Physics I – Final Exam – Fall 2002
Answer Key
Pages 2-4 – Part A –Multiple Choice
1. C
2. C
3. A
4. A give +2 points for choice B
5. A
6. D
7. D
8. D
Page 5 – Part B – B-1 & B-2
B-1
2
v 2  v0  2 a (y  y0 )
v  2  9.8  78.4  39.2
or
v  a t  9.8  4.0  39.2
t  2  78.4  9.8  4.0
( y  y 0 )  v 0 t  12 a t 2
v = 39.2 m/s +2 points, –1 for wrong units (39 is close enough, don’t worry about – )
B-2
Fnet  m a
T cos()  F
m
+2 points, –1 for wrong units (1.4 is close enough)
T cos()  F  m a
a = 1.38 m/s2
a
Page 6 – Part B – B-3 & B-4
B-3
2r 24
v2
3.14 2

 3.14
or
Fm
 25 
 61.7
T
8
r
4
2 2


 0.785
F  m 2 r  25  0.785 2  4  61.7
T
8
F = 61.7 N
+2 points, –1 for wrong units (62 is close enough)
v
B-4
2
0.785 2


 0.0491 or
  0  2  (  0 )
2 (   0 ) 2 (2 )
2
2
 0.785
 0.0491
(   0 )  12 (0  ) t t  1 
 2 T  16   
t
16
1 2

2
2
T
Can also get t = 16 by reasoning that average speed would be half max speed so time
during acceleration to make one revolution is twice the period at full speed.
 = 0.0491 rad/s2
+2 points, –1 for wrong units (0.05 is close enough)
Can also do this in rev/s, in which case
 = 7.8125 x 10-3 rev/s2
+2 points, –1 for wrong units (0.0078 is close enough)
2
2
Page 7 – Part B – B-5 & B-6
B-5
 
+2 points
W  Fd
Can use any consistent coordinate system, like N = +Y and E = +X.
 
F  d  Fx d x  Fy d y  250  (30)  150  90  7500  13500  6000
W = 6000 J +2 points, –1 point for incorrect units.
B-6
J
m
Area = 4500+15000+31500 = 51000
v = 34 m/s +2 points, –1 point for incorrect units.
No credit if trying to solve with F = m a unless using proper calculus to solve the
problem with non-constant a. If they use concept of average force, that’s OK.
J  Area  P or v 
Page 8 – Part B – B-7 & B-8
B-7
vcm = 10 m/s +2 points, –1 point for incorrect units.
B-8
GM
R2
g = 1.62 m/s2 +2 points, –1 for incorrect units (1.6 is close enough)
g
Page 9 – Part B – B-9 & B-10
B-9
V  E d or equivalent (don’t worry about signs)
E = 60/0.05 = 1200 V/m or N/C
+2 points, –1 for incorrect units
B-10
(Can also do it the hard way by calculating Q = 4.44 x 10-10 and using that to calculate E.)
E = 100/0.04 = 2500 V/m or N/C
+2 points, –1 for incorrect units
Page 10 – Part C – C-1
V0x = 196 cos(30) = 170 m/s
V0y = 196 sin(30) = 98 m/s
A: t  V0 y / 9.8  10 then y  98  10  0.5  9.8  10 2  490
V0 y 2
 490
2  9.8
x  V0x 10  1700
Vertical height = 490 m
+2 points, –1 point for incorrect units
Horizontal distance = 1700 m +2 points, –1 point for incorrect units
If the student mixed up sin/cos
Vertical height = 1470 m
+2 points, –1 point for incorrect units
Horizontal distance = 1700 m +2 points, –1 point for incorrect units
Or B: y 
Page 11 – Part C – C-2
Ike: –500 + T = m a
Mike: +514 – T = m a
+2 points for correct acceleration of Ike = 0.1 m/s2
+2 points for correct tension on rope = 507 N
+2 points for Mike’s acceleration the same as Ike’s, right or wrong.
Page 12 – Part C – C-3 & C-4
C-3:
+1 points for check mark on F
+1 points for check mark on N
+1 points for check mark on W
+1 points for no check on C
C-4: D. +4 points
Page 13 – Part C – C-5 & C-6
C-5: A. +4 points
C-6:
+4 points for each equation, no partial credit.
OK to multiply whole equation by –1.
OK to put W = M g.
X:
Y:
 N sin( )  F cos()  M (a )
N cos()  F sin( )  W  0
Page 14 – Part C – C-7
  
l  rp
The distance is r  (4) 2  3 2  5 .0 x 10–12 m.
 3 
The angle is 180  arctan 
  143.13 º. (quadrant II)
4
l  5  10 12  9.1 10 31  1.5  10 7  sin( 143.13)  4.1 10 35
+2 for magnitude = 4.1 x 10-35 kg m2/s, –1 point for incorrect units
+1 for 0 as X component
+1 for 0 as Y component
+1 for putting magnitude into Z component (same answer)
+1 for putting – sign in front of Z component (whatever it is)
Note: Can’t do this problem by calculating L = I  without using calculus to find  by
taking the time derivative of the angle of the electron. This can be done to get the right
answer, but it is beyond the capability of almost everyone taking Physics 1:
y
  2 v (–Z direction)
I  mr2
r
Page 15 – Part C – C-8
+4 for correct magnitude = 5.76 x 1013 N/C or V/m, –1 point for incorrect units
+2 for correct X component = –4.608 x 1013, or –0.8 times magnitude,
–1 point for no – sign
+2 for correct Y component = +3.456 x 1013, or 0.6 times magnitude,
–1 point for putting a – sign
+2 for correct Z component = 0 or can leave it blank (gift)
Page 16 – Part C – C-9 & C-10
C-9
+2 for knowing to multiply result of C-8 by q = –1.6 x 10-19
The numerical results for this are based on the answers to C-8.
+1 for correct magnitude based on C-8, must have units = N
if C-8 magnitude is correct, this is 9.2 x 10-6 N
ignore – sign on magnitude if someone makes a mistake and puts one
+1 for correct X component; need opposite sign from C-8
if C-8 is correct, this is +7.37 x 10-6
+1 for correct Y component; need opposite sign from C-8
if C-8 is correct, this is –5.53 x 10-6
+1 for correct Z component = 0 or can leave it blank (gift)
C-10
1 q1
1 q1 q 2
and U = q2 V or use directly U 
4  0 r
4  0 r
+2 for using correct q1 in formula (a gift)
+2 for using correct q2 in formula, must have – sign indicated somewhere
+2 for using correct distance, not squared
+2 for correct answer = –288 eV or  4.608  10 17 J, must have – sign, –1 for no units
+2 for using V 
Page 17 – Part C – C-11
+4 for knowing to use K + U = 0
+2 for knowing U initial was from C-10 or calculate it again
+2 for knowing U final is zero
+2 for knowing K  21 m v 2 initial and final
+2 for correct answer
K final  K initial  U initial
v final = 1.11 x 107 m/s