Mechanics – AP Physics C – 1995
1. J  F  t   F (t ) dt
(a) Area = 4 + 4 + 4 = 12 Ns
(b) J  p (Use J=12 on the 5kg ball...)
26 m/s
0.5kg
|
5kg
2m
12  5v f  5(0)  v f  2.4 m s
1.2m
(c) For the 0.5kg cube in collision with...
p0 = pf (Using conservation of momentum for the 2-mass system...)
.5(26) + 5(0) = .5vcf + 5(2.4) (i ) 13  .5vcf  12  vcf  2 m s (ii ) right
(d) K  K f  Ki 

1 (.5)262
2

1 (.5)22
2

 12 5(2.4)2 = 169  1  14.4  = 153.6 J
So the kinetic energy lost was: 153.6 J
(e) Time to floor (using d = ½at2 or 1.2 = 4.9t2 or 5t2 with g = 10m/s2...): t = .495 or .490 s.
Now (using d = rt): dcube = 2(.49) =0.98m vs dball = 2.4(.49) = 1.176m.
Subtracting we get a difference in landing points of: .196m
2. The key to these potential energy graph/function problems is: 
dU
 F ( x) (a biggie!)
dx
dU
  12 k  2 x   kx  Fs ( Hooke ' s Law) .
dx
ab a
Well, getting back to our problem (where r = x)...
U (r )  ba r 
 b r  abr 1
r
(a) Potential Energy (U) is at a minimum when r = r0 but in terms of 'a' and 'b'... ???
Well, let's do some calculus and see what we get...
dU a
a ab
(i)
  ab r 2  0   2  r 2  b2  r0  b (Graph shows us the minimum.)
dr b
b r
a
ab
(ii) U (r0 )  U (b)  (b) 
 a  a  2a
b
(b)
dU
a

 a ab  ab a
    ab r 2      2   2   F (r ) (Now graph!)
(b) Now Fnet = 
dr
b
b

b r  r
1
This is a rational function such as y = 1/x2 shifted down say 3 as in y = 2  3
x
ex/ Recall Uspring = ½ kx2 and notice: 
Notice the vertical asymptote: x = 0
and the horizontal asymptote: y = -a/b
and the x-intercept of r0
r0
y = -a/b
#2 continued on the next page...
Mechanics – AP Physics C – 1995
p. 2
2. (continued)
(c) We'll use the Work-Energy Theorem: Wnet  K where the work done will equal the
decrease in potential energy or W  U  (U f  Ui )  Ui  U f with:
Ui  U
   U       ab  a2  2a  52a
r0
2
b
2
a b
b 2
b
2
(where above, we found r0 = b)
U f  U  r0   2a (where above, we found U0 = 2a with U0 is Uf here)
5
1
So the work done is: W  U i  U f  a  2a  a (Wait! in terms of U0???)
2
2
Oh yeah, U0 = 2a, so a = ½U0 so W = 14 U 0
(d) Using the Work-Energy Thm again, we notice that since we started from rest...
To come to rest again we'll need Wnet  K = 0J or Uf = Ui .
ab
ab
, so we can equate these to get: 52 a  ba r 
r
r
(e) The particle accelerate to the right until r = r0. Then acceleration will be back to the
left bringing the particle rest when U = 52 a again. It returns and repeats the process forever.
3. In terms of Ra and Rb and T (the same for both masses)...
v2
(a) Using Fc  m  mac ... and Newton's Universal Law of Gravity on mass 'A'...
R
M bM a
Mb

M

a

a

G
We get: G
a
c
c
 Ra  Rb 2
 Ra  Rb 2
Ui =
5
a
2
and U (r )  ba r 
4 2
2
v2
 2 
2
(b) ac   R and  
, we get: ac  R 
  R 2 and
R
T
T
 T 
2
Mb
4

R
using the boxed result above... G
a
T2
 Ra  Rb 2
Notice that Ra + Rb is the force distance and Ra is the orbital radius or distance for Ma
2
M b  Ra
Solving for Mb ...
4 2  Ra  Rb 
2
GT 2
(c) Similar to (b) but with Mb's orbital radius Rb (not Ra), we'll get:
G
Ma
 Ra  Rb 2
4
4 2
 Rb 2 and solve for Ma ... M a  Rb
T
(d) I  M a Ra2  M b Rb2

(e) L  I  M a Ra2
     M R   
2
T
2
b b
2
T
2
 Ra  Rb 2
GT 2
= 2Mb