Physics 205 Ock Test 2 Form 2 Worksheet With Answers - Dr. Huerta, 2014
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6Dr. Huerta Phy 205 oCK Test 2 FORM 2 ANSWER KEY March 3, 2014
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Name:
Idnumber:
1
THIS IS THE ANSWER KEY TO FORM 2
MARK YOUR DISCUSSION SECTION BELOW
2
[ ] Dr. Nepomechie 5O, 9:30 - 10:20 a.m.
[ ] Dr. Huerta 5P, 11:00 - 11:50 a.m.
[ ] Dr. Huerta 5Q, 12:30 - 1:20 p.m.
[ ] Dr. Alvarez 5R, 2:00 - 2:50 p.m.
Do problem [I] which consists of 15 multiple choice questions
worth 5 points each
Also do the long problem [II] worth 25 points
TO GET PARTIAL CREDIT IN PROBLEM [II]
YOU MUST SHOW GOOD WORK.
dr(t)
dv(t)
d v
v(t) =
, a(t) =
, a =
r(t) =
v(t)dt, v(t) =
a(t)dt
dt
dt
dt
For motion with constant acceleration a
x
with v
(t = 0) = v
and x(t = 0) = x
, we have:
x
0x
0
1
2
v
(t) = v
+ a
t,
x(t) = x
+ v
t +
a
t
,
x
0x
x
0
0x
x
2
x
x
v
+v
2
2
f
i
i
f
v
= v
+ 2a
(x
x
) and v
=
=
.
x
f
i
av x
f
i
t
t
2
f
i
v
In circular motion the centripetal acceleration is a =
r
t
dp
f
p = m v, F =
, J =
F(t) dt = p
p
f
i
dt
t
i
m
r
= m
r
+m
r
+m
r
+.., m
v
= m
v
+m
v
+m
v
+..
total
CM
1
1
2
2
3
3
total
CM
1
1
2
2
3
3
dp
total
external
p
= m
v
, F
=
total
total
CM
total
dt
In a collision J
0, so then p
= p
.
total
f
i
total
total
Physics 205D TEST 2 FORM 2
March 3, 2014
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