MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term Exam 2 Solutions

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 801 Physics Fall Term 013 Problem 1 of 4 (5 points) Exam Solutions Answers without work shown will not be given any credit A block of mass m is released from rest at a height 3R above the ground and slides down a frictionless vertical track When it reaches a height R above the ground it enters a semi-circular frictionless track of radius R At the bottom of the semi-circular track it enters a horizontal track starting at x = 0 The coefficient of kinetic friction along the horizontal track is non-uniform and varies according to µ(x) = bx, where b is a positive constant The block slides and comes to rest at x = d The magnitude of the gravitational acceleration is g a) What is the normal force of the semi-circular track on the block when the block makes an angle θ as shown in the figure above? Express your answer in terms of m, R, θ, and g as needed b) Calculate the distance d that the block slides along the horizontal track before coming to a rest Express your answer in terms of m, R, b, and g as needed Solutions: a) (15 points) The free body diagram for the block when it reaches the angle θ is shown in the figure below Newton s Second Law in the radial direction is N θ + mg cosθ = mv θ R N = mg cosθ + mv θ θ R (11) 1

2 The energy state diagrams for the initial state of the block (state i ) and when the block reaches the angle θ (state θ ) are shown in the figure below Choose the ground for the zero point for the gravitational potential energy The mechanical energy in the initial state is E i = mg3r (1) The mechanical energy when the block reaches the angle θ is E θ = mgr(1 cosθ) + 1 mv θ (13) There is no non-conservative work done on the block when it moves from (state i ) to state θ Therefore E θ E i = 0 mgr(1 cosθ) + 1 mv θ = mg3r mv θ R = 4mg + mg cosθ (14) In order to determine the normal force on the block when the block reaches the angle θ, we substitute Eq (14) into Eq (11) and determine that N θ = 4mg + 3mg cosθ (15)

3 b) (10 points) We show the energy state diagram for when the block comes to rest on the horizontal (state f ) in the figure below The final mechanical energy is E f = 0 (16) The non-conservative frictional work done on the block when it moves from (state i ) to state f is f W i f = f k d x =d x =d x =d mgbd s = f k d x = µ( x )mgd = mgb x d x = (17) i x =0 x =0 x =0 The energy principle then becomes W i f = E f E i then becomes mgbd = 3mgR, (18) which we can solve for the distance than the block traveled on the horizontal surface d = 6R b (19) 3

4 Problem of 4 (5 points) Answers without work shown will not be given any credit Two astronauts are playing catch in a zero gravitational field Astronaut 1 of mass m 1 is initially moving to the right with speed v 1 Astronaut of mass m is initially moving to the right with speed v > v 1 Astronaut 1 throws a ball of mass m with speed u relative to herself in a direction opposite to her motion Astronaut catches the ball The final speed of astronaut 1 is v f,1 and the final speed of astronaut is v f, before during after a) What is the speed v f,1 of astronaut 1 after throwing the ball? Express your answer in terms of m, m 1, m, u, and v 1, as needed b) What is the required speed u of the ball (relative to astronaut 1) such that the final speed of both astronauts are equal v f,1 = v f,? Express your answer in terms of m, m 1, m, v 1, and v, as needed 4

5 Solutions a) (10 points) Because the momentum of astronaut does not change between the before state and the during state, we choose as our system the astronaut 1 and the ball We choose as a reference frame the frame in which astronaut 1 and the ball are moving with speed v 1 before the ball is thrown We pick the positive î -direction to point in the direction of their velocity (to the right) When the ball leaves the astronaut 1 it is traveling with speed u v f,1 in the negative î -direction where v f,1 is the speed of the astronaut 1 after the ball has left her hand The momentum diagrams are shown in the figure below There are no external forces acting on the system consisting of the astronaut 1 and the ball Therefore the momentum of the system is constant and we have that (m + m 1 )v 1 = m 1 v f,1 m(u v f,1 ) v f,1 = v 1 + mu m + m 1 (110) b) (15 points) We can find the relative speed u that the ball was thrown two different ways Our first approach will be to consider the system consisting of astronaut and the ball and our two states are the during state when the ball has not yet reached astronaut and the after state when astronaut catches the ball The momentum diagrams are shown in the figure below There are no external forces acting on the system consisting of the astronaut and the ball Therefore the momentum of the system is constant and we have that 5

6 m v m(u v f,1 ) = (m + m )v f, (111) We are told that v f,1 = v f,, therefore Eq (111) becomes m v mu = m v f,1 u = m m (v v f,1) (11) We now substitute Eq (110) into Eq (11) yielding u = m m v v 1 + mu m + m 1 (113) We can now solve Eq (113) for the relative speed of the ball u m + m + m 1 m + m 1 = m (v v ) 1 m u = m (m + m )(v v ) (114) 1 1 m(m + m 1 + m ) An alternative approach is to apply the momentum principle to the before state and after state We choose as our system the ball and both astronauts 1 and The momentum state diagrams are shown in the figure below The momentum of the system is constant (no external forces) and we have that Set v f,1 = v f, in Eq (115) yielding m v + (m + m 1 )v 1 = (m + m )v f, + m 1 v f,1 (115) v f,1 = m v + (m + m 1 )v 1 (m + m 1 + m ) Now substitute Eq (110) into Eq (116) yielding (116) 6

7 v 1 + We now solve Eq (117) for the relative speed of the ball which agrees with Eq (114) u = m + m 1 m = m + m 1 m mu = m v + (m + m )v 1 1 (117) m + m 1 (m + m 1 + m ) v 1 m v + (m + m 1 )v 1 (m + m 1 + m ), (118) m (v 1 v ) (m + m 1 + m ), 7

8 Problem 3 of 4 (5 points) Answers without work shown will not be given any credit A two-stage rocket in a zero gravitational field starts from rest and burns fuel The fuel is ejected at a speed u relative to the rocket After all the fuel has been burned, explosive bolts release the first stage and push it backwards with a speed v 1 relative to the second stage The mass of the first stage before any fuel is burned is m 1 m 0 + m f, where m f is the mass of the fuel The mass of the second stage is m The total mass of the rocket before any fuel is burnt is m 1 + m The goal of this problem is to find the speed of the second stage after the separation a) When the rocket is traveling at speed v, derive a relation between the differential of the speed of the rocket dv, and the differential of the mass of the rocket, dm r, and the speed u relative to the rocket of the ejected fuel You must show your work in order to receive credit b) What is the speed v f of the rocket immediately after all the fuel has been burned but before the second stage is released? Express your answers in terms of u, m f, m 0, and m as needed c) What is the speed v of the second stage immediately after it has been released? Solution: Express your answers in terms of v 1, v f, m f, m, and m 0 as needed a) (10 points) Choose as a system the rocket and fuel (mass m r ) that is not burned during the interval [t,t + dt], and the infinitesimally small amount of fuel (mass dm f ) that is burned mass during this interval The momentum state diagrams for time t and t + Δt are shown in the figures below There are no external forces so the momentum principle becomes 8

9 0 = dm f (v + dv u) + m r (v + dv u) (dm f + m r )v (119) 0 = dm f dv dm f u + m r dv In the limit as dt 0, the term dm f dv is much smaller than the other two terms We also know that dm f = dm r Using these two facts, and so Eq (119) becomes dv = dm r u m r (10) b) (10 points) The initial mass of the two stages is m 0 + m f + m When all the fuel is burned the mass of the two stages of the rocket is m 0 + m The integral version of Eq (10) becomes Integration then yields v =v f v =0 d v = u m r =m 0 +m d m r (11) mr =m 0 +m f +m m r m v f = uln 0 + m = uln m + m + m m 0 f m 0 + m f + m m 0 + m = uln 1+ f m 0 + m (1) c) (5 points) After all the fuel has been burned choose as our system the two stages of the rocket The momentum state diagrams for before and after the explosive bolts are fired are shown in the figure below No external forces are acting on the system so the momentum principle becomes 0 = m 0 (v v 1 ) + m v (m 0 + m )v f (13) We can solve Eq (13) for the speed of stage two immediately after the bolts have been fired 0 = m 0 (v v 1 ) + m v (m 0 + m )v f v = v f + m v (14) 0 1 m 0 + m 9

10 Problem 4 of 4 (5 points): Concept Questions (Parts A through E) Part A (5 points) The potential energy function U (x) = (1/ )kx energy E is shown below for a particle with total mechanical The particle was released moving in the positive x -direction at the point x = x 1 at t = 0 The x -component of the velocity of the particle at t = 0 is 1) v x (t = 0) = 0, ) v x (t = 0) = + E m, 3) v x (t = 0) = + m E 1 kx 1, 4) v x (t = 0) = kx 1 m, 5) None of the above Answer 3 Kinetic energy K(x 1 ) = E U (x 1 ) 1 m(v x (t = 0)) = E 1 kx 1 Therefore 10

11 v x (t = 0) = + m E 1 kx 1, where we take the positive square root because we are told that the object is moving in the positive x -direction at t = 0 11

12 Part B (5 points) Consider two carts, of masses m and 4m, at rest on an air track If you first push both carts with an equal force F an equal distance d, then the change in momentum of the light cart is 1) one-fourth ) one-half 3) equal to 4) twice 5) four times the momentum of the heavy cart Solution The work done is the same on both objects therefore W = ΔK 1 = ΔK Where object 1 is the lighter than object Because the objects started from rest, the change in kinetic energies are related to the change in momentum by ΔK 1 = Δp 1 and ΔK m = Δp 1 m 1 Therefore using the given information that m 1 = 1, we have that m 4 Δp 1 = Δp m Δp m 1 m 1 = Δp 1 = 1 m Δp 1

13 Part C (5 points) A worker is unloading the last car (mass m c ) from a barge (mass m b ) tied to a pier by a short taut cable Starting at rest, he drives the car with a constant acceleration until it reaches the end of the barge at time t 0 What is the tension in the cable during the interval 0 < t < t 0? 1) m c a c ) (m c + m b )a c 3) m c m c + m b a c 4) m b m c + m b a c 5) None of the above Solution 1 Choose the car and barge as the system The external force is the tension in the rope Then m ac T = m Acm T = (m c + m b ) c + m ab b m c + m b = m c because the barge is at rest so a b = 0 Therefore the magnitude of the tension in the rope is m c a c Alternatively there is a force F b,c on the car due to the interaction between the car and the barge (frictional force) Newton s Second Law on the car is then F b,c = m c a c The barge is not accelerating so T F c,b = 0 T = F c,b By Newton s Third Law F c,b = F b,c Therefore T = m c a c a c 13

14 Part D (5 points) Suppose water is leaking out through the hole of the bottom of a cart that is moving with speed v on a frictionless surface You may ignore all the effects of air resistance While the water is leaking out, the speed of the cart 1) increases but is independent of the rate that the water is leaking out of the cart ) increases but depends on the rate that the water is leaking out of the cart 3) does not change 4) decreases but is independent of the rate that the water is leaking out of the cart 5) decreases but depends on the rate that the water is leaking out of the cart Solution 3 Suppose at time t we take as our system the cart, the sand in the cart that does not fall out during the time interval [t,t + Δt], and the sand that does fall out during that time interval Because the sand falls down vertically, the momentum of the sand does not change There are no external forces acting in the horizontal direction, so the momentum of the system is constant If the momentum of the sand falling out does not change then the momentum of the cart and the rest of the sand inside cannot change as well Because the mass of the cart and the sand in the cart that does not fall out during the time interval [t,t + Δt] does not change, the speed of the cart does not change 14

15 Part E (5 points) A point-like object of mass m is attached to one end of a massless string that is fixed at the point S The object moves in a vertical circle of radius r centered at S with constant angular speed ω in a zero gravitational field Which of the vectors 1-6 best describes the velocity, acceleration, and net force acting on the object at the moment illustrated in the figure above? Solution 3 The velocity is directed upwards, the speed is constant so the acceleration is towards the center of the circle Therefore there must be a force directed towards the center of the circle Answer: 15

Exam 2: Equation Summary

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term 2012 Exam 2: Equation Summary Newton s Second Law: Force, Mass, Acceleration: Newton s Third Law: Center of Mass: Velocity