NAME (Print)______Me____Hulan E. Jack Jr_____________________________May 25, 2010__
Borough of Manhattan Community College
Course Physics 210/Sec101
Instructor: Dr. Hulan E. Jack Jr.
Date May 20, 2010
Final Exam – My Solutions
INSTRUCTIONS: Do any 6 of the 8 problems below. Check the 6 you want graded, otherwise I grade the first 6
problems. NUMERICAL CALCULATIONS ARE NOT NEEDED FOR GRADE!
1. Each of the displacement vectors A and B shown in Figure P3.4 has a magnitude of 3.00 m.
Graphically sketch (a) 4pts, (b) 4 pts, ( c) 4 pts, and (d) 5pts on listed figures.
(b) A-B
(d ) A-2B
(a) A+B
A+
B
A
-B
-B
B
-B
B
A-2
AB
-B
A
-B
-B
(c)
A
B-A
-A
B -A
-------------------------------------------------------------------------------------------------------------
0
So, for both to reach the same height, they must have the same initial y velocity, vy0 .
But, for the ball thrown straight has vx =0, the on thrown at 30o has vx = v0cos30o .
Speed for straight up ball v1 = v0y , for 30o ball is v2= sqrt(v0y 2+ v02sin30o 2)
Page 1 of 4

Height
2. A ball is thrown straight upward and returns to the thrower's hand after 3.00 s
y
in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At
top
upv
what speed must the second ball be thrown so that it reaches the same height as
v0
the one thrown vertically? (HINT: It’s very simple. Think a bit before you leap)
v
0
17 pts
v
x
1st throw 2nd throw
st
o
Given: time up and return t = 3.00s , angle of 1 ball = 0 , angle 0 of second ball straight
at an angle
up
=30.0o.
Find: v0 for each ball.
Projectile motion is
1. free fall in the vertical direction. Taking up as the positive direction a = -g (9.8m/s2on earth), vy= vy0 gt , vy0 = ?, and y =v0yt – ½gt2.
2. constant velocity in horizontal direction: vx = v0x = constant, so x= v0xt .
NAME (Print)______Me____Hulan E. Jack Jr_____________________________May 25, 2010__
Borough of Manhattan Community College
Course Physics 210/Sec101
Instructor: Dr. Hulan E. Jack Jr.
Date May 20, 2010
Final Exam – My Solutions
INSTRUCTIONS: Do any 6 of the 8 problems below. Check the 6 you want graded, otherwise I grade the first 6
problems. NUMERICAL CALCULATIONS ARE NOT NEEDED FOR GRADE!
velocity, v
3.
When a Highway curves, one way to supply the force to maintain the
into the paper
centripetal acceleration is by friction between the vehicle and the road.
N
=Ncos

Ca

v
Another way is to bank the road at an angle so that the normal force, N,
r
2
ac = v /R
between the vehicle and the road supplies this force and also supports the
weight of the vehicle. Find the bank angle, B for v = 50m/s and R = 400m.
Nh=Nsin
(a) Set up the Newton’s Laws for this situation. 6 pts.
v
O
ac
Car
From Newton’s 2 Law
F=ma
N
So, in the horizontal direction Fh = Nh = mac . Hence NsinB = v2/R
But, the car in equilibrium in the vertical direction. Hence a = 0, so Fv =Nv – mg =
0.
(Numerical: Nh = mcarv2/R = mcar(50m/s)2 /400m= mcar 6.25 AND Nv = mcarg.
(b) Solve for the tangent of the angle B. 11 pts
B
W=mg
R
nd
Looking at the expressions for Nv and Nh on the figure, the angle qB follows as
top view of road
Nh/Nv = NsinB/NcosB = sinB/cosB = tan B .
Hence, = arctan (Nh/Nv ) = arctan(v2/gR) = arctan(6.25/9.8) = arctan(0.6377) = 36.5o.
-------------------------------------------------------------------------------------------------------------4. A uniform plank of length , L=2.00 m and mass, Wplank= 300 N is supported by three ropes, as
indicated by the blue vectors in Figure P8.21. Find the tension in
each rope when a Wperson= 700-N person, = x= 0.500 m from the
left end.
Given: Lplank =L=2.00m, Wplank=300N, Wperson = 700N, position
x=0.500m, = 40o . Find: T1, T2, T3 .
(a) Sketch the Free Body Diagram of the plank. 5 pts
The only things needed are the addition of the weights of the
person, mpersong , and of the plank, mplankg. The horizontal and
vertical components was added for clarity.
(b) Very briefly explain why taking the point at T2 and T3 as the
axis of rotation for torque. 2 pts
Only one unknown, T1, has a torque about this axis.
(c) Solve for the T’s. 10 pts
ccw+O = 0 ; LT1 - ½LWplank - (L-x)Wperson =0; T1 = [½LWplank - (L-x)Wperson ]/L
Fh = 0; T1cos - T3 = 0,
= 40o .;
T3 = cos40o [½LWplank - (L-x)Wperson ]/L
Fv = 0; T2 + T1sin -Wperson - Wplank = 0 .
T2 = Wperson + Wplank - sin40o [½LWplank - (L-x)Wperson ]/L
Page 2 of 4
NAME (Print)______Me____Hulan E. Jack Jr_____________________________May 25, 2010__
Borough of Manhattan Community College
Course Physics 210/Sec101
Instructor: Dr. Hulan E. Jack Jr.
Date May 20, 2010
Final Exam – My Solutions
INSTRUCTIONS: Do any 6 of the 8 problems below. Check the 6 you want graded, otherwise I grade the first 6
problems. NUMERICAL CALCULATIONS ARE NOT NEEDED FOR GRADE!
5.
A light string is tightly wrapped around the spool, of mass M=5.00kg and
radius R= 0.600m , so that it does not slip as it unwinds. The other end is attached to
the bucket of mass m=3.00 kg. All starting from rest the bucket falls 4.00 m.
Using Conservation of Energy, what is the angular speed of the spool after the bucket
has fallen 4.00 m.
17 pts.
Given: height h = 4.00m, m = 3 kg, R= 0.600 m, M = 5.00 kg, start from restv0 = 0.
Find: f .
 =Rv , v=/R , IO = ½MR2 for uniform disk (given on board)
Initial
Final
Conservation of Energy


=0
f=vf /R
PEi + KEtransi + KEroti = PEf + KEtransf +KErotf
i
2
2
2
2
mgh + ½mv0 + ½IO0 = mg*0 + ½mvf + ½IOf ,
M
M
mgh +
0
+
0 = at bottom+ ½mR2f2 + ½*½MR2 f2,
O
f2 = mgh /([½mR2 + ½*½MR2]) = mgh /(½ R2[ m+½M])
=2*mgh /( R2[ m+½M])
f = sqrt(2*mgh /( R2[ m+½M])).
= sqrt(2*3.0kg*9.8m/s2*4.0m /( 0.62[ 3+5/2]kg))
=sqrt(2*3.0*9.8*4.0m /( 0.36* 5.5)= sqrt( 235./1.99) = sqrt( 235./1.99)
= sqrt( 118.7) = 10.9 rad/sec.
.
O
R
R
m
h
mg
m
v
Ar
ea
Area
------------------------------------------------------------------------------------------------------------6. Water flowing through a garden hose of diameter 2.74 cm fills a 25.0-L bucket in 1.50 min.
(a) What physical principle applies the this problem. 3 pts
Given: Dhose D=2.74cm, Flow rate Q = Vol/time =25.0 L/1.50min,
D v
Find: flow velocity from hose vhose = v.
Continuity Equation or Equation of Continuity
Flow rate Q =Av = constant.
nozzle
D/3
(b) What is the speed of the water leaving the end of the hose? 8pts
Flow rate Q =Ahosevhose. So, vhose = Q/Ahose = (Vol/time)/Ahose = (Vol/time)/[(D/2)2]
(c) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the
hose, what is the speed of the water leaving the nozzle? 6 pts
Given: Dnozzle = Dhose/3 = 2.74/3 cm, c = 1/3
Find: vnozzle =V.
Ahose vhose = Anozzle vnozzle = Q. So,
v nozzle =
Page 3 of 4
Q
A nozzle
Vol
Vol
Vol
time
= time = time 2 =
A nozzle
D nozzle
(cD hose ) 2
π
π
4
4
v
NAME (Print)______Me____Hulan E. Jack Jr_____________________________May 25, 2010__
Borough of Manhattan Community College
Course Physics 210/Sec101
Instructor: Dr. Hulan E. Jack Jr.
Date May 20, 2010
Final Exam – My Solutions
INSTRUCTIONS: Do any 6 of the 8 problems below. Check the 6 you want graded, otherwise I grade the first 6
problems. NUMERICAL CALCULATIONS ARE NOT NEEDED FOR GRADE!
7.
A steel rod with initially at length, L0= 4.00m, cross-sectional area A=4.00x10-4m2, and T0 = o
5 C is heated to a temperature Tf = 45oC. But, it is not allowed to expand.
Thermal Expansion
What is the thermal stress in the rod.
T
(a) Calculate the Lthermal if the rod had been allowed to expand. 9 pts
LOW
Given: L0= 4.00m, T0 = -5oC , Tf = 45oC, steel = 11x10-6/oC.
Find: Lthermal
The Coefficient Linear Thermal Expansion of Steel, steel = 11x10-6/oC.
Lthermal =  L0T = steel L0(Tf - T0) = 11x10-6/oC*4.00m *(45-(-5))oC.
= 11x10-6/oC*4.00m *(45+5_)oC = 2.20x10-3 .
THIGH
L0
L
Mechanical Pushback at THIGH
F
F
THIGH
L0
L
F
F
(b) An axial stress put on the rod to reduce back it original length, L0 .This
is the thermal stess. Young’s Modulus for steel, YSteel = 20x1010 Pa (N/m2). 8 pts
Given: A=4.00x10-4m2, YSteel = 20x1010 Pa, 1st approximation set L0= 4.00m.
Find: thermal stress .
 = YL/ L0 = YSteel L/ L0 = YSteel Lthermal/ L0
= YSteel steel L0(Tf - T0)/L0 = YSteel steel (Tf - T0)
= 20x1010 Pa* 11x10-6/oC * (45+5)oC = 1.10x108 Pa.
-----------------------------------------------------------------------------------------------------------8.
What mass of water at 25.0oC must be allowed to come to thermal equilibrium with a 1.85-kg
cube of aluminum initially at 1.50 x 102 oC to lower the temperature of the aluminum to 65.0°C? Assume
any water turned to steam subsequently recondenses.
Given: Cwater = 1cal/gmoC, CAl = 0.215cal/gmoC, mAl = 1.85 kg, Twater0 =25.0 oC, TAl0=1.50x102oC,
Twaterf = TAlf =Tf= 65.0oC (all are in thermal equilibrium). ( Cwater
Initial
Final
and CAl on board)
Water
Water
o
Find: mwater.
25 C
65
C
Q
AL
AL
(a) What physical principle applies here. 4 pts
150 C
Conservation of Energy in the form for heat Q.
Heat ,Q, flows out
Qfinal
Qinitial =
of AL into water
CAlmAl TAl0 + mwater CwaterTwater0 = mwater Cwater Twaterf + mAl CAlTwaterf
(1)
o
o
(b) Solve the problem . 13 pts
From Eq.(1) get mwater terms on left side and all else on the right,take care with signs,gives
mwater CwaterTwaterf - mwater Cwater Twater0 = CAlmAl TAl0 -mAl CAl Twaterf
mwater Cwater(Twaterf - Twater0) = CAlmAl TAl0 -mAl CAl Twaterf
so,
mwater = [CAlmAl (TAl0 -Twaterf) ]/ [Cwater(Twaterf - Twater0)]
mwater = [0.215cal/gmoC*1.85 x103gm (150-65)oC ]/ [1cal/gmoC (65 -25)oC]
=[0.215*1.85 x103*85]/ [40] = 8.45x102 gm.
Page 4 of 4