midterm exam 1 - siena collegemmccolgan/gp130f12/exams/mid… · · 2012-10-09midterm exam 1...

Physics 130 General Physics - McColgan Fall 2012

Midterm Exam 1October 2, 2012

Name:

Instructions

1. This examination is closed book and closed notes. All your belongingsexcept a pen or pencil and a calculator should be put away and yourbookbag should be placed on the floor.

2. You will find one page of useful formulae on the last page of the exam.

3. Please show all your work in the space provided on each page. If youneed more space, feel free to use the back side of each page.

4. Academic dishonesty (i.e., copying or cheating in any way) willresult in a zero for the exam, and may cause you to fail theclass.

In order to receive maximum credit,each solution should have:

1. A labeled picture or diagram, if appropriate.2. A list of given variables.3. A list of the unknown quantities (i.e., what you are solving for).4. Graphical representations of the motion (e.g., position, velocity, acceler-

ation vs. time), if appropriate.5. A labeled 1D or 2D coordinate axis system, if appropriate.6. The full equation or equations needed to solve the problem

(e.g., appropriate equations of motion).7. An algebraic solution of the unknown variables in terms of the

known variables.8. A final numerical solution, including units, with a box around it.9. An answer to additional questions posed in the problem, if any.

1


1. One game at the amusement park has you push a puck up a long, frictionless ramp. Youwin a stuffed animal if the puck, at its highest point, comes to within 10 cm of the endof the ramp without going off. You give the puck a push, releasing it with a speed of6.0 m{s when it is 9 m from the end of the ramp. The puck’s speed after traveling 2.5 mis 5.0 m{s. Are you a winner?

Solution:

This is a 1D kinematics question. We are given the puck’s initial velocity vDi, andit’s velocity after it has traveled 2.5 m. We are also given the length of the ramp,xf , and we are asked to find whether the puck stops within 10 cm of the end of theramp. To solve this problem we need the 1D kinematic equation relating velocities,distance, and acceleration. We need to find the acceleration from the two velocitiesand the two speeds at the start of the puck’s trip up the slide:

v2f � v2i � 2apxf � xiq (1)

2apxf � xiq � v2i � v2f (2)

a � v2i � v2f2pxf � xiq (3)

a � p6 m{sq2 � p5 m{sq22p2.5q (4)

a � �2.2 m{s2 (5)

(6)

We can use the same equation again, using the starting and ending velocity and theacceleration that we just found, where vf � 0:

v2f � v2i � 2apxf � xiq (7)

2


0 � v2i � 2apxf � xiq (8)

2apxf � xiq � v2i (9)

pxf � xiq � v2i2a

(10)

� p6 m{sq22p2.2 m{s2q (11)

� 8.2 m (12)

(13)

Unfortunately, you do not win! The puck stops well before the winning distance. Towin, the puck has to stop at a distance between 8.9 and 9 m.

3


2. As a science project, you drop a watermelon off the top of the Empire State Building,350 m above the sidewalk. It so happens that Superman flies by at the instant yourelease the watermelon. Superman is headed straight down with a constant speed of40 m{s. How fast is the watermelon going when it passes Superman?

Solution:

This is a 1D kinematics question. We are given the watermelon and Superman’sstarting position at 350 m and Superman’s constant velocity of 40 m{s. We alsoknow that the acceleration of gravity is 9.8m{s2.To solve this problem we need use the 1D kinematic equations for the watermelonand Superman:

xSf � xSi � vSi∆t� 1

2aSp∆tq2 � xSi � vSi∆t, (14)

xWf � xWi � vWi∆t� 1

2aW p∆tq2 � xWi � 1

2aW p∆tq2. (15)

We’re trying to find the speed of the watermelon just as it passes Superman, or whenthe xSf � xWf . Setting these two equations equal to each other, we have:

xSi � vSi∆t � xWi � 1

2aW p∆tq2 (16)

(17)

Since the initial positions are the same, xSi � xWi, and this equation becomes:

vSi∆t � 1

2aW p∆tq2 (18)

(19)

4


We can factor ∆t, and substitute aW � g, and the equation simplifies to:

vSi � 1

2g∆t (20)

(21)

Solving for the time, ∆t:

∆t � 2vSig

� 2p�40 mq�9.8 m{s2 � 8.2 sec (22)

(23)

To find the velocity of the watermelon:

vf � vi � a∆t (24)

vf � �g∆t (25)

vf � �9.8 m{s2p8.2 secq � �80.4 m{s (26)

(27)

5


3. Let ~A � 9ı̂� 3̂, ~B � �7ı̂� 5̂, and ~D � 3 ~A� 2 ~B.

(a) Write vector ~D in component form.

(b) Draw a coordinate system and on it show vectors ~A, ~B, and ~D.

(c) What are the magnitude and direction of vector ~D?

Solution:

(a) To write ~D in component form we first multiply ~A by 3 and ~B by 2.

3 ~A � 3 � 9ı̂� 3 � 3̂ � 27ı̂� 9̂ (28)

2 ~B � 2 � p�7q̂ı� 2 � 5̂ � �14ı̂� 10̂ (29)

(30)

Next we add the x- and y-components:

~D � 3 ~A� 2 ~B (31)

� p27ı̂� 9̂q � p�14ı̂� 10̂q (32)

� p27 � 14q̂ı� p�9 � 10q̂ (33)

� 41ı̂� 19̂. (34)

(b)

(c) The magnitude of the vector ~D is given by

| ~D| �a

412 � p�19q2 (35)

� 45.2, (36)

6


and the direction θ (measured positive counter-clockwise from the x-axis) is equal to

θ � tan�1

�Dy

Dx

� tan�1

�19

41

� �24.9�or 335.1�. (37)

7


4. The figure shows three ropes tied together in a knot. One of your friends pulls on arope with 3.0 units of force and another pulls on a second rope with 5.0 units of force.How hard and in what direction must you pull on the third rope to keep the knot frommoving?

Solution:

This is a vector problem that uses x- and y- components of vectors. The three vectorsmust be balanced. This means that the x- and y - components of the three vectorsmust be balanced.

¸Fx � 0 (38)¸Fy � 0 (39)

(40)

The 3 unit force vector is only in the x-direction. The 5 unit force vector has x-and y- components. The x-component of the 5 unit force vector is found from SOH-CAH-TOA:

cos θ � Fx

5(41)

Fx � 5 cos θ � 5 cosp60q � 2.5 units (42)

sin θ � Fy

5(43)

Fy � 5 sin θ � 5 sinp60q � 4.33 units (44)

(45)

8


Now we add the x-components:¸Fx � �2.5 � 3 � 0.5 (46)

(47)

For the unknown vector to balance the other forces on the knot,

Fx � �0.5 (48)

Fy � �4.33 (49)

(50)

To find the magnitude of the force,

| ~D| �ap�.5q2 � p4.33q2 (51)

� 4.36 units (52)

and the direction θ (measured positive counter-clockwise from the x-axis) is equal to

θ � tan�1

�Dy

Dx

� tan�1

�4.33

.5

� 83.4� S of W or θ � 263.4�. (53)

9


5. A stunt-man drives a car at a speed of 35 m{s off a 25 m high cliff. The road leading tothe cliff is inclined upward at an angle of 15�.

(a) How far from the base of the cliff does the car land?

(b) What is the car’s impact speed?

Solution:

This is a 2D projectile motion problem. We are given the initial height of the carabove the ground, yi � 25 m, its initial velocity, vi � 35.0 m{s, and the angle abovethe horizontal with which the car leaves the cliff, θ � 15�. We are asked about thefinal distance from the base of the cliff and the car’s impact speed.

a) To solve the problem we first determine the initial velocity of the ball in thehorizontal (x) and vertical (y) dimensions:

vxi � vi cos θ � p35 m{sq � cos p15�q � 33.8 m{s, (54)

vyi � vi sin θ � p35 m{sq � sin p15�q � 9.1 m{s. (55)

Next, we use the vertical kinematic equation of projectile motion to find the time ittakes the car to hit the ground, resulting in yf � 0:

yf � 0 � yi � vyi∆t� 1

2gp∆tq2, (56)

1

2gp∆tq2 � vyi∆t� yi � 0, (57)

4.9∆t2 � 9.1∆t� 25 � 0 (58)

(59)

10


Using the quadratic equation and substituting in values,

∆t � �b�?b2 � 4ac

2a(60)

∆t � 9.1 �ap�9.1q2 � 4p4.9qp�25q2p4.9q (61)

∆t � 3.4 sec (62)

(63)

Next, we use the horizontal kinematic equation of projectile motion to find thedistance the car lands from the bottom of the cliff:

xf � xi � vxi∆t (64)

� vxi∆t (65)

� p33.8 m{sqp3.4 secq (66)

� 114 m (67)

(68)

b)To find the impact speed, we need to find the final x- and y- components of velocity.The x-component of the final velocity is:

vfx � vix � 33.8 m{s (69)

(70)

The y-component of the final velocity is:

vfy � viy � g∆t � 9.1 m{s � p9.8 m{s2qp3.4 secq (71)

� �24.2 m{s (72)

(73)

To find the magnitude of the velocity or the impact speed,

|~v| �ap33.8q2 � p�24.2q2 (74)

� 41.6 m{s (75)

(76)

11


6. You are watching an archery tournament when you start wondering how fast an arrowis shot from the bow. Remembering your physics, you ask one of the archers to shootan arrow parallel to the ground. You find the arrow stuck in the ground 50 m away,making a 2� angle with the ground. How fast was the arrow shot?

Solution:

This is a 2D projectile motion problem. We are told that the arrow is pointed in thehorizontal direction initially, that the range, or horizontal distance the arrow travelsis xf � 50 m, and that the angle at which the arrow hits the ground with respect tothe horizontal is θ � 15�. We are asked to find the initial speed of the arrow.

a) To start the problem, let’s use the kinematic equation for x where the initialposition, xi � 0:

xf � xi � vix∆t � vix∆t (77)

(78)

Then, let’s use the kinematic equation for y where the initial velocity viy � 0:

vyf � vyi � g∆t � �g∆t (79)

(80)

Since the velocity in the x-direction is a constant, let’s set vix � vfx � vx.

Now, use the angle at which the arrow hits the ground to relate vx and vy.

tan θ � vyvx

(81)

vx � vytan θ

(82)

(83)

12


From the equation above, we know that vy � �g∆t, which gives

vx � vytan θ

� �g∆t

tan θ(84)

(85)

Plugging this into the equation for xf , we get

xf � vx∆t � �g∆t

tan θ∆t (86)

� �g∆t2

tan θ(87)

(88)

Find the time by rearranging the equation to get

∆t2 � xf tan θ

g(89)

∆t �dxf tan θ

g∆t �

dp50 mq tanp2q

9.8 m{s2 ∆t � 0.4 sec (90)

To find the initial velocity,

vx � vytan θ

� g∆t

tan θ(91)

vx � vytan θ

� p9.8 m{s2qp0.4 secqtanp2q (92)

vx � 112 m{s (93)

(94)

Note: We drop the negative sign on g because we’re keeping track of the positiveand negative values, and we know that vx is positive. The negative in the equationwould result from the fact that the velocity is in the negative y-direction and theangle is in the 4th quadrant.

13


7. As the earth rotates, what is the speed of (a) a physics student in Miami, Florida, atlatitude 25�, and (b) a physics student in Fairbanks, Alaska, at latitude 66�? Ignore therevolution of the earth around the sun. The radius of the earth is 6390 km.

Solution:

This is a uniform angular acceleration problem in which we are given the radius ofthe earth, re � 6390 km, and the latitude (angle θ with respect to the equator) ofeach physics student.

(a) From the geometry of the problem, the linear distance of the student from therotational axis of the earth is

r � re sin p90 � θq (95)

� p6390 kmq � sinp65�q (96)

� 5791 km. (97)

The speed of the student in Miami is just equal to the circumference divided by theperiod, T , which is just one day:

v � 2πr

T(98)

��

2π � 5791 km

1 day

��

1 day

24 � 3600 sec

(99)

� 0.421 km s�1 (100)

� 421 m{s. (101)

(b) This solution proceeds just like above but with θ � 66�. In this case the radiusis 2600 km, and v � 0.189 m{s � 189 km s�1.

14


8. A rock stuck in the tread of a 56.0 cm diameter bicycle wheel has a tangential speed of2.80 m{s. When the brakes are applied, the rock’s tangential deceleration is 1.30 m{s2.(a) What are the magnitudes of the rock’s angular velocity and angular acceleration at

t � 1.70 s?

(b) At what time is the magnitude of the rock’s acceleration equal to g?

Solution:

This is a non-uniform angular acceleration problem. We are given the size of thewheel, r � 28 cm � 0.28 m, the initial tangential speed of the rock, vi � 2.80 m{s, andthe tangential deceleration of the rock when the brakes are applied, aT � 1.30 m{s2.

(a) In order to determine the final angular velocity, ωf of the rock after t � 1.70 sec,we first have to determine the initial angular velocity:

wi � vir� 2.8 m{s

0.28 m� 10 rad{sec. (102)

The angular acceleration is a constant and given by:

α � aTr� �1.30 m{s2

0.28 m� �4.64 rad{sec2. (103)

Finally, using the appropriate rotational kinematic equation and substituting, weobtain

ωf � ωi � α∆t (104)

� 10 rad{sec � p4.64 rad{sec2q � p1.7 secq (105)

� 2.11 rad{sec. (106)

15


(b) In order to find the time at which the magnitude of the angular acceleration ofthe rock equals g, we first need to find the linear velocity of the rock at that time.We want |~a| � g �

aa2T � a2r, where aT is given in the problem and ar � v2{r.

Substituting and solving for v we get

g �da2T �

�v2

r

2

(107)

g2 � a2T �v4

r2(108)

ñ v4 � r2pg2 � a2Tq (109)

v � r1{2pg2 � a2Tq1{4 (110)

� p0.28 mq1{2 � rp9.8 m{s2q2 � p1.30 m{s2q2s1{4 (111)

� 1.65 m{s. (112)

To find the time at which the rock reaches this velocity, we use the 1D kinematicequation for velocity and solve for the time:

vf � vi � aT∆t (113)

ñ ∆t � vi � vfaT

(114)

� 2.80 m{s � 1.65 m{s1.30 m{s2 (115)

� 0.88 sec. (116)

16


Kinematics and Mechanics

xf � xi � vxi∆t� 1

2axp∆tq2

vxf � vxi � ax∆t

v2xf � v2xi � 2axpxf � xiq

yf � yi � vyit� 1

2ayt

2

vyf � vyi � ayt

v2yf � v2yi � 2aypyf � yiq

θf � θi � ωi∆t� 1

2α∆t2

ωf � ωi � α∆t

ωf2 � ωi

2 � 2α∆θ

s � rθ

c � 2πr

v � 2πr

Tvt � ωr

ar � v2

r� ω2r

at � rα

17

midterm exam 1 - siena collegemmccolgan/gp130f12/exams/mid… · · 2012-10-09midterm exam 1...

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