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001
10.0 points
Assume
an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 3
.
1 m
/
s
strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 1
.
2 m
/
s at an angle of
θ
with
respect to its original line of motion.
3
.
1 m
/
s
1
.
2 m
/
s
θ
φ
Before
After
Find the eight ball’s speed after the colli
sion.
Correct answer: 2
.
85832 m
/
s.
Explanation:
Let :
v
q
i
= 3
.
1 m
/
s
and
v
q
f
= 1
.
2 m
/
s
.
Given
m
q
=
m
e
=
m
,
vectorp
e
i
= 0,
vectorp
≡
mvectorv
,
and
vectorp
·
vectorp
≡
p
2
. Conservation of energy (and
multiplying by 2
m
) gives
p
2
q
f
+
p
2
e
f
=
p
2
q
i
.
(1)
Conservation
of
momentum
(specifically,
vectorp
q
f
+
vectorp
e
f
=
vectorp
q
i
, and squaring) gives
(
vectorp
q
f
+
vectorp
e
f
)
·
(
vectorp
q
f
+
vectorp
e
f
) =
vectorp
q
i
·
vectorp
q
i
.
Carrying out the scalar multiplication term
by term gives
vectorp
q
f
·
vectorp
q
f
+
vectorp
e
f
·
vectorp
e
f
+ 2
vectorp
q
f
·
vectorp
e
f
=
vectorp
q
i
·
vectorp
q
i
.
Rewriting in a simplified form
p
2
q
f
+
p
2
e
f
+ 2
vectorp
q
f
·
vectorp
e
f
=
p
2
q
i
.
(2)
Subtracting the Eq. 1 for conservation of en
ergy we have
2
vectorp
q
f
·
vectorp
e
f
= 0
.
(3)
Dividing Eq. 3 by 2
m
vectorv
q
f
·
vectorv
e
f
= 0
,
(4)
yields three possibilities
1)
vectorv
e
f
= 0, where
m
q
misses
m
e
.
2)
vectorv
q
f
= 0, where a headon collision results.
3)
vectorv
e
f
⊥
vectorv
e
f
;
i.e.
,
θ
+
φ
= 90
◦
.
This third possibility agrees with the condi
tions shown in the figure.
Note:
Most pool players already know that
the queue ball and the target ball scatter at
90
◦
to oneanother after a twobody collision
(to a close approximation).
3
.
1 m
/
s
1
.
2 m
/
s
2
.
86 m
/
s
67
.
2
◦
22
.
8
◦
90
◦
Before
After
Equation 1 gives us
1
2
m v
2
q
i
=
1
2
m v
2
q
f
+
1
2
m v
2
e
f
,
rewriting
v
2
e
f
=
v
2
q
i
−
v
2
q
f
,
then
v
e
f
=
radicalBig
v
2
q
i
−
v
2
q
f
=
radicalBig
(3
.
1 m
/
s)
2
−
(1
.
2 m
/
s)
2
= 2
.
85832 m
/
s
,
and
θ
= arctan
parenleftbigg
v
e
f
v
q
f
parenrightbigg
= arctan
parenleftbigg
2
.
85832 m
/
s
1
.
2 m
/
s
parenrightbigg
= 67
.
226
◦
,
also
v
e
f
=
v
q
i
sin
θ
= (3
.
1 m
/
s) sin(67
.
226
◦
)
=
2
.
85832 m
/
s
,
and
v
q
f
=
v
q
i
cos
θ
= (3
.
1 m
/
s) cos(67
.
226
◦
)
= 1
.
2 m
/
s
,
finally
oldmidterm3 01 – Turner – (58185)
1
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hinojosa (jlh3938) – oldmidterm3 01 – Turner – (58185)
2
φ
= 90
◦
−
θ
= 22
.
774
◦
002
10.0 points
Sand from a stationary hopper falls on a mov
ing conveyor belt at the rate of 2
.
9 kg
/
s, as
shown in the figure. The belt is supported by
frictionless rollers and moves at 1
.
11 m
/
s un
der the action of a horizontal external force
supplied by the motor that drives the belt.
F
ext
Find the frictional force exerted by the belt
on the sand.
Correct answer: 3
.
219 N.
Explanation:
Let :
˙
m
= 2
.
9 kg
/
s
and
v
= 1
.
11 m
/
s
.
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 Spring '08
 Turner
 Force, Friction, Kinetic Energy, Mass, Momentum, Correct Answer, kg

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