Physics I – Final Exam – Spring 2003
Answer Key
Pages 2-3 – Part A –Multiple Choice
1. A
2. A
3. B
4. E Give +2 points for choice F (sign error).
5. A Give +2 points for choice B (sign error).
6. B Give +2 points for choice A (sign error).
7. A Give +2 points for choice E (sign error).
Page 4 – Part B – B-1, B-2, B-3
B-1
v1  0  a (2)
v 2  30  v1 
x 1  0  0 (2)  12 a (2) 2  10
a
(8)  6 a
a  5 m / s2
2
x 2  x 1  12 (v1  v 2 )(8)  10  4 (10  30)  170 m
B-2
Total time is twice the time to drop from maximum height to Jack’s hand (or vice versa).
2h
2  19.6
2
t1 

2
Total Time = 4 seconds.
h  12 g t 1
g
9 .8
B-3
The answer is B. The time depends only on maximum height, not horizontal velocity.
Page 5 – Part B – B-4 & B-5
B-4

   

p i  m v i  (2) (5 î )  10 î
 p  p f  p i  J  18 î  6 ĵ

 


p f  p i  J  10 î  18 î  6 ĵ  8 î  6 ĵ
v f  p f / m  (4 î  3 ĵ ) m / s
B-5
Constant force:
 
W  F  d  Fx d x  Fy d y  (8  3)  (6  4)  24  24  0 J
1
Page 6 – Part B – B-6 & B-7
B-6
There are no external forces so momentum is conserved.
We will call the initial direction of Puck A the +X direction.



Pafter  Pbefore  m v 0  12 m î
Total mass of the system is M = 3 m.


v cm  Pafter / M  12 m î / (3 m)  4 î
The answer is the speed or magnitude, which is 4 m/s.
B-7
f  0
0  10 2

 0.5 rev / s 2
2 
2  100
Accept either + or – answer. (Asked for magnitude.)
If student gives answer in rad/s2:  = – rad/s2.
2
f  0  2  
2
2
2

Page 7 – Part B – B-8 & B-9
B-8
I total  I1  I 2  I 3  121 M1L2  M 2 r 2  M 3 r 2 
1
12
 0.6  0.5 2  0.5  0.25 2  0.5  0.25 2  0.075 kg m 2
B-9
Step 1: Use Conservation of Angular Momentum to find new rotation speed:
I1 1  1.25 12  15  I 2 2  0.75 2
2  15 / 0.75  20
Step 2: Use Work-Kinetic Energy Theorem to find work done:
2
2
W  K 2  K1  12 I 2 2  12 I11  0.5  0.75  20 2  0.5  1.25  12 2  150  90  60 J
Page 8 – Part B – B-10 & B-11
B-10
We are only interested in magnitude.
F  q v B sin( )  1.6  10 19  5.6  10 6  2.5  10 4  sin( 120)  1.94  10 16 N
B-11
mv
r
qB
m v  q B r  1.6  10 19  1.0  0.5  8.0  10 20 kg m / s
2
Page 9 – Part B – B-12 – Matching Problem
_12_
The Impulse-Momentum Theorem
_22_
The Work-Kinetic Energy Theorem
_26_
Change in Mechanical Energy from Non-Conservative Forces
_25_
Potential Energy of an Ideal Spring
__6_
Newton’s 2nd Law of Motion
_46_
Newton’s Law of Universal Gravitation
_47_
Coulomb’s Law
_54_
The Magnetic or Lorentz Force on a Moving Charged Particle
_18_
Finding the Scalar or Dot Product of Two Vectors
_35_
Finding the Vector or Cross Product (Magnitude) of Two Vectors
_44_
Collision in Two Dimensions with Conservation of Momentum
_45_
Elastic Collision in One Dimension
_39_
Definition of Torque
_41_
Definition of Angular Momentum for a Particle
Page 10 – Part C – C-1
K  U  (K f  K i )  ( U f  U i )  Wnoncons
K f  K i  U f  U i  Wnoncons
K f  0  0  12 k x 2  Ffriction d  0.5 *125 * (0.02) 2  0.0475 * 0.10  0.02025 J
2 Kf
2 * 0.02025

 9m/s
m
0.0005
2h
2  1 .0
t

 0.45 s
The time to drop is given by h  12 g t 2
g
9 .8
The horizontal distance = 9 t = 4.066 m.
T
Page 11 – Part C – C-2
1
2
m v2  Kf
v

Q E  T sin( )  0
T cos()  M g  0
T sin( ) Q E
tan( ) 

T cos() M g
T cos()
QE
T sin()
3
 QE 
 735 
1  2  10
  tan 1
  tan 
  tan 
 Mg 
 0.3  9.8 
1
   26.565
1
2
Mg
3
Page 12 – Part C – C-3 & C-4
C-3
Since the two particles are equal mass and the CM is (0,0), the positron location is
negative the electron position: X = –4.0 x 10–12 m, Y = –3.0 x 10–12 m.
C-4
Since the two particles are equal mass and the velocity of the CM is 0, the positron
velocity is negative the electron velocity: Vx = –6.0 x 10+6 m/s, Vy = 0.0 m/s.
Page 13 – Part C – C-5
   
 

Angular momentum: l  r  p  r  (m v)  m r  v
Direction by right-hand rule is into the page (–Z).
Angle between r and v is  = tan-1(3/4).
Magnitude:

l  m r v sin( )  9.1  10 31  5.0  10 12  6.0  10 6  sin(tan
1 3
4
( ))  1.64  10 35 kg m 2 / s
X and Y components are zero. Z component is –1.64 x 10-35 kg m2/s
Page 14 – Part C – C-6
The problem is solved with conservation of energy at two positions: y = d and y = h.
Since sphere A is not moving at the initial and final positions, we have no kinetic energy
and two forms of potential energy: electrostatic and gravitational.
q2
q2
k
 Wd  k
 Wh
where W = m g and k is the electrostatic constant.
d
h
Note that h = d is a trivial solution and not what we want, but useful to know.
Turn this into a quadratic equation for unknown h, multiplying by h and d:
W d h 2  (k q 2  W d 2 ) h  k q 2 d  0
There are two ways to solve this, the pedestrian way and the elegant way.
The pedestrian way: Use the quadratic formula with numbers:
(0.036) h 2  (0.03744) h  0.00144  0
0.03744  0.03744 2  4  0.036  0.00144
h
1 m
2  0.036
The elegant way is to realize that we can factor using (h – d) and get
(h  d) ( W d h  k q 2 )  0
The non-trivial solution is the second root or
k q 2 0.036
h

1 m
W d 0.036
4