Physics I – Exam 1 – Fall 2006
Answer Key
Part A – 1: C, 2: D, 3: G, 4: C, 5: A, 6: A
4 pts each.
Give credit for 6 if it agrees with 5 even if 5 is wrong.
B-1 20 Points
The key to this problem is the Impulse-Momentum Theorem and also Newton’s Second
Law in the form of Equation 11. There are 3 sections to the curve (or two counting the
upward and downward part on the right as one), each a section of a parabola.
Looking For
curves are sections of parabolas.
maximum at t = 4 sec.
maximum value = 2 kg m/s.
value at t = 2 sec is half the maximum value (whatever it was).
symmetric curve around t = 4 sec for the interval (2,6) seconds.
Fnet (N)
1
0
t (sec)
2
4
6
2
4
6
-1
p ( kg m/s )
2
1
0
t (sec)
B-2 24 Points
This problem is an application of conservation of momentum, equations 44x and 44y.
The total momentum in the X direction is +1.2 kg m/s and in the Y direction is 0.
Looking For
trying to use conservation of momentum in
some form, right or wrong.
using conservation of momentum separately
in X and Y directions.
VBx flat where VAx is flat and a sloped line
where VAx is a sloped line.
VBx final value = 3 m/s.
VBy flat where VAy is flat and a sloped line
where VAy is a sloped line.
VBy final value = –1 m/s.
VAx (m/s)
VAy (m/s)
4
4
2
2
0
t (msec)
1.0
0
2.0
-2
t (msec)
1.0
2.0
1.0
2.0
-2
VBx (m/s)
VBy (m/s)
4
2
t (msec)
0
1.0
t (msec)
0
2.0
-0.5
-1
Part C
Must show work to receive credit.
C-1 32 points
Step 1: X and Y components of initial velocity: V0 X  V0 cos(), V0 Y  V0 sin( )
V0 X  150.00 m / s, V0 Y  50.00 m / s
Step 2: Find time to reach impact point using X equation (no acceleration):
t impact  5.000  (1500.0)  (150.00)  15.00 seconds after the target drops.
Step 3: Find Y posiitons of target and cannonball at the time of impact:
2
y target  h  0  t impact   12 (g)t impact   1112.5  (4.9)(15) 2  10.0 m
y cannonball  0  V0 Y t impact  5.000  12 (g)t impact  5.000  500.0  (4.9)(10) 2  10.0 m
2