1

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym

NAME: _________________________________ STUDENT ID: _______________________ INSTRUCTION

1. This exam booklet has 13 pages. Make sure none are missing 2. There is an equation sheet on page 13. 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. Non-Programmable calculators are allowed 5. Programmable calculators are NOT ALLOWED.

2

Hint: Draw a FBD on composite (A+B+C) then calculate acceleration. Then draw FBD on C, and use Newton’s second law.

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) Hanging over the railing of a bridge, you drop an egg (no initial velocity) as you throw a second egg downward. The two curves that represent the dropped and thrown eggs are:

2. (2.5 point) Two automobiles are 150 kilometers apart and traveling toward each other.

One automobile is moving at 60 km/h and the other is moving at 40 km/h. In how many hours will they meet ? a) 1.25 h b) 1.5 h c) 1.75 h d) 2.0 h e) 2.5 h

3. (2.5 point) A stone is thrown horizontally and follows the path XYZ shown. The direction of the acceleration of the stone at point Y is:

4. (2.5 point) Two forces, one with a magnitude of 3 N and the other with a magnitude of 5

N, are applied to an object. For which orientation of the forces shown in the diagrams is the magnitude of the acceleration of the object the least ?

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 tension of the rope connecting B and C is:

a) 0 b) !F / 3 c)

!F / 2 d) 2

!F / 3 e)

!F

A) B) C) D) E)

A) Curve A and Curve B, respectively B) Curve C and Curve E, respectively C) Curve D and Curve G, respectively D) Curve D and Curve E, respectively E) Curve E and Curve G, respectively

A) I B) II C) III D) IV E) V

3

6. (2.5 point) A 400-N block is dragged along a horizontal surface by an applied force !F as

shown. The coefficient of kinetic friction is uk = 0.4 and the block moves at constant velocity. The magnitude of

!F is closest to:

A) 100 N B) 150 N C) 200 N D) 290 N E) 400 N

7. (2.5 points) At time t = 0 a 2-kg particle has a velocity of 4m / s( ) i ! 3m / s( ) j . At t = 3

s its velocity is 2m / s( ) i + 3m / s( ) j . During this time the work done on it was:

A) 4 J B) −4 J C) 4J( ) i + 36J( ) j D) −12 J E) −40 J 8. (2.5 points) Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while

block B, with a mass of 8.0 kg, is moving in the opposite direction with a speed of 3.0 m/s. The momentum of the center of mass of the two-block system is: a) 12 m/s in the same direction as B b) 16 m/s in the same direction as B c) 32 m/s in the same direction as A d) 60 m/s in the same direction as A e) 16 m/s in the same direction as A

9. (2.5 points) In the angular velocity (! ) vs. time (t) graph below, ! > 0 corresponds to a counterclockwise (CCW) rotation. Which point corresponds to a CCW rotation with decreasing magnitude, ! ?

ω II III IV t I V 10. (2.5 points) A wheel initially has an angular velocity of 36 rad/s but after 6.0s its angular

velocity is 24 rad/s. If its angular acceleration is constant the value is: a) 2.0 rad/s2 b) −2.0 rad/s2 c) 3.0 rad/s2 d) −3.0 rad/s2 e) 6.0 rad/s2 11. (2.5 points) In the diagram below, a hollow cylinder ( I = MR2 , M = 0.1 kg, R = 0.1m)

rolls without slipping and moves 2.4 m from point A to point B. If its speed at point A is vA = 5m / s , its speed at point B, vB is closest to:

A) I B) II C) III D) IV E) V

A) 6.38 m/s B) 1.22 m/s C) 6.06 m/s D) 6.97 m/s E) 3.64 m/s

vB = ?

HINT: 1) The weight of box is mg = 400 N; 2) May be easiest to substitute the answers to see which one obeys Newton’s law.

4

12. (2.5 points) In the figure below a square with length of the side 0.180 m, is acted on by three forces:

!F1,!F2 ,!F3 . Calculate the torque due to the force F3about an axis of rotation

perpendicular to the origin. Take counterclockwise to be positive. F1 = 18.0 N F2 = 26.0 N F3 = 14.0 N 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) A soccer ball is kicked from the ground with an initial speed of 17.7 m/s at an upward angle of 42.7! . A player 53.3 m away in the direction of the kick starts running to meet the ball at that instant.

a) Draw a diagram of the path of the trajectory (or a motion diagram) of the soccer ball, which shows the direction of the velocity and acceleration at the following points: i) the instant it leaves the ground; ii) its maximum height; iii) the instant when it hits the ground. Calculate the x and y component of the initial velocity (when it first leaves the ground).

b) Calculate the time it takes for the ball to hit the ground. Hence, calculate the range (the horizontal distance, before it hits the ground) of the ball.

A) 162 N•m B) −2.5 N•m C) 2.34 N•m D) 1.78 N•m E) −2.34 N•m

5

Question 13 (continued) c) What must the average speed of the player who is 53.3 m away be, in order to meet the ball before it hits the ground?

14. (10 points) In the figure, the mass of the box is m = 8.7 kg and the angle θ = 37.0° above the horizontal. There is no friction between the box and the incline.

a) Draw a free-body diagram of all the forces on the box. Calculate the normal force, FN, on the box. Calculate the tension in the cord that connects the wall and the box.

b) If the cord is cut, draw the free-body diagram of all the forces on the box. Use Newton’s second law to calculate the magnitude of the box’s acceleration.

6

15. (10 points) In the diagram below a horizontal force, F = 10 N is applied to a 5 kg crate moving up a 36.9! incline, with friction coefficientµk = 0.1and µs = 0.2 .

!v

F = 10 N 36.9!

a) Draw free-body diagram that includes all forces on the crate. Calculate the normal force, and the friction force on the crate. Use Newton’s second law to find the magnitude and direction of acceleration of the crate.

b) Assume that the sled is initially traveling at 32 m/s up the incline. What is its velocity after 1.2 seconds?

7

16. (10 points) In the figure, a 2.9 kg block is accelerated from rest by a spring of spring constant k = 630 N/m. The block leaves the spring at the spring’s relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction µk = 0.255. The friction force stops the block in distance D = 7.3m.

a) Calculate the work done by friction, Wf . Hence determine the increase in thermal energy, Eth = !Wf . b) Use the work-energy theorem or conservation of energy to calculate the original compression distance of the spring.

8

17. (10 points) A hockey puck of mass m = 2 kg traveling at 4.5 m/s along the x axis hits another identical hockey puck at rest.

A) If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, calculate the final velocity of the first puck?

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

9

P

18. (10 points) A yoyo is pulled as shown in the figure, by a force of magnitude F = 2 N. It moves linearly to the left with an acceleration,

!a , and rotate without slipping, with a counterclockwise (CCW) angular acceleration

!! .

!a

!!

F = 2 N r1 r2

a) Briefly explain why the force of the ground on the wheel at point P is a static force of friction,

!fS . Deduce the direction of

!fS . Draw

!fS at point P. Briefly justify your answer.

b) Use Newton’s second law for linear and angular (rotational) motion to find the magnitude of the linear and angular acceleration. Write

!a and !! in unit vector form with

+x as right, +y as up, and +z as out of the page. c) Calculate the magnitude of friction, fS . Calculate the minimum value of the coefficient of static friction, µS , in order for the yoyo to rotate without slipping.

DATA Mass: M = 0.2 kg Moment of Inertia: I = 8.2 !10

"4 kgim2 . Inner radius: r1 = 0.05m Outer radius: r2 = 0.25m

10

19. (10 points) In the figure below, box A and B are connected by a rope-pulley system. Box A moves up and B moves down, with linear acceleration, a. The rope moves over the pulley 1 and 2 without slipping, and both pulleys have clockwise (CW) angular acceleration α1 and α2, respectively. The data are shown on and beside the figure.

Pulley 1 T12 T12 Pulley 2 α1 α2 TA TB TA TB a a a) The linear acceleration of box A and B is a = 1.7mis!2 . Draw a free body diagram of all the forces on box B (mB = 4 kg). Use a = 1.7mis!2 and Newton’s second law (linear motion) to find the tension, TB , in the segment of rope connecting pulley 2 and box B.

b) Draw a free body diagram of all the forces on pulley 2 (data above figure). Use Newton’s second law (for rotation) to find the tension, T12, in the segment of rope connecting pulley 1 and 2.

c) Use the method of part a or part b to find the tension, TA, in the segment of rope connecting pulley 1 and box A.

mB = 4.0kg mA = 2.0kg

Data Pulley 1: left, Mass M1 = 6 kg, Radius R1 = 0.2 m Pulley 2: right, Mass M2 = 5 kg, Radius R2 = 0.15 m Both pulleys are solid cylinders: I = 1 / 2( )MR2

11

20. (10 points) This problem considers the same system as in problem 19, which refers to the figure with box A, box B, pulley 1, and pulley 2, all connected by a rope. Assume the rope rotates the pulleys without slipping.

a) Calculate the change in gravitational potential energy, !U =U final "Uinitial , of the system (A, B, pulley 1 and 2), after box B has fallen 1.5m.

b) Assume that the system start from rest. Use conservation of mechanical energy (!K = "!U or an equivalent equation), or the work-energy theorem to calculate the linear velocity of box A and of box B, after box B has fallen 1.5 m. WARNING: Solving this part by using kinematics equations method will give you a MARK of ZERO! c) Calculate the angular velocity of pulley 1 (ω1) and 2 (ω2) after box B has fallen 1.5 m. Briefly comment on why the values of ω1 and ω2 are different. Hence calculate the rotational kinetic energy of pulley 1 and 2.

12

21. (10 points) In figure below, a 2.0 g bullet is fired into a block (mass 0.5 kg) attached to a rod of length 0.60 m, and moment of inertia (about A) Irod = 0.060kg •m

2 . The speed of the bullet just before collision is vbullet = 500m / s . The system (bullet+block+rod) rotates about an axis of rotation passing through point A, perpendicular to the page.

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

!Lbullet , with respect to point A? What is the direction of

!Lbullet ? Directions (+x, +y, +z, -

x, -y, -z) are as indicated in the above figure. b) After the collision calculate the moment of inertia of the system (bullet+block+rod) about an axis through point A, perpendicular to the page. Treat the block and bullet as point particles. HINT: Use Itotal = Irod +mass1 ! distance( )2 +mass2 ! distance( )2 .

c) Use the appropriate conservation law to calculate the angular speed ! final of the bullet+block+rod about axis through point A, after the collision. What is the direction and the angular velocity,

!! final ? Use same direction definition as for part a).

d) Using result of part c calculate the magnitude and direction of the angular momentum of the bullet+block+rod about point A. Compare this to the answer of part a) and make a one or two-sentence comment.

+y

+x

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

13

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

2

Also !Emech = !U + !K = U f "Ui( ) + K f " Ki( ) = 0# !K = "!U

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

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)#t2 ,! 2 =!0

2 + 2" # $#0( ) Linear and angular variables: s = r! , v = r! , atan = R! (tangential) ,arad = v

2 / 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.