force is negative, F < 0 . When the spring is x compressed x < 0 , the force on the object points in the positive ˆi , hence the xcomponent of the force is positive, Fx > 0 .
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Part II Analytic Problems: (80 points). Answers without work shown will not be given any credit. Good luck! Problem 1: Circular Motion (20 points) A car of mass M traveling at a constant speed v enters a curve of radius r that is banked at an angle ! towards the center of the curve. The coefficient of static friction between the tires and the road is µ .
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a) (8 points) Suppose µ tan ! < 1 . What is the maximum speed vmax with which the car can enter the curve so that it does not slide off the road? (Answer in terms of the quantities M , ! , g , r , and µ , although you may not need them all.)
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Problem 3: Walking the Plank (20 points) A uniform plank of length l and mass m is pivoted at one end and supported by a vertical rope at the other end. (Express your answers in terms of the quantities m , g , and l , although you may not need them all.)
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A person of mass 4m walks across the plank and stops so that the center of mass of the person is a distance 3 l / 4 from the pivot point. . (Express your answers in terms of the quantities m , g , and l , although you may not need them all.)
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Problem 4: Towing a Sled (20 points) A mother tows her daughter on a sled on level ice. The friction between the sled and the ice is negligible, and the tow rope makes an angle of ! to the horizontal. The combined mass of the sled and the child is M . The sled has an acceleration in the horizontal direction of magnitude a . As we will learn to justify in a few weeks, the child and sled can be treated in this problem as if they comprised a single particle. Express your answers in terms of the given variables M , ! , g ,and a .
a) (6 points) Calculate the tension, T , in the rope.
b) (4 points) What is the magnitude of the normal force, N , exerted by the ice on the sled?
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The mother then tows her daughter up a hill that is inclined by an angle ! with respect to the horizonal. The friction between the sled and the ice is negligible, and the tow rope makes an angle of ! to the hill. The sled has the same magnitude of acceleration, a . Express your answers in terms of the given variables M , ! , ! , g ,and a .
c) (6 points) Calculate the new tension, T ! , in the rope.
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d) (4 points) What is the magnitude of the new normal force, N ! , exerted by the ice on the sled?
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