PY 211 Mid Term Exam I

February 13, 2007

The Formulae Sheet

Problem 1

Three forces are applied to a circular ring in such a waythat the ring remains in equilibrium. If the forces, inNewtons, are applied to the ring as shown below, findthe magnitude and direction of F3 so that the ringremains in equilibrium.

!

r F

1= 20ˆ i +10 ˆ j

r F

2= "10 ˆ j

The Math

!

r F " =

r F

1+

r F

2+

r F

3= 0

r F = 20ˆ i + 0 ˆ j +

r F

3= 0"

!

r F

3= "20ˆ i

Questions?

Problem 22. A hunter aims his rifle at a 10 degree angle with respect to the horizontal at a coconut in the top of

a palm tree. At the instant he pulls the trigger, a monkey drops the coconut which falls toward the

ground. If the palm tree is 100 m from the hunter and the hunter’s bullet hits the falling coconut

0.25 seconds later, what is the velocity of the bullet as it leaves the barrel of the rifle?

The Math

!

xcoco(t) = x

0+ v

0,xt +1

2axt2

x0

= xcoco

v0,x

= 0

ax

= 0

xcoco(t) = x

coco

!

ycoco(t) = y0

+ v0,yt +

1

2ayt

2

y0

= h

v0,y = 0

ay = "g

ycoco(t) = h "1

2gt

2

!

xb(t) = x

0+ v

0,xt +1

2axt2

x0

= 0

v0,x

= v cos"

ax

= 0

xb(t) = v cos(")t

!

yb (t) = y0 + v0,yt +1

2ayt

2

y0 = 0

v0,y = v sin"

ay = #g

yb (t) = v sin(")t #1

2gt

2

At t=0.25, boththe bullet and thecoconut are at thesame place inspace!

!

xcoco

= xb

100 = v cos(")t

v =100

cos(10°)(0.25)= 407

Questions?

!

r v = v cos"ˆ i + v sin"ˆ j r v = 401ˆ i + 71ˆ j

Problem 3

• A car makes a sharp turn to the left and it isobserved that a mug of coffee initially at reston the dashboard slide to the right until it ispressed against the frame of the car. Write aparagraph which explain how each ofNewton’s three laws of motion apply to thissituation. You may uses sketches andequations if you wish, but your answer shouldbe a paragraph.

Newton’s Laws

• Objects in motion stay in motion and atconstant velocity, lest a force act uponthem.

• The vector sum of the forces equals tothe product of mass and acceleration.

• For every action there is an equal andopposite reaction.

Solution

An observer in an inertial frame notes that, from Newton I, thecup is moving in a straight line and will continue to do so unlessacted on by a non-zero net force. When the car turns, therewould need to be a force acting on the cup such that F = ma(Newton II). The friction force is not sufficient to provide this, sothe cup does not turn with the car, but slides out tangentiallyuntil it hits the frame. The cup then exerts a contact forceagainst the frame equal (in magnitude) to the force exerted bythe frame on the cup. This force then provides the centripetalacceleration such that the cup undergoes circular motion.

Problem 44. An object moves along the x-axis according to the equation,

3( ) (3 2 6)x t t t= ! + meters. Find

the average speed of the object between t = 2 s and t = 3 s. Find the instantaneous speed and

instantaneous acceleration at t = 2 s. (3 answers required)

!

v =x(t

2) " x(t

1)

t2" t

1

!

x(3) = 3(3)3" 2(3) + 6 = 81

x(2) = 3(2)3" 2(2) + 6 = 26

#x = 55

#t =1

v =#x

#t= 55

!

v(t) =dx(t)

dt= 9t

2" 2

v(2) = 36 " 2 = 34

!

a(t) =d2x(t)

dt2

=dv(t)

dt=18t

a(2) =18(2) = 36

Questions?

Problem 55. A spaceship with its rocket engine turned off initially is traveling through space with a constant

velocity of 100 m/s relative to an observer on a nearby planet (fixed reference system). A special

rocket engine on the spaceship is turned on at time t = 0, providing an increasing rate of acceleration,

a(t) = 10t m/s3 , where t is the time since the engine was turned on. What is the velocity of the

rocket 10 seconds after turning on the engine? How far did the rocket travel in those 10 seconds?

The MathCaution! The acceleration is NOT constant so the kinematicequations will NOT work!

!

v(t) " v(0) = a(t')dt't= 0

t

#

a(t) =10t

v(t) = 10t'dt '= 5t2

0

t

# +100

v(10) = 5(10)2

+100 = 600

!

x(t) " x(0) = v(t')dt't= 0

t

#

v(t) = 5t2

+100

x(t) = 5t'2dt '=

5

3t3

0

t

# +100t + x0

x(10) " x0 =5

3(10)

3+100(10) =

5,000

3+1,000 $ 2,670

Questions?

Problem 66. Two pennies are resting on the turntable of a record player, with one of them 5.0 cm from the

center and the other 10.0 cm from the center. The coefficient of static friction between a penny and

the surface of the turntable is s = 0.5. If the turntable makes one complete revolution every 0.5 s,

what happens to each of the pennies? Support your answer with the appropriate calculation.

The Math

!

v = r" = 2#rf =2#r

$

!

Fc

=mv

2

r=m(4" 2

r2)

r#2

=4" 2

mr

#2

!

Ff

max

= µFN =1

2mg

!

1

2mg > ? <

4" 2mr

#2

!

1

2mg =

4" 2mr

#2

r =g#

2

8" 2

r =(9.8)(0.5)

2

8" 2$ 0.031m

r $ 3.1cm

Questions?So, they both slide off the turn table!

Problem 77. Two masses (m1 and m2) are joined by a massless rope as shown in the figure below. A vertical

force, F = 30 N is applied to the upper mass which gives the system of masses an upward

acceleration of 3.2 m/s2. If the tension in the rope joining the masses is 18 N, what are the values

of the two masses in kilograms?

Free Body Diagrams

m2m1

!

Fg

!

Fg

!

T!

T

!

F

The Math

!

F = ma"F #T # Fg = m

2a

F #T #m2g = m

2a

!

F" = ma

T # Fg = m1a

T #m1g = m

1a

!

F "T = m2(a + g)

m2

=F "T

a + g

!

T = m1(a + g)

m1

=T

a + g

Questions?!

m2

=30 "18

3.2 + 9.8=12

13# 0.923

!

m1

=30

3.2 + 9.8=30

13" 2.31

Problem 88. An object of mass m = 3 kg is attached to a string 0.5 m long. The other end of the string is

looped over a n ail in the center of a flat table as shown below. The object is given an initial

velocity of v1 = 2.5 m/s causing it to slide in a circular motion on the table top. The coefficient

of kinetic friction between the object and the table is k =0.03. What is the tension in the string

after 2 seconds?

Free Body Diagram

!

FN

!

Fg

!

Ff

!

Tx

y

z

The Math

!

Fc

= mac= m

v2

R"

T =mv

2

R

!

v(t) = v0

+ at

v(t) = v0"µgt

!

T(2) =3[2.5 " (0.03)(9.8)(2)]

2

0.5# 22

!

Ftan

= matan

= "Ff#

atan

="Ff

m

!

F"z

= FN #mg = 0

FN = mg

!

Ff = µFN

Ff = µmg

atan

= "µg

!

T(t) =mv(t)

2

R=m(v

0"µgt)2

R

Questions?

Problem 99.) The two masses below are sliding on a frictionless table surface due to the 750 N applied force as

shown below. Find the acceleration of the system of masses. Identify and label all of the action –

reaction force pairs in this situation.

Free Body Diagrams

80 kg 210 kg

!

N

!

N

!

Fg

!

Fg

!

F

!

F12

!

F12

!

F21

!

F21and are the action reaction pair.

Free Body Diagram and Math

290 kg

!

F = 750

!

F = MTOT" a

a =750

290# 2.59

Questions?

Problem 1010. Two masses (m1 = 2 kg) and (m2 = 3 kg) are connected by a rope looped over a frictionless pulley

as shown below. Mass m1 rests on an inclined plane which makes an angle = 37° with respect to the horizontal. The coefficient of friction between mass m1 and the incline is k =0.25. Find the magnitude and direction of the acceleration of the system of masses.

Free Body Diagrams

m2

!

T

!

Fg

m1

!

FN

!

Fg = mg

!

Ff

!

Fgx'

!

Fgy'

!

"

!

T

We know friction points opposite tension because, in the absence of friction,the mass on the inclined plane would slide up the plane as m2 > m1.

The Math

!

Fy = m2a"

m2g #T = m

2a

!

Ff = µFN = µm1gcos"

!

m1a = m2g "m2a "m1g(µcos# " sin#)

a(m1 + m2) = g(m2 "m1µcos# + m1 sin#)

!

a =g(m2 "m1µcos# + m1 sin#)

(m1 + m2)$ 2.74

!

Fx' = m1a = T " Ff "m1gsin#$

Fy' = m2a = FN "mgcos# = 0$

FN = m1gcos#

!

m1a = T "mg(µcos# " sin#)

T = m2(g " a)

Questions?The block slides up the incline.

Statistics

Fin!