Midterm Exam 1 October 2, 2012

Midterm Exam 1

October 2, 2012

Name:

Instructions

1. This examination is closed book and closed notes. All your belongings except a pen or pencil and a calculator should be put away and your bookbag 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 you

need more space, feel free to use the back side of each page.

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

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. One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its highest point, comes to within 10 cm of the end of the ramp without going off. You give the puck a push, releasing it with a speed of 6.0 m{s when it is 9 m from the end of the ramp. The puck’s speed after traveling 2.5 m is 5.0 m{s. Are you a winner?

Solution:

This is a 1D kinematics question. We are given the puck’s initial velocity vDi, and it’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 the ramp. To solve this problem we need the 1D kinematic equation relating velocities, distance, and acceleration. We need to find the acceleration from the two velocities and the two speeds at the start of the puck’s trip up the slide:

v2_f v_i22apxf xiq (1)

2apxf xiq vi2v

2

f (2)

a v

2

i vf2 2pxf xiq

(3)

a p6 m{sq

2 _{p5 m{sq}2

2p2.5q (4)

a 2.2 m{s2 (5)

(6) We can use the same equation again, using the starting and ending velocity and the acceleration that we just found, where vf 0:

0 v_i22apxf xiq (8)

2apxf xiq vi2 (9)

pxf xiq

v2

i

2a (10)

p6 m{sq2

2p2.2 m{s2q (11)

8.2 m (12)

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

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 you release the watermelon. Superman is headed straight down with a constant speed of 40 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’s starting position at 350 m and Superman’s constant velocity of 40 m{s. We also know that the acceleration of gravity is 9.8m{s2_.

To solve this problem we need use the 1D kinematic equations for the watermelon and Superman:

xSf xSi vSi∆t 1

2aSp∆tq

2

xSi vSi∆t, (14)

xW f xW i vW i∆t 1

2aWp∆tq

2

xW i 1

2aWp∆tq

2

. (15)

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

xSi vSi∆t xW i 1

2aWp∆tq

2 ₍₁₆₎

(17) Since the initial positions are the same, xSi xW i, and this equation becomes:

vSi∆t 1

2aWp∆tq

2

(18) (19)

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 2vSi

g

2p40 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)

3. Let A~ 9ˆı3ˆ, B~ 7ˆı 5ˆ, and D~ 3A~2B~. (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 multiplyA~ by 3 and B~ by 2.

3A~ 39ˆı33ˆ27ˆı9ˆ (28) 2B~ 2 p7qˆı 25ˆ 14ˆı 10ˆ (29) (30) Next we add the x- and y-components:

~

D3A~2B~ (31)

p27ˆı9ˆq p14ˆı 10ˆq (32) p27 14qˆı p910qˆ (33)

41ˆı19ˆ. (34)

(b)

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

|_D~_| a₄₁2 p₁₉q2 ₍₃₅₎

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

θtan1

Dy

Dx

tan1

19 41

24.9or 335.1. (37)

4. The figure shows three ropes tied together in a knot. One of your friends pulls on a rope 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 from moving?

Solution:

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

¸

Fx 0 (38)

¸

Fy 0 (39)

(40) The 3 unit force vector is only in the 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:

cosθ Fx

5 (41)

Fx 5 cosθ 5 cosp60q 2.5 units (42) sinθ Fy

5 (43)

Fy 5 sinθ 5 sinp60q 4.33 units (44) (45)

Now we add the x-components: ¸

Fx 2.5 30.5 (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~_| a_p.5q2 p_4.33q2 ₍₅₁₎

4.36 units (52)

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

θ tan1

Dy

Dx

tan1

4.33

.5

83.4 S of W or θ263.4. (53)

5. A stunt-man drives a car at a speed of 35 m{s off a 25 m high cliff. The road leading to the 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 car above the ground, yi 25 m, its initial velocity,vi 35.0 m{s, and the angle above the horizontal with which the car leaves the cliff, θ 15. We are asked about the final 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 the horizontal (x) and vertical (y) dimensions:

vxi vicosθ p35 m{sq cosp15q 33.8 m{s, (54)

vyi visinθ p35 m{sq sinp15q 9.1 m{s. (55) Next, we use the vertical kinematic equation of projectile motion to find the time it takes the car to hit the ground, resulting in yf 0:

yf 0yi vyi∆t 1 2gp∆tq

2_{, (56)}

1 2gp∆tq

2_v

yi∆tyi 0, (57)

4.9∆t29.1∆t250 (58)

Using the quadratic equation and substituting in values, ∆t b

?

b2_4ac

2a (60)

∆t 9.1

a

p9.1q2₄p_4.9qp₂₅q

2p4.9q (61)

∆t3.4 sec (62)

(63) Next, we use the horizontal kinematic equation of projectile motion to find the distance 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:

vf xvix 33.8 m{s (69)

(70) The y-component of the final velocity is:

vf y viyg∆t9.1 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)

6. You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an 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 the horizontal direction initially, that the range, or horizontal distance the arrow travels is xf 50 m, and that the angle at which the arrow hits the ground with respect to the horizontal is θ15. 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 initial position, xi 0:

xf xi vix∆tvix∆t (77)

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

vyf vyig∆t g∆t (79)

(80) Since the velocity in the x-direction is a constant, let’s set vix vf xvx.

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

tanθ vy vx

(81)

vx

vy

tanθ (82)

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

vx

vy tanθ

g∆t

tanθ (84)

(85) Plugging this into the equation for xf, we get

xf vx∆t

g∆t

tanθ ∆t (86)

g∆t2

tanθ (87)

(88) Find the time by rearranging the equation to get

∆t2 xftanθ

g (89)

∆t

d

xftanθ

g ∆t

d

p50 mqtanp2q

9.8 m{s2 ∆t0.4 sec (90) To find the initial velocity,

vx

vy tanθ

g∆t

tanθ (91)

vx

vy tanθ

p9.8 m{s2_{qp0.4 secq}

tanp2q (92)

vx 112 m{s (93)

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

7. As the earth rotates, what is the speed of (a) a physics student in Miami, Florida, at latitude 25, and (b) a physics student in Fairbanks, Alaska, at latitude 66? Ignore the revolution 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 of the earth, re 6390 km, and the latitude (angle θ with respect to the equator) of each physics student.

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

r resinp90θq (95)

p6390 kmq sinp65q (96)

5791 km. (97)

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

v 2πr

T (98)

2π5791 km 1 day

1 day 243600 sec

(99)

0.421 km s1 ₍₁₀₀₎

421 m{s. (101)

(b) This solution proceeds just like above but with θ 66. In this case the radius is 2600 km, and v 0.189 m{s189 km s1_.

2.80 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 the wheel,r 28 cm0.28 m, the initial tangential speed of the rock,vi 2.80 m{s, and the tangential deceleration of the rock when the brakes are applied, aT1.30 m{s2.

(a) In order to determine the final angular velocity,ωf of the rock aftert1.70 sec, we first have to determine the initial angular velocity:

wi

vi

r

2.8 m{s

0.28 m 10 rad{sec. (102) The angular acceleration is a constant and given by:

α aT

r

1.30 m{s2

0.28 m 4.64 rad{sec

2_{. (103)}

Finally, using the appropriate rotational kinematic equation and substituting, we obtain

ωf ωi α∆t (104)

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

2.11 rad{sec. (106)

(b) In order to find the time at which the magnitude of the angular acceleration of the rock equals g, we first need to find the linear velocity of the rock at that time. We want |~a| g aa2

T a2r, where aT is given in the problem and ar v2{r. Substituting and solving for v we get

g

d

a2 T

v2

r

2

(107)

g2 a2_T v

4

r2 (108)

ñv4 r2pg2 a2_Tq (109)

v r1{2pg2a2_Tq1{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 kinematic equation for velocity and solve for the time:

vf viaT∆t (113)

ñ∆t vivf aT

(114) 2.80 m{s1.65 m{s

1.30 m{s2 (115)

xf xi vxi∆t 1

2axp∆tq

2

vxf vxi ax∆t

v2_xf v_xi2 2axpxf xiq

yf yi vyit 1 2ayt

2

vyf vyi ayt

v2_yf v2_yi 2aypyf yiq

θf θi ωi∆t 1 2α∆t

2

ωf ωi α∆t

ωf2 ωi2 2α∆θ

srθ c2πr v 2πr

T vtωr

ar

v2

r ω

2

r at rα