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FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003

NAME: _________________________________ STUDENT ID: _______________________ INSTRUCTION

1. This exam booklet has 14 pages. Make sure none are missing 2. There is an equation sheet on page 14. You may tear the equation

sheet off. 3. There are two parts to the exam:

• Part I has twelve multiple choice questions (1 to 12), where you must circle the one correct answer (a,b,c,d,e). Rough work can be done on the backside of the sheet opposite the question page

• Part II includes nine full-answer questions (13 to 21). Do all nine questions. All works must be done on the blank space below the questions. If you run out of space, you may write on the backside of the sheet opposite the question page.

4. Calculators are allowed

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PART I: MULTIPLE CHOICE QUESTIONS (question 1 to 12) For each question circle the one correct answer (a,b,c,d or e).

1. (2.5 point) Which of the following five acceleration versus time graphs is correct for an object moving in a straight line at a constant velocity of 20 m/s?

a) I b) II c) III d) IV e) V

2. (2.5 point) The coordinate of an object is given as a function of time by x = 4t2 – 3t3,

where x is in meters and t is in seconds. Its average acceleration over the interval from t = 0 to t = 2 s is: a) -13 m/s2 b) 4 m/s2 c) 10 m/s2 d) -10 m/s2 e) -8 m/s2

3. (2.5 point) A cannon fires a projectile as shown. The dashed line shows the trajectory in

the absence of gravity; points MNOP correspond to the position of the projectile at one second intervals. If g = 10 m/s2, the lengths X,Y,Z are:

4. (2.5 point) A horizontal force of 12 N pushes a 0.50-kg book against a vertical wall. The

book is initially at rest. If the coefficients of friction are µs = 0.60 and µk = 0.50 which of the following is true.

A) The block will start moving and accelerate B) If started moving downward, the block will accelerate C) The frictional force is 4.9 N D) The frictional force is 7.2 N E) The normal force is 4.9 N

A) 5 m, 20 m, 45 m B) 0.2 m, 0.8 m, 1.8 m C) 10 m, 20 m, 30 m D) 5 m, 10 m, 15 m E) 10 m, 40 m, 90 m

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5. (2.5 point) Three blocks (A, B, C), each having the same mass M, are connected by strings as shown. Block C is pulled to the right by a force that causes the entire system to accelerate. Neglecting friction, the net force acting on block B is:

a) 0 b) !F / 3 c)

!F / 2 d) 2

!F / 3 e)

!F

6. (2.5 point) A toy cork gun contains a spring whose spring constant is 10.0 N/m. The spring is compressed 5.00 cm and then used to propel a 6.00-g cork. The cork, however, sticks to the spring for 1.00 cm beyond its unstretched length before separation occurs. The muzzle velocity of this cork is:

A) 1.02 m/s B) 1.41 m/s C) 2.00 m/s D) 2.04 m/s E) 4.00 m/s

7. (2.5 points) A 700-N man jumps out of a window into a fire net 10 m below. The net

stretches 2 m before bringing the man to rest and tossing him back into the air. The maximum potential energy of the net, compared to its unstretched potential energy, is:

A) 300 J B) 710 J C) 850 J D) 7000 J E) 8400 J

8. (2.5 points) Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. The boys pull themselves together along the rod. When they meet the 40-kg boy will have moved what distance? a) 4 m b) 5 m c) 6 m d) 10 m e) need to know the forces they exert

9. (2.5 points) A 2.5-kg stone is released from rest and falls toward Earth. After 4.0 s, the magnitude of its momentum is: a) 98 kg·m/s b) 78 kg·m/s c) 39 kg·m/s d) 24 kg·m/s e) 0 kg·m/s

10. (2.5 points) At time t = 0 s, a wheel has an angular displacement of zero radians and an angular velocity of +18 rad/s. The wheel has a constant acceleration of -0.55 rad/s2. In this situation, the time t (after t = 0 s), at which the kinetic energy of the wheel is twice the initial value, is closest to:

a) 46 s b) 57 s c) 69 s d) 33 s e) 79 s 11. (2.5 points) Which of the following statement regarding angular momentum is correct?

a) A particle moving in a straight line with constant speed necessarily has zero angular momentum. b) If the torque acting on a particle is zero about an arbitrary origin, then the angular momentum of the particle is also zero about that origin. c) The angular momentum of a moving particle depends on the specific origin with respect to which the angular momentum is calculated. d) If the speed of a particle is constant, then the angular momentum of the particle about any specific origin must also be constant. e) Consider a planet moving in a circular orbit about a star. Even if the planet is spinning it is not possible for its total angular momentum to be zero.

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12. (2.5 points) A light triangular plate OAB is in a horizontal plane. Three forces, F1 = 3 N,

F2 = 1 N, and F3 = 9 N, act on the plate, which is pivoted about a vertical axes through point O. In Figure below, the magnitude of the torque due to force F1 about the axis through point O is closest to:

PART II: FULL ANSWER QUESTIONS (question 13 to 21) Do all nine questions on the provided area below the questions. Show all works.

13. (10 points) The position of a particle moving in an xy plane is given by

!r = 5t3 ! 6t( ) i + 5 ! 2t4( ) j , with

!r in meters and t in seconds.

a) In unit-vector notation, calculate the position, !r , velocity,

!v , and acceleration, !a , at t = 3 s.

b) What is the angle between the positive direction of the +x axis and a line tangent (i.e. !v ) to

the particle's path at t = 3 s? Give your answer in the range of (-180o; 180o).

a) 1.1 N•m b) 1.4 N•m c) 0.90 N•m d) 1.8 N•m e) 1.6 N•m

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14. (10 points) In the figure, a stone is projected at a cliff of height h with an initial speed of 46.0 m/s directed at an angle θ0 = 59.0° above the horizontal. The stone strikes at A, 5.85 s after launching.

a) Find the height, h, of the cliff.

b) Find the speed of the stone just before impact at A.

c) Find the maximum height reached, H, above the ground. What is the speed of the stone at the maximum height?

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15. (10 points) A loaded penguin sled weighing 65 N rests on a plane inclined at angle θ =

23° to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.26, and the coefficient of kinetic friction is 0.17.

a) Draw free-body diagrams that includes all forces on the sled. Assume that the magnitude of the force

!F is enough to accelerate the sled up the incline.

b) Assume that the sled is initially at rest, what is the minimum magnitude of the force !F that will accelerate the sled up the incline? c) For the force calculated in part b, what is the acceleration of the box?

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16. (10 points) A 6.6 kg brick moves along an x axis. Its acceleration as a function of its position is shown in the figure below.

a) Use graphical integration to calculate the net work done on the brick as it moves from x = 2m to x = 8m . b) If the speed of the brick is 2.2 m/s when it is at x = 2m , calculate its speed when it is at x = 8m .

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17. (10 points) In the figure, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle θ1 = 69.0° and the oxygen nucleus recoils with speed 1.60 × 105 m/s and at angle θ2 = 52.0°.

a) Using conservation of momentum calculate the initial and final speed of the alpha particle.

b) Calculate the change in kinetic energy, !K . Is the collision elastic? Briefly explain.

Alpha Particle mass m1 = 4u = 6.68 !10

"27 kg Oxygen mass m2 = 16u = 2.672 !10

"26 kg

1u ! atomic mass unit = 1.67 "10#27 kg

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18. (10 points) In the figure below, a lawn roller in the form of a solid cylinder ( I = 12MR2 )

is being pulled horizontally by a horizontal force, B, applied to an axle through the center of the roller, as shown in the sketch. The roller has radius R = 0.68 meters and mass M = 68 kg. The roller rolls without slipping on a rough surface with µs = 0.3 , and has a

linear acceleration of a = 2.5 ms2

. After it rolls 1.0 m it falls down an icy slope.

1.0 m icy smooth slope (no friction) h = 3.0 m

a) For the part when the roller is still on the flat surface. Draw a free-body diagram that includes all forces acting on the roller. The diagram must show the direction of the linear and angular acceleration. You must briefly explain the reason for the direction of force of friction, as well as the reason why you use static and not kinetic friction. Use Newton’s law for translation and rotation to find the magnitude of the force B.

b) Using the diagram from part a) determine the minimum coefficient of static friction, µs , in order for the roller to roll without slipping. Your answer must be smaller than 0.3, which is the value for the surfaces of the problem.

Rough surface µ > 0

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18. Continued c) Find the linear and angular speed of the roller at the edge of the slope. Assume roller starts from rest without any rotational speed. Assume that initially, v0 = 0 and !0 = 0 .

d) Use conservation of energy to find the linear (v) and angular speed (ω) when it reaches the bottom of the slope h = 3.0 m below. Assume that after it reaches the edge, there is no force acting on it (i.e. B = 0). 19. (10 points) In the figure below, box A and B are connected by a rope-pulley system. Box

A moves to the right, and the rope moves over the pulley without slipping, and the pulley has a clockwise angular velocity, ! . The data are shown in the figure.

v TA TA ω mass of pulley is MP = 3.0 kg

f TB I = 12MPR

2 solid cylindrical pulley

R = 0.2 m, radius of cylinder Coefficient of Friction µk = 0.5 TB

a) Draw a free body diagram of the boxes showing all the forces acting on it. Draw a free body diagram of the pulley showing all the forces (including those due to the hinge) acting on it. The diagrams should include the direction of linear acceleration, a, of mA and mB, and the angular acceleration α of the pulley.

mA = 10.0 kg

mB=4.0kg

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19. Continued b) Use Newton’s law for translation and rotation to find the linear acceleration (magnitude and direction)

!a of mA and mB, and the angular acceleration (magnitude and sense of rotation), α, of the pulley. The no-slip condition is useful. Assume the rope does not slip on the pulley. Note thatTA ! TB . Also you may want to assume that Box B accelerates up.

20. (10 points) In figure below, a carousel has a radius of 3.0 m and a moment of inertia of

IC = 8000kg •m2 , for rotation about axis perpendicular to the its center. The carousel is rotating unpowered and without friction with an angular velocity of 1.2 rad/s. An 80-kg man runs with a velocity of 5.0 m/s, on a line tangent to the rim of the carousel, overtaking it. The man runs onto the carousel and grabs hold of a pole on the rim.

a) Before the collision, what is the magnitude of the angular momentum of the rotating carousel,

!LC , with respect to the center of the carousel? What is the direction of

!LC ?

Directions (+x, +y, +z, -x, -y, -z) are as indicated in the above figure.

+y

+x

Direction perpendicular to x-y plane ! indicates +z out of the page ! indicates –z into page

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20 continued b) Before the collision, what is the magnitude of the angular moment of the running 80-kg man,

!LM , with respect to the center of the carousel? What is the direction

of !LM ?

c) After the collision when the man is on the carousel, what is the magnitude of the final angular velocity of the carousel (with the man on it), ! fC ? What is the direction of the

final angular velocity !! fC ? Note: Itotal = IC + mR

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21. (10 points) A diving board of length 3.00 m is supported at a point (P) 1.00 m from the end, and a diver weighing 520 N stands at the free end. The diving board is of uniform

cross section and weighs 300 N.

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21 continued a) Find the force of support at point P Hint: Do the torque part first! b) Find the force of support at point L, at the left-hand end.

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Useful Equations Kinematics x = x0 + v0xt + (1 / 2)axt

2 , vx = v0x + axt , vx2 = v0x

2 + 2ax x ! x0( ) , vx = dx / dt ;

ax = dvx / dt ; !v = vxi + vy j + vzk ;

!a = axi + ay j + azk ; average speed savg = (total distance)/(total

time); average velocity (x-com)vavg,x = x2 ! x1( ) / t2 ! t1( ) , average acceleration (x-com)

aavg,x = v2x ! v1x( ) / t2 ! t1( ) . Newton’s Laws !Fnet =

!Fi! = 0 (Object in equilibrium);

!Fnet = m

!a

(Nonzero net force); Weight: Fg = mg , g = 9.8m / s2 ; Centripetal acceleration arad =v2

r;

Friction fs ! µsFN , fk = µkFN . Hooke’s Law Fx = !kx . Work and Energy

W =!F •!d = F cos!( )d = F||d ; W net = !K = (1 / 2)mvf

2 " (1 / 2)mvi2 (valid if netW is the net or

total work done on the object);W grav = !mg yf ! yi( ) (gravitational work),

W el = ! (1 / 2)kx f2 ! (1 / 2)kxi

2( ) (elastic work) Conservation of Mechanical Energy (only conservative forces are present) Emech =U + K W net = !"U = ! U2 !U1( ) = "K = K2 ! K1 ,U1 + K1 =U2 + K2 ,Ugrav = mgy ,Uel = (1 / 2)kx

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Non-Conservative Forces Wexternal = !Emech + !Eth (Wext work done by external forces, and we set !Eint = 0 ), where !Eth = fkd (thermal energy or negative work done by friction). Using !Emech = !U + !K = U f "Ui( ) + K f " Ki( ) , U f + K f =Ui + Ki +Wext ! fkd

Work due to variable force 1D: W = Fx dxxi

x f! " area under Fx vs. x, from x = xi to xf

POWER: averagePavg =W / !t = Fd / !t = Fvavg ; instantaneous P =!F • !v = Fvcos!

Momentum: !P = m!v ,

!J =

!F

t1

t2! dt =!Fav t2 " t1( ) ,

!J = !

!P =!P2 "

!P1 . Newton’s Law in Terms of

Momentum !Fnet = d

!p / dt . For !Fnet = 0 , d

!p / dt = 0 gives momentum conservation: !P ! constant.

Rotational Kinematics Equations:!avg = "2 #"1( ) / t2 # t1( ) , !avg = "2 #"1( ) / t2 # t1( ) For ! z = constant, ! =!0 +"t , ! = !0 +"0t + (1 / 2)#t

2 ,! 2 =!02 + 2" # $#0( )

Linear and angular variables: s = r! , v = r! , atan = R! (tangential) ,arad = v2 / r =! 2r (radial)

Moment of Inertia and Rotational Kinetic Energy

I = miri2

i=1

N

! , Krot = (1 / 2)I!2 . Center of

Mass (COM) !rcom = mi

!ri / mi!! . Torque and Newton’s Laws of Rotating Body: rigid

body! = Fr" , !! net = "

!! iext = I# , r! -moment arm about axis; point

! ! =! r "! F about origin O.

Combined Rotation and Translation of a Rigid BodyK = (1 / 2)Mvcom2 + (1 / 2)Icom!

2 ,

!Fnet = M

!acom , !! net = Icom

!" . Rolling without slipping

s = R! , vcom = R! , acom = R! . Angular Momentum L = I! (solid object) where I is the moment of inertia about the axis of rotation.

!L = !r ! !p" L = mvr sin# , valid for point particle about an origin O.

Newton’s Second Law of rigid body in terms of angular momentum !! net = "

!! iext = d

!L / dt( ) .

For !! net = 0 ,

d!L / dt( ) = 0 and angular momentum is conserved,

!L ! constant.

Equilibrium conditions !Fnet =

!Fiezt = 0! about all object,

!! net =

!! iext = 0" about any point.