Physics 110

Fall 2009

Exam #1

September 30, 2009

Name______________

In keeping with the Union College policy on academic honesty, it is assumed that you will neither accept nor provide unauthorized assistance in the completion of this work.

Part Multiple Choice / 10

Problem #1 /45 Problem #2 / 45

Total / 100

Part I: Free Response Problems Please show all work in order to receive partial credit. If your solutions are illegible

no credit will be given. Please use the back of the page if necessary, but number the problem you are working on. Each subpart of a problem is worth 9 points.

1. The 89th Airlift Wing, otherwise known as the Presidential Airlift Wing, is

responsible for flying the President of the United States to all points of the world. Suppose, as a week or so ago, that Air Force One (the code name of the plane when the President is aboard) is on the runway awaiting takeoff clearance from the Albany International Airport tower. Once the plane receives clearance the pilots apply full power to the 4 engines and this accelerates the aircraft, from rest, down the runway. Air Force One needs to be traveling at a minimum speed of 80 m/s (~ 108 mi/hr) before it can become airborne and this takes approximately 50s.

a. What minimum acceleration is needed to get Air Force One to 80 m/s?

€

v fx = vix + axt→ ax =v fxt

=80 m

s

50s=1.6 m

s2 or 1.6 m/s2 in the +x-direction.

b. Albany International Airport has two runways, a main runway (called runway

1/19) and a crosswind runway (called runway 10/28). Runway 1/19 is 8500 feet long (2592m) and points approximately due north (10o) to due south (190o). Will Air force One be able to become airborne in this time? That is, what is the total displacement of Air Force One and is this displacement less than 2592m?

€

v fx2 = vix

2 + 2axΔx→Δx =v fx2

2ax=80 m

s( )2

2 ×1.6 ms2

= 2000m which is greater than 2592m so

yes it makes it off of the runway.

c. Air Force One is a Boeing 747-200 model aircraft with 4 General Electric engines with a total mass (fuel + aircraft) of 400,000 kg. The total thrust from all 4 engines is 1500 kN and this is what propels the aircraft forward. When the aircraft wants to take-off, the pilot rotates the aircraft’s nose skyward at an angle of 12o (with respect to the horizontal) and the plane climbs away from the ground at this angle. Ignoring the rotational motion during lift-off and suppose that the acceleration of the aircraft in flight is the same magnitude as in part a. (If you could not determine a value for the acceleration, assume a value of 2 m/s2.) In addition, there are several forces that act on an aircraft in flight. The thrust, Fthrust, acts parallel to the direction of flight while the drag (FD – due to friction) opposes this motion. Perpendicular to the wings is a force called lift FL, which makes the aircraft rise. On the picture at the top of the next page, carefully label the forces given above using the coordinate system shown on the picture.

d. What is the magnitude of the lifting force, FL?

€

Fy : FL − FW cosθ = may = 0→∑ FL = FW cosθ = 4 ×105kg × 9.8 ms2 × cos12 = 3.8 ×106N

e. What is the magnitude of the drag force, FD?

€

Fx : Fthrust − FD − Fw sinθ = ma→∑ FD = Fthrust − Fw sinθ −ma =1500 ×103N − 4 ×105kg × 9.8 ms2( )sin12 − 4 ×105kg×1.6 m

s2( ) = 4.5 ×104N

y

x

FL

FThrust FD

FW

2. A tennis ball is loaded into a launcher and is fired at an angle θ, with respect to the horizontal, at a building. The ball leaves the launcher at vi = 20 m/s at the unknown angle θ. The ball eventually hits the building at a horizontal distance of xf = 6.1m and yf = 15.2m vertically from the launch point.

a. What is the minimum angle that the projectile is launched from? (Hint, you will

need to solve a quadratic equation that involves angles and sin2θ + cos2θ = 1.)

€

x f = vixt = vi cosθ( )t→ t =x f

vi cosθ

y f = viyt − 12 gt

2 = vi sinθ( )t − 12 gt

2 = x f tanθ −gx f

2

2vi2 cos2θ

= x f tanθ −gx f

2

2vi2 tan

2θ +1( )

∴0 = y f +gx f

2

2vi2

− x f tanθ +

gx f2

2vi2 tan

2θ → 0 =15.7 − 6.1tanθ + 0.46tan2θ

tanθ =9.83.5

→θ =

84o

74o

b. What is the time of flight of the projectile from launch to hitting the building?

€

t =x f

vi cosθ=

6.1m20 m

s cos74=1.1s

c. Will the projectile be rising or falling when it hits the building? Justify your

answer with calculations.

€

v fy = viy − gtrise → trise =viyg

=vi sinθg

=20 m

s sin749.8 m

s2=1.96s Since the time of flight is less

than the time to get to maximum height, the projectile strikes the building while rising.

d. What will be the projectile’s impact velocity?

€

v fx = vix = vi cosθ = 20 ms cos74 = 5.5 m

s

v fy = viy − gt = vi sinθ − gt = 20 ms sin74 − 9.8 m

s2×1.1s = 8.4 m

s

∴v f = v fx2 + v fy

2 @θ = tan−1v fyv fx

→ v f =10.0 m

s@θ = 56.8o

e. The ball will collide with the wall at the velocity in part d. Suppose that the ball

bounces off of the wall with the same magnitude and direction of the impact velocity as it strikes the wall, as shown in the diagram. What is the average horizontal force on the projectile from the wall if the ball has a mass of 0.2kg and is in contact with the wall for 0.2s? (If you cannot get a value in part d, assume that the impact velocity is 15 m/s at 45o.)

€

F = m ΔvΔt

= m−2vimpact cosθ

Δt

= 0.2kg ×

−2 ×10 ms × cos56.80.2s

= −11N

vimpact

vimpact

θ

θ

Part II: Multiple-Choice Circle the best answer to each question. Any other marks will not be given credit. Each multiple-choice question is worth 2 points for a total of 10 points.

1. Assume that you weigh 2000 N and that you are falling down through the air (in the

presence of air resistance) at constant speed, the force of the air against your body is a. zero. b. greater than 0 but less than 2000 N. c. 2000 N. d. more than 2000N.

2. Suppose that you unfortunately drop a brick on your toe (OUCH!). The greater force

of impact will be

a. on the brick. b. on your toe. c. same force on both. d. unable to be determined with the information given.

3. Suppose that an object travels in the negative x-direction in three steps. In the first

step the object covers 100m in 12 seconds then the second step has the object travel at a constant velocity of 12 m/s for 24 seconds and third step has the object come to rest after traveling a further 250m in 10s. For the entire trip, the average acceleration of the object is

a. -4.7 m/s2 b. 0 m/s2 c. 9.8 m/s2 d. 13.9 m/s2

4. Suppose that a 10,000kg block is sitting on a horizontal frictionless surface. A

0.01kg washer is connected to the block by a string passing over a pulley and is attached to the block. If the washer is released from rest, the acceleration of the block is a. equal to 0 m/s2 b. 0 m/s2< a < 9.8 m/s2 c. greater than 9.8 m/s2 d. cannot tell from the information given.

5. Two marbles sitting on a tabletop, are flicked off, one just falling vertically (marble A) and the other shot out horizontally off the table (marble B). Compared to marble B, the vertical component of marble A’s impact velocity is a. the same. b. greater. c. less. d. unable to be determined.

Useful formulas:

€

g = 9.8m s2 G = 6.67×10−11 Nm 2

kg2

NA = 6.02×1023 atomsmole kB = 1.38×10−23 J Kσ = 5.67×10−8 W m 2K 4 vsound = 343m s

Motion in the r = x, y or z-directions Uniform Circular Motion Geometry /Algebra

Vectors Useful Constants

Linear Momentum/Forces Work/Energy Heat

Rotational Motion Fluids Simple Harmonic Motion/Waves Sound