Boston U Physics Midterm

Physics 130 General Physics - Moustakas Fall 2012

Midterm Exam 3

November 29, 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 being asked to find).4. One or more free-body or force-interaction diagrams, as appropriate,

with labeled 1D or 2D coordinate axes.5. Algebraic expression for the net force along each dimension, as appro-

priate.6. Algebraic expression for the conservation of energy or momentum equa-

tions, as appropriate.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. A 25 g ball of clay traveling east at 2.5 m{s collides with a 35 g ball of clay traveling25˝ south of west at 1.5 m{s. What are the speed and the direction of the resulting 60 gball of clay?

Solution:

This is an inelastic two-dimensional conservation of momentum problem. Applyingconservation of momentum in the x-direction, the final momentum in the x-directionis

pfx “ pix (1)

“ m1pvixq1 ` m2pvixq2 (2)

“ p0.025 kgqp2.5 m{sq ´ p0.035 kgqp1.5 m{sq cos 25˝ (3)

“ 0.0149 kgm{s (4)

Applying conservation of momentum in the y-direction, the final momentum in they-direction is

pfy “ piy (5)

“ m1pviyq1 ` m2pviyq2 (6)

“ p0.025 kgqp0 m{sq ´ p0.035 kgqp1.5 m{sq sin 25˝ (7)

“ ´0.0222 kg m{s (8)

Next, we find the magnitude of the final momentum by combining the x- and y-components using the Pythagorean theorem:

pf “a

pfx ` pfy (9)

“a

p0.0149 kg m{sq2 ` p´0.0222 kg m{sq2 (10)

“ 0.0267 kg m{s (11)

2


Using the final momentum, we can solve for the final speed:

pf “ pm1 ` m2qvf (12)

ñ vf “ pf

m1 ` m2

(13)

“ 0.0267 kg m{s0.025 kg ` 0.035 kg

(14)

“ 0.45 m{s (15)

(16)

The direction is given by

θ “ tan´1

ˆ

pfy

pfx

˙

(17)

“ tan´1

ˆ´0.0222 kg m{s0.0149 kg m{s

˙

(18)

“ ´56˝ (19)

3


2. Fred (mass 70 kg) is running with the football at a speed of 7.0 m{s when he is methead-on by Brutus (mass 130 kg), who is moving at 5.0 m{s. Brutus grabs Fred in atight grip, and they fall to the ground. Which way do they slide, and how far? Thecoefficient of kinetic friction between football uniforms and Astroturf is 0.25.

Solution:

This is an inelastic collision in which Fred and Brutus are moving toward each other.Assume that Brutus is moving in the positive x-direction and Fred is moving in thenegative x-direction. Because momentum is conserved, we have

pix “ pfx (1)

mBpviqB ´ mF pviqF “ pmB ` mF qpvf q (2)

Solving for the final velocity of Brutus and Fred,

vf “ mBpviqB ´ mF pviqFpmB ` mF q (3)

“ p130 kgqp5.0 m{sq ´ p70 kgqp7.0 m{sqp130 kg ` 70 kgq (4)

“ 0.80 m{s (5)

Now that we know that Brutus and Fred moved to the right after the collision, wecan find the distance that they traveled. Friction slows them down, and thereforethe force of friction must act in the negative x-direction. The net force is equal tothe total mass (the mass of both players) times the acceleration.

ÿ

Fx “ ´fk “ pmB ` mF qa (6)

´µn “ pmB ` mF qa (7)

´µpmB ` mF qg “ pmB ` mF qa (8)

ñ a “ ´µg (9)

“ ´p0.25q ˆ p9.8 m{s2q (10)

“ ´2.45 m{s2 (11)

4


Finally, we can use the kinematic equation to find the distance traveled with vf “ 0:

v2f “ v2i ` 2ad (12)

ñ d “ ´ v2i2a

(13)

“ ´p0.80 m{s2q22 ˆ p´2.45 m{s2q (14)

“ 0.13 m (15)

“ 13 cm (16)

5


3. A roller coaster car on the frictionless track shown below starts from rest at height h.The track’s valley and hill consist of circular-shaped segments of radius R.

(a) What is the maximum height hmax from which the car can start so as not to fly offthe track when going over the hill? Give your answer as a multiple of R. Hint Firstfind the maximum speed for going over the hill.

(b) Evaluate hmax for a roller coaster that has R “ 12 m.

Solution:

(a) When the car is in the valley and on the hill it can be considered to be in circularmotion. We start at the top of the hill and determine the maximum speed sothat the car stays on the track. We can use the equations for circular motion.The acceleration is given by

a “ ´v2

r, (1)

where the minus sign means that the centripetal acceleration points inward,toward the center of the circular track.

The free-body diagram for the car at the top of the hill illustrates the two forcesacting on the car. Using Newton’s second law, we know

ÿ

F “ ma “ ´mv2

R(2)

FN ´ Fg “ ´mv2

R(3)

When the normal force is greater than zero, the car is still touching the track.Therefore, by setting the normal force equal to zero we can find the maximumspeed that the car can have and still stay on the track.

Fg “ mv2max

R(4)

6


mg “ mv2max

R(5)

ñ vmax “a

gR (6)

Now that we know the maximum velocity at the top of the hill, we can useconservation of energy to find the starting height.

Kf ` Uf ` Ki ` Ui (7)

1

2mv2max ` mgyf “ 1

2mv2i ` mgyi (8)

1

2mv2max ` mgyf “ 0 ` mgyi (9)

We’ll use yf “ R for the height at the top of the hill and factor the mass term,

1

2v2max ` gR “ 0 ` gyi (10)

yi “ v2max

2g` R (11)

We can now plug in our results for vmax from above

yi “ p?gRq22g

` R (12)

“ R

2` R (13)

“ 3R

2(14)

(b) Plugging into the above equation, hmax for a roller coaster that has R “ 12 mwe get

yi “ 3R

2“ 3 ˆ 12 m

2“ 18 m (15)

7


4. An 7.0 kg crate is pulled 6.0 m up a 25˝ incline by a rope angled 15˝ above the incline.The tension in the rope is 110 N, and the crate’s coefficient of kinetic friction on theincline is 0.3.

(a) How much work is done by tension, by gravity, and by the normal force?

(b) What is the increase in thermal energy of the crate and incline?

Solution:

(a) Find work done by tension, gravity and the normal

WT “ T ¨ ∆r “ T∆x cosp15˝q (1)

“ p110 Nq ˆ p6.0 mq ˆ pcos 15˝q (2)

“ 640 J (3)

Wg “ Fg ¨ ∆r “ mg∆x cosp90˝ ` 25˝q (4)

“ p7 kgq ˆ p9.8 m{s2q ˆ p6.0 mq ˆ pcos 115˝q (5)

“ ´174 J (6)

WN “ 0 J (7)

(b) Find the increase in thermal energy

∆Eth “ fk ¨ ∆r (8)

“ Fx∆x (9)

“ µkFn∆x (10)

To find the normal force,ÿ

Fy “ 0 (11)

Fn ´ Fg cos 25˝ ` T sin 15˝ “ 0 (12)

ñ Fn “ Fg cos 25˝ ´ T sin 15˝ (13)

“ p7.0 kgq ˆ p9.8 m{s2q ˆ pcos 25˝q ´ p110 Nq ˆ psin 15˝q (14)

“ 33.703 N (15)

8


Plugging this result into the equation for the thermal energy, we find

∆Eth “ µkFn∆x (16)

“ p0.3q ˆ p33.703 Nq ˆ p6.0 mq (17)

“ 61 J (18)

(19)

9


5. A 6.0 kg cat leaps from the floor to the top of a 85 cm high table. If the cat pushesagainst the floor for 0.25 s to accomplish this feat, what was her average power outputduring the pushoff period?

Solution:

The average power output during the push-off period is equal to the work done bythe cat divided by the time the cat applied the force. Since the force on the floor bythe cat is equal in magnitude to the force on the cat by the floor, work done by thecat can be found using the workkinetic-energy theorem during the push-off period:Wnet “ Wfloor “ ∆K. We do not need to explicitly calculate Wcat, since we knowthat the cat’s kinetic energy is transformed into its potential energy during the leap.In other words,

∆Ug “ mgpy2 ´ y1q (1)

“ p6.0 kgqp9.8 m{s2qp0.85 mq (2)

“ 49.98 J. (3)

Thus, the average power output during the push-off period is

P “ Wnet

t(4)

“ 49.98 J

0.25 s(5)

“ 200 W. (6)

10


6. You need to determine the density of a ceramic statue. If you suspend it from a springscale, the scale reads 26.0 N. If you then lower the statue into a tub of water, so that itis completely submerged, the scale reads 15.0 N. What is the statue’s density?

Solution:

The buoyant force, FB, on the can is given by Archimedes’ principle.

The free-body diagram sketched above shows that the gravitational force of thestatue, FG,statue is balanced by the spring force, Fsp, and the buoyant force, FB

exerted by the water. Since the statue is in static equilibrium, we know that thenet force is zero. The rest of the solution follows through a series of substitutions,recalling that the density of water is ρw “ 1000 kg m´3:

FB ` Fsp ´ FG,statue “ 0 (1)

ñ FB “ FG,statue ´ Fsp (2)

ρwVstatueg “ FG,statue ´ Fsp (3)

Vstatue “ FG,statue ´ Fsp

ρwg(4)

mstatue

ρstatue“ FG,statue ´ Fsp

ρwg(5)

(6)

ñ ρstatue

mstatue

“ ρwg

FG,statue ´ Fsp

(7)

ρstatue “ ρwmstatueg

FG,statue ´ Fsp

(8)

ρstatue “ ρwFG,statue

FG,statue ´ Fsp

(9)

“ p1000 kg m´3q ˆ p26.0 Nqp26.0 N ´ 15.0 Nq (10)

“ 2360 kg m´3. (11)

11


7. A hurricane wind blows across a 5.0 m ˆ18.0 m flat roof at a speed of 125 km/hr.

(a) Is the air pressure above the roof higher or lower than the pressure inside the house?Explain.

(b) What is the pressure difference?

(c) How much force is exerted on the roof? If the roof cannot withstand this muchforce, will it “blow in” or “blow out”?

Solution:

(a) The pressure above the roof is lower due to the higher velocity of the air.

(b) Bernoulli’s equation with yinside « youtside is

pinside “ poutside ` 1

2ρairv

2 (1)

ñ ∆p “ 1

2ρairv

2 (2)

“ 1

2ˆ p1.28 kg{m3q ˆ

ˆ

125 km

hrˆ 1000 m

kmˆ 1 hr

3600 s

˙2

(3)

“ 770 Pa. (4)

The pressure difference is 0.77 kPa.

(c) The force on the roof is

Froof “ p∆pqA (5)

“ p770 Paq ˆ p5.0 m ˆ 18.0 mq (6)

“ 6.9 ˆ 104 N. (7)

The roof will blow up, because the pressure inside the house is greater than thepressure on the top of the roof.

12


8. A 120 g ice cube at ´12˝ C is placed in an aluminum cup whose initial temperature is65˝ C. The system comes to an equilibrium temperature of 15˝ C. What is the mass ofthe cup?

Solution:

There are two interacting systems: aluminum and ice. The system comes to thermalequilibrium in four steps: (1) the ice temperature increases from ´12 ˝C to ´0 ˝C;(2) the ice becomes water at 0 ˝C; (3) the water temperature increases from 0 ˝C to15 ˝C; and (4) the cup temperature decreases from 65 ˝C to 15 ˝C.

Since the aluminum and ice form a closed system, we have

Q “ Q1 ` Q2 ` Q3 ` Q4 “ 0. (1)

Each term is as follows:

Q1 “ Mice cice ∆T (2)

“ p0.120 kgq ˆ r2090 J{pkg Kqs ˆ p12 Kq“ 3010 J.

Q2 “ Mice Lf (3)

“ p0.120 kgq ˆ p3.33 ˆ 105 J{kgq“ 40, 000 J.

Q3 “ Mice cwater∆T (4)

“ p0.120 kgq ˆ r4190 J{pkg Kqs ˆ p15 Kq“ 7540 J.

Q4 “ MAl cAl∆T (5)

“ MAl ˆ r900 J{pkg Kqs ˆ p´50 Kq“ ´p45, 000 J{kgqMAl.

Inserting everything into equation (1), we get

50, 550 J ´ p45, 000 J{kgqMAl “ 0

ñ MAl “ 1.1 kg. (6)

13


Kinematics and Mechanics

xf “ xi ` vxit ` 1

2axt

2

vxf “ vxi ` axt

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

vt “ ωr

ar “ v2

r“ ω2r

v “ 2πr

T

Forces

~Fnet “ Σ~F “ ma

~Fnet “ Σ~Fr “ ma “ mv2

r

Fg “ mg

0 ă fs ă“ µsFN

fk “ µkFN

~FAonB “ ´~FBonA

Momentum

~pi “ ~pf

~p “ m~v

pvfx q1 “ m1 ´ m2

m1 ` m2

pvix q1

pvfx q2 “ 2m1

m1 ` m2

pvix q1

Energy

Kf ` Ugf “ Ki ` Ugi

∆K ` ∆U ` ∆Eth “ ∆Emech ` ∆Eth “ ∆Esys “ Wext

Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext

K “ 1

2mv2

U “ mgy

∆Eth “ fk∆s

W “ ~F ¨ ~dP “ ~F ¨ ~vP “ ∆E{∆t

Fluids and Thermal Energy

P “ F

A

ρ “ m

VQ “ vA

v1A1 “ v2A2

p1 ` 1

2ρv2

1 ` ρgy1 “ p2 ` 1

2ρv2

2 ` ρgy2

Qnet “ Q1 ` Q2 ` ... “ 0

Q “ MLf for freezing/melting

Q “ Mc∆T

Constants

g “ 9.8 m{s2

Mearth “ 5.98 ˆ 1024 kg

rearth´sun “ 1.50 ˆ 1011 m

ρwater “ 1000 kg{m3

ρair “ 1 .28 kg{m3

cice “ 2090J

kg K

cwater “ 4190J

kg K

cAl “ 900J

kg K

Lf “ 333 , 000J

kgice to water

14

Boston U Physics Midterm

Documents

Transcript of Boston U Physics Midterm