Version 061 – Midterm 3 – florin – (58140)
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001
10.0 points
A triangular wedge 5 m high, 14 m base
length, and with a 11 kg mass is placed on
a frictionless table. A small block with a 5 kg
mass (and negligible size) is placed on top of
the wedge as shown in the figure below.
11 kg
14 m
5 m
5 kg
Δ
X
wedge
11 kg
14 m
5 m
5 kg
All surfaces are frictionless, so the block
slides down the wedge while the wedge slides
sidewise on the table. By the time the block
slides all the way down to the bottom of the
wedge, how far Δ
X
wedge
does the wedge slide
to the right?
1. 5.2
2. 5.29412
3. 5.4
4. 5.85
5. 4.8
6. 5.71429
7. 3.82353
8. 4.375
9. 4.57143
10. 5.6
Correct answer: 4
.
375 m.
Explanation:
Let :
M
= 11 kg
,
m
= 5 kg
,
L
= 14 m
,
and
H
= 5 m
.
Consider the wedge and the block as a two
body system.
The
external forces
acting on
this system — the weight of the wedge, the
weight of the block and the normal force from
the table — are all vertical, hence the net
hor
izontal momentum
of the system is conserved,
P
wedge
x
+
P
block
x
= constant
.
Furthermore, we start from rest =
⇒
center
ofmass is not moving, and therefore the
X
coordinate of the centerofmass will remain
constant while the wedge slides to the right
and the block slides down and to the left,
X
cm
=
m X
block
+
M X
wedge
m
+
M
= constant
.
Note:
Only the
X
coordinate of the center
ofmass is a constant of motion;
i.e.
, the
Y
cm
accelerates downward because the
P
y
compo
nent of the net momentum is not conserved.
Constant
X
cm
means Δ
X
cm
= 0 and there
fore
m
Δ
X
block
+
M
Δ
X
wedge
= 0
.
Note that this formula does not depend on
where the wedge has its own centerofmass;
as long as the wedge is rigid, its overall dis
placement Δ
X
wedge
is all we need to know.
Finally, consider the geometry of the prob
lem: By the time the block slides all the way
down, its displacement
relative to the wedge
is equal to the wedge length
L
, or rather

L
because the block moves to the left of the
wedge. In terms of displacements relative to
the inertial frame of the table, this means
Δ
X
block

Δ
X
wedge
=

L .
Consequently,
0 =
m
Δ
X
block
+
M
Δ
X
wedge
=
m
(

L
+ Δ
X
wedge
) +
M
Δ
X
wedge
and therefore
Δ
X
wedge
=
m
m
+
M
L
=
(5 kg)
(5 kg) + (11 kg)
(14 m)
=
4
.
375 m
.
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Version 061 – Midterm 3 – florin – (58140)
2
002
10.0 points
Assume
an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 3
.
3 m
/
s
strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 1
.
1 m
/
s
.
3
.
3 m
/
s
1
.
1 m
/
s
θ
φ
Before
After
Find the eight ball’s angle
φ
with respect to
the original line of motion of the queue ball.
1. 20.1713
2. 33.367
3. 30.7723
4. 34.8499
5. 26.3878
6. 23.1998
7. 18.8765
8. 35.2344
9. 16.3827
10. 19.4712
Correct answer: 19
.
4712
◦
.
Explanation:
Let :
v
q
i
= 3
.
3 m
/
s
and
v
q
f
= 1
.
1 m
/
s
.
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 Spring '10
 SVHETS
 Mass, Moment Of Inertia, Irod

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