Physics I – Exam 1 – Spring 2006
Answer Key
Part A – 1: C, 2: C, 3: B, 4: A, 5: B, 6: B, 7: D, 8: B
4 pts each.
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 4 sections to the curve, each a section of a
parabola.
Looking For
curves are sections of parabolas.
maximum at t = 4 sec.
maximum value = 4 kg m/s.
value at t = 2 sec is half the maximum value (whatever it was).
symmetric curve around t = 4 sec.
Fnet (N)
2
1
0
-1
t (sec)
2
4
6
8
-2
p (kg m/s)
4
3
2
1
0
t (sec)
2
4
6
8
B-2 24 Points
Vx and Vy (m/s)
This problem is an application of equations 1-5 for twodimensional motion.
Looking For
Vx is constant = +9.8 m/s
Vy is straight line.
Vy starts at 0 and ends at –19.6 m/s.
X starts at 0 and ends at 19.6 m.
X must be a straight line.
Y is a parabola opening down.
Y = 14.7 at t = 1 sec.
9.8
Vx
0
t (s)
-9.8
1.0
Vy
2.0
-19.6
X (m)
19.6
9.8
t (s)
0
Part C
Must show work to receive credit.
C-1 24 points
This is a straightforward conservation of momentum

  
problem in two dimensions. P  10 m i  p1  p 2



p1  9.24 cos( 22.5) m i  9.24 sin( 22.5) m j



p1  8.54 m i  3.54 m j
 



p 2  P  p1  (10  8.54) m i  (0  3.54) m j




v 2  p 2  m  1.46 i  (3.54) j
1.0
2.0
1.0
2.0
Y (m)
19.6
14.7
9.8
4.9
t (s)
0