Exam #1-Solutions
Physics I
Fall 2000
Part A-Multiple Choice (4 points each, no partial credit)
1.C
2.C
3.C
4.C
5.B
6.C
7.E
8.C
Part B
1.
Turn around point
+
velocity
time
0
Turn around point
+
acceleration
time
0
2. Acceleration looks just like force.
1.5
1
Force
0.5
0
0
2
4
6
-0.5
-1
-1.5
Time
8
10
12
3. a) p=J=(.14)(39)-(.14)(-39)=2(.14)(39)=10.92 kg-m/s
b) p=J=10.92 kg-m/s=Fave(time) so Fave=10.92/(1.2x10-3)=9100 N
Part C
1.
a) Voy= 2.3Cos41o=1.74 m/s
(taking down as positive and the origin at the start point)
y-yo=voyt+1/2(ay)t2
Height=1.74(1.2)+(.5)(9.8)(1.2)2=9.14 m
b)
Vox= 2.3Sin41o= 1.51 m/s
(taking left as positive and the origin at the start point)
x-xo=voxt+1/2(ax)t2
Distance=1.51(1.2)+(.5)(0)(1.2)2=1.81 m
c)
V=2.3 m/s
41o
2.
a)
1.
2.
3.
4.
5.
Push from the person’s weight down on the sled
Weight of the sled (gravitational force)
Normal force from the ice on the sled
Tension in the rope connecting the sleds
Tension in the rope to the horse (pull from horse)
b)
1. Weight (gravitational force)
2. Normal force from the sled on the package
c)
1. Weight (gravitational force)
2. Normal force on from the ice on the sled
3. Tension in the rope connecting the sleds
4. Push from the package on the sled
d)
e)
Nice on sled
N sled on package
T
Wsled
Wpackage
Fpackage on sled
f)
Nice on sled
T2
T
Wsled
Fperson on sled
g) see diagrams
h) Taking down negative and left negative
T2Sin+N-W-Fperson on sled =0
-T+T2Cos=Ma