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Unformatted text preview: 1999 Physics (3 Solutions Distribution of points
Mesh. l (15 points)
(a) 5 points
For cemervatien of momentum 1 point
my0 r: (m + [W0 )1)
v m me)O
m +MO
For conservation of mechanical energy l point
K +1] 2 + U” or é—MWM m Mmm gh
Fer calculating I: in terms of E I point
11 m [(1  COS 69)
For substituting for Mmml and v in the energy equafjon 1 point:
2
5(1‘7’1 + A40 3 (m + .MO W COS For the correct: answer 1 point;
m +M
yo 7: 0 .lZngwCOSG)
m
(b) 4 points
For Newton’s seeend law (not awarded if net force was set equal to zero) 1 point;
If“: = ma
For any equatien that indicated that the tension minus the weight is £191; zero l point
Y "” Nltotalg : Muﬁala
For an expression fer the centripetal foree '1 point
a l. 2: or 313.
w r 3?
U2
T ”"‘ A4ng 3 Mtoml T "‘
For curreetly substituting for Mom; and 1) Ll. point w 2
1
("2+MO)?(W> =T—(m+M0)g Solving for T
T : (i?1+M0)g(3m2COSI9) 1999 Physics C Solutions
Mech. 1 (continued)
(c) 4 points For all of the following:
‘1) A practical procedure that uses some or all of the apparatus
listed and would work
2) Recognition of any assumptions that must be made
3) Indication of the proper mathematical computation using the
variables measured Two points were awarded if the description of the procedure was not complete
but it would work, or if the mathematical work did not clearly specify
the variables used, or any combination of the above. No points were awarded if the procedure would not be feasible in a laboratory
situation with the apparatus listed, or if the procedure was merely a repeat
of that outlined in part (a) (d) 2 points Fm mma mwbo Ii ., . . a , t do
For expressing the acceleration as the time derivative of the speed, a = —« t) l’ 212
no 0
'2) hr
in —» a: « 
1)“ m
U m Uggwbt/m or a general expression for the length of the dart in the block as a function
of time or for the expression for the total distance L
M mun wet/m
i? —. Mb (I u— e ) “.2332”
L“ 32 Distribution
of points 4 points 1 point 1 point 1999 Physics C Solutions Distribution
0f points Mach” 1 (continued)
((11) (continual)
Alremate Solution 1 Alternate paints Fur indicating that work equals the: change in kinetic energy I point
If? dx 2 1' m2); 2
m £339., n) 4.4 average L A
b) q
Iii dxzimvg
2 2 0 Zn) 1 »
wiLx—mﬁ 2 2
For the correct expression f‘nr the: total distance L I paint mp0 3:: L: Alternate SG/z'ution 2 Alternate poinm Fm At a Ap
For the, above: expmssien on f ~51) d: = ~va
0 I paint d ~ 1 . .
Using 1) :3— and the fact that 3 goes from zero to L as nme goes dt
from zero to inﬁnity \ L
j ~22 (IIS :2 mm
0 «('71) = «mu0
Fm the: conect expression for the total distance L I paint
ml)0 b Lm 1999 Physics C Solutions Distribution of points Mean. 2 (15 points)
(a) 3 points For indicating that the equation for gravitational force is applicable 11 point F x m Gmélm2 r
For using the proper expression for the mass of the planet enclosed by the radius 1 point
Gm pim3
17‘ m .— 3
P2
For proper cancellation of terms to ShOW the ﬁnal result 1 point
4 v F x  ~3~ mirsz r (b) 4 points
F
S For drawing a straight line from the origin to a distance R, and not going past R 1 point For having the maximum magnitude occur 21th 1 point For having the curve from R to SR decreasing in magnitude with proper curvature and appearing to reach an asymptote 1 point For recognizing that the force is always negative, Le. the graph is always below
the xaxis 1 point 1999 Physics C Solutions Distribution of points
Mech. 2 (continued)
to) 3 points
In this and all subsequent parts, either C or i;— nGpm could be used.
For indicating the integral that needs to be calculated to determine the work 1 point
W 5 f1? air = j~Cr dr
For using the proper limits on the integral (R to zero, not 7') l point
0
W a j ~01» dr
R
2 O
W 3 ~61
2 R
For the correct answer ll point
C 2
W = R
2
Alternate Solution Alternate points
For recognition that the work is the area under the curve, which is triangular 1 point
For using the corrects limits (zero to R) 1 point
For the correct answer 1 point
2
W, :3 CR
2
(d) 2 points
For using conservation of energy or workcenergy relationship '1 point
AK 2 AU :2 W n
l 2 CR2
—mv =
2 2
For the correct answer 1 point
[CR2
1) z
m
An alternate solution indicating the potential energy as that of a harmonic
oscillator also received full credit:
(e) 2 points
For indicating that the ball will move from the center to the surface of the planet 1 point
For indicating that the ball will stop at the surface, return to the center,
and continue oscillating in this manner, with no damping i point Describing the motion as simple harmonic oscillation with no damping earned full, credit ...
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This note was uploaded on 01/22/2012 for the course PHYS 2425 taught by Professor Park during the Spring '09 term at CyFair College.
 Spring '09
 Park
 Physics

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