Physics I – Exam 1 – Spring 2004
Answer Key
Part A – 1. 12, 2. D, 3. A, 4. B 4 points each, no partial credit.
x (m)
B-1 18 Points
Force is constant, and so a is constant.
The cart goes –0.5 m in 1.0 s, and so the
average v is –0.5 m/s. Since the cart
momentarily stops at x = 0.5, the initial
v must be –1.0 m/s. This is all we need
to know to draw the graphs.
Many combinations of equations 1-5
could be used for these calculations.
1.0
0.5
0
t (sec)
1.0
2.0
v (m/s)
+1
t (sec)
0
1.0
2.0
1.0
2.0
-1
a (m/s2)
+1
t (sec)
0
Xcm (m)
B-2 12 Points
The cart + track has no external X force.
X momentum is constant = 0.5 kg m/s.
Vcm is constant = 0.5/(0.5+2.0) = 0.2 m/s.
Xcm starts at 0.0, then Xcm = Vcm * t.
2
1
t (sec)
0
Vcm (m/s)
2
4
6
8
10
0.2
0.1
t (sec)
0
2
4
6
8
10
Part C
Must show work to receive credit.
C-1 16 points
 y  V0 y t  12 g t 2  310 m  V0 y  80 m / s
V0 y  V0 sin( )  V0  160 m / s
V0 x  V0 cos()  V0 x  139 m / s
d  Vox t  1390 m (1400 OK for 2 sig digits)
Y
Y
N2
N1
C-2 18 points
(a) 6 points: T  m1 g sin( )  m1 a
(b) 6 points: m2 g sin( )  T  m2 a
m 2 sin( )  m1sin( )
g
(c) 6 points: a 
m1  m 2
a
X
m1
T
X
m2

m1 g

m2 g
C-3 20 points
Use conservation of momentum in X & Y directions.
Py  0
Before:
Px  M 1000
2M
2M
M
M
Vx Py   (1000.61) sin( 2) 
Vy
After: Px  (1000.61) cos( 2) 
3
3
3
3
Solving using before = after: Vx  1000.00 and Vy  17.46
  arctan( Vy / Vx )  1.0003
This angle is just barely enough. What did you expect, it’s a B grade SciFi movie!
Full credit for an angle close to the correct answer.