exam 2 phys 1600: engineering physics 1 (sec....

Exam 2 – Version 1 1

EXAM 2 PHYS 1600: Engineering Physics 1 (Sec. 021-026)

October 3, 2014 – Fall Semester, 2014 SOLUTIONS

Multiple choice V1 V2 1 d b 2 b e 3 d c 4 c b 5 a d

Part 1: Problems 1 through 5 (6 pts each – 30 pts. total) Multiple Choice Problems Circle your answer AND write your choice (a-e) on the front page 1) A ball is thrown horizontally off of a tall building at 6 m/s. It takes a total of 1.6 seconds to hit the ground. The acceleration vector after it has fallen for ½ of the total time is: ANSWER: 0.0 m/s2 i – 9.8 m/s2 j 2) Two ropes are used to pull a 4 kg object across a smooth, flat surface. The two forces exerted by the ropes are: F1 = 4 i + 5 j [N] and F2 = 3 i – 2 j [N]. The magnitude of the net force is: Sum over x- and y-components: Fnet-x = F1x + F2x and Fnet-y = F1y + F2y Magnitude is then: Fnet = [Fnet-x

2 + Fnet-y2]1/2

3) Consider the system shown below. The mass “a” is pulling masses “b” and “c” along a rough (with friction) inclined plane. How many forces are acting on mass “b”?

Mass b: 5 forces Mass c: 4 forces

a" b

c"


4) A TARDIS, lifted by a cable, is moving downward and slowing. Which is the correct free-body diagram? The arrows represent the tension in the cable (upward arrow) and weight of the object (downward arrow).

(a) (b) (c) (d) 5) A set of keys on the end of a string is swung steadily in a horizontal circle. In one trial, it moves at speed v in a circle of radius r. In a second trial, it moves at a slower speed v/2 at a smaller radius r/2. In the second trial, how does the acceleration of the keys compare to the first trial? (a) ½ as large (b) 2 times larger (c) 4 times larger (d) ¼ as large (e) the same


[For all of the remaining problems, show your algebraic steps towards a solutions.] Problems 6 to 8: Short Answer Problems: (10 pts each) 6) A 100 N box is being pulled with a force of 45 N by a rope along a frictionless, flat surface. The rope is at an angle of θ = 30˚ above the horizontal. The normal force on the box is:

Fy∑ = may = 0 = N +T sinθ −mg

N = mg −T sinθ

7) A 500 kg car has an engine that can produce a force of 800 N forward. At some instant, while the car is traveling at 12 m/s along a straight road, the car has experiences a resistive force (due to drag, friction, etc.) of 950 N. What is the speed of the car after 15 seconds?

One-dimensional system:Fnet = ma = Fengine − Fresistive

"# $%

Solve for the acceleration:

a =Fengine − Fresistive"# $%

mThen, use the acceleration, initial velocity and time to get the final speed:v f = vi + at

8) A girl traveling in a car is moving at 6 m/s in the +x-direction. A boy is traveling in another car at 7 m/s in the –x-direction. The boy throws a ball behind himself at 2 m/s. What is the speed of the ball relative to the girl? All in units of [m/s]

V1 V2

v-‐girl/road 6 4 v-‐boy/road -‐7 -‐5 v-‐ball/boy 2 2 v-‐ball/road = v-‐boy/road+v-‐ball-‐boy -‐5 -‐3 v-‐ball/girl 11 7

T"θ"

V1 V2

T [N] 45 60 W [N] 100 85 θ 30 30 N=W-‐Tsin(θ) 77.5 55

V1 V2

m [kg] 500 400 F-‐forward [N] 800 600 F-‐resist [N] -‐950 -‐800 a=(Fnet/m) [m/s2] -‐0.3 -‐0.5 t [s] 15 15 vi [m/s] 12 10 vf = vi+a*t [m/s] 7.5 2.5


Extended Problems: These problems at 20 points each. You need to choose 2 out of the 3 problems. Indicate which two problems you want to have graded using the box AND on the front of the exam. Scoring: invalid or no units [-1/each]; no calculus/algebraic work shown [-3] Extended Problem 1: [Parabolic trajectory] A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a cliff. The quick stop causes a number of watermelons to fly off the truck. One melon leaves the hood of the truck with an initial speed vi = 12 m/s in the horizontal direction. The cross section of the bank has the shave of the bottom half of a parabola, with its vertex at the initial location of the projected watermelon and with the equation y2 = 10x, where x and y are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank? [horizontal components – 4, vertical components – 4, connecting equations – 8, solution – 4]

Horizontal

x f = xi + vixt +12

axt2 [xi ,ax → 0]

x = vixt

Vertical

y f = yi + yixt +12

ayt2 [yi ,viy → 0; ay →−g]

y = − 12

gt2 = −12

gt2

Where: y2 =Cx [C is the coefficient in the problem]

y2 =Cx ⇒C vixt( ) = 14

g 2t4 ⇒C vix( ) = 14

g 2t3

Therefore, one can solve for the time: t =4Cvix

g 2

$

%&

'

()

1/3

And, then get the x and y positions using: x = vixt and y = − 12

gt2

V1 V2

Horizontal xi [m] 0 0

xf [m] ? ? vix [m/s] 12 10 vfx [m/s] 12 10 ax [m/s2] 0 0

Vertical yi [m] 0 0

yf [m] ? ? viy [m/s] 0 0 vfy [m/s] ? ? ay = -‐g [m/s2] -‐9.8 -‐9.8

Coeff 10 8 T [sec] 1.71 1.49 X [m] 20.5 14.9 Y [m] -‐14.3 -‐10.9


Extended Problem 2: [Two moving objects] Two blocks connected by a rope of negligible mass are being dragged by a force, F = 60 N at an angle of 37˚ above the horizontal. Let m1 = 15 kg and m2 = 6 kg and there is a coefficient of kinetic friction of 0.10 between each block and the surface. a) Make free body diagrams for each mass. [6 pts] b) Find force equations for each object. [6 pts] c) Find the tension in the rope. [4 pts] d) If the blocks have an initial velocity of 2 m/s in the +x-direction, what is the magnitude of the velocity of the system after 3 seconds. [4 pts]

Mass 2:

Fy = m2ay = 0 = N2 + F sinθ −m2g ⇒ N2 =∑ m2g − F sinθ

Fx = m2ax = F cosθ −T − f f 1 = F cosθ −T − µk N2 = F cosθ −T − µk m2g − F sinθ( )∑

Mass 1:

Fy = m1ay = 0 = N1 −m1g ⇒ N1 =∑ m1g

Fx = m1ax =T − f f 2 =T − µk N1 =T − µkm1g∑ ⇒ T = m1ax + µkm1g

Combine these two results:

m2ax = F cosθ −T − µk m2g − F sinθ( ) = F cosθ − m1ax + µkm1g( )− µk m2g − F sinθ( )m1ax +m2ax = F cosθ − µkm1g − µkm2g + µk F sinθ

a x=F cosθ+ µk sinθ( )− µk g m1+m2( )

m1+m2

Finally: v fx = vix + axt

Solve for acceleration, then for T, and then for the final velocity after 3 seconds.

N2#

T#

ff1#

m2g#

N1#

m1g#

ff2#

T#


V1 V2

m1 [kg] 15 12 m2 [kg] 6 8 angle 37 53 F [N] 60 60 µk 0.1 0.1 ax [m/s2] 1.47 1.07 Tension [N] – mass1 36.81 24.54 Tension(alt) – mass2 36.81 24.54 vix [m/s] 2 2 time [s] 3 3 vfx [m/s] 6.42 5.20 * Note, the two tension calculation are from:

Mass 1:T = m1ax + µkm1g

Mass 2: [alternative calculation]T = F cosθ −m2ax − µk m2g − F sinθ( )


Extended Problem 3: [Two dimensional motion in vector format] One side of the roof of a house slopes up at 37˚. A roofer kicks a round, flat rock that is on the roof. The rock slides straight up the incline with an initial speed of 16 m/s. The coefficient of kinetic friction between the rock and the roof is 0.30. The rock slides 10 m up the roof to its peak. It crosses the ridge and goes into projectile motion above the far side of the roof with negligible air resistance. Determine the maximum height of the rock above the point where it was kicked. [Drawing – 4; Motion along incline – 10; Motion in air – 6]

Fy = may = 0 = N −mg cosθ ⇒ N =∑ mg cosθ

Fx = max = −mg sinθ − f fk = −mg sinθ − µk N = −mg sinθ − µkmg cosθ∑⇒ ax = −g sinθ + µk cosθ( )

Use this acceleration to get the speed of the rock at the top of the roof:

vroof − f

2 = vroof −i2 +2a ∆ x( ) ⇒ vroof − f = vroof −i

2 +2a ∆ x( )

With: vroof-i – given and ∆x = 10 m. Determine the y-component of the velocity at the top of the roof in order to work out the maximum vertical height of the rock. vyi = vroof-f sin(θ) Then, maximum height of the rock above the top of the roof will be obtained from the vertical component of the motion:

θ" N"

mg"f&"


vyf2 = vyi

2 +2ay ∆ y - where: v fy = 0 m/s and ay = −g

∆ y =vyi

2

2g - this is the maximum height of the rock above the top of the roof

The height of the top of the room from the initial position of the rock is: H = Lroof sin(θ) So: total height is: h = H + ∆y

V1 V2

vix [m/s] 16 13 µk 0.3 0.2 Angle, θ 37 30 ax [m/s2] -‐8.25 -‐6.60 v-‐roof [m/s] 9.54 6.09 For projectile motion

vix = vroof cos(θ) [m/s] 7.62 5.27 viy = vroof sin(θ) [m/s] 5.74 3.04 ∆y (from top of roof) [m] 1.68 0.47 Determine final height Lroof (length of roof) [m] 10 10 H (height of roof) [m] 6.02 5.00 h = H + ∆y [m] 7.70 5.47

exam 2 phys 1600: engineering physics 1 (sec....

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