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Physics 112 – HW #9 Solutions
Fall 2005
1
S
A
v
si
v
Ai
S
A
v
A
v
s
θ=37
φ=23
837
[Collision on a Frozen Pond]
Conservation of momentum gives:
m
s
v
si
= m
s
v
s
cos
θ
+ m
a
v
A
cos
φ
m
A
v
Ai
= m
s
v
s
sin
θ
 m
a
v
A
sin
φ
which give:
v
si
=
(6.0 m/s) cos 37
°
+
50.0kg
80.0kg
(9.0 m/s) cos 23
°
=
9.97m/s
v
Ai
=
80.0kg
50.0kg
(6.0 m/s) sin 37
°
+
(9.0 m/s) sin 23
°
= 2.26m/s
The energy loss is:
E
f
 E
i
=
½ m
s
(v
s
2
 v
si
2
) +
½ m
A
(v
A
2
 v
Ai
2
)
= ½ (80 kg) {(6.0 m/s)
2
 (9.97 m/s)
2
)}
+
½ (50 kg) {(9.0 m/s)
2
(2.26 m/s)
2
}
=
640 J
838
[Blocks with Spring Bumpers]
(a) At the instant when the spring is under maximum compression, the blocks are moving
together with the same velocity, V.
Use conservation of momentum to find the velocity of the blocks:
m
A
v
A,i
= (m
A
+ m
B
)V
V
=
(m
A
v
A,i
)
m
A
+ m
B
=
(2.00 kg)(2.00 m/s)
2.00 kg + 10.0 kg
=
0.333 m/s
Now use conservation of energy to find how much energy U
s
is stored in the spring at that
instant:
K
A,i
=
K
A+B
½ m
A
v
a,i
2
= U
s
+ ½ (m
A
+ m
B
)V
2
U
s
= ½ m
A
v
a,i
2
 ½ (m
A
+ m
B
)V
2
= ½ (2.00 kg)(2.00 m/s)
2
 ½ (12.00 kg)(3.33 m/s)
2
= 3.33 J
(b) Since the spring bumpers are “ideal”, the collision is elastic
.
For an elastic collision, we can use
Equs 8.24 and 8.25 in your text:
v
A,f
=
m
A
 m
B
m
A
+ m
B
v
A,i
=
2.00 kg  10.00 kg
2.00 kg + 10.00 kg
(2.00 m/s) = 1.33 m/s
(moving to left)
v
B,f
=
2m
A
m
A
+ m
B
v
A,i
=
2(2.00 kg)
2.00 kg + 10.00 kg
(2.00 m/s) = 0.67 m/s
(moving to right)
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View Full Document Physics 112 – HW #9 Solutions
Fall 2005
2
8.56
[Rocket in Space]
(a)
v(t) – v
o
= v
ex
ln
m
m
0
,
so
m
0
m
= exp
v  v
0
v
ex
= exp
8.00 x 10
3
m/s  0 m/s
2100 m/s
=
45.1
865
[Railroad Handcar]
(a)
Since the mass is thrown sideways out of the car in the car's reference frame, it's momentum
component along the car's direction of motion doesn't change.
So the car's momentum and
speed don't change.
v
car f
=
5.00 m/s
East
(b) The mass's final velocity with respect to the track is 0.
Use Cons. of Momentum:
Mv
car o
=
m(0) + (M  m)v
car f
⇒
v
car f
=
Mv
car o
M  m
=
(200 kg)(5.00 m/s)
200 kg  25 kg
=
5.71 m/s
East
(c)
Cons. of Momentum again:
Mv
car o
 mv
throw
=
(M + m)v
car f
⇒
v
car f
=
Mv
car o
 mv
throw
M + m
=
(200 kg)(5.00 m/s)  (25 kg)(6.00 m/s)
200 kg + 25 kg
=
3.78 m/s
East
894
[Walking in a Canoe]
The keys here are: (1) the fact that the CM of
the system doesn't move relative to the water;
and (2) symmetry.
Relative to the canoe, the woman's final
position in the canoe's right side is
symmetrically located about the canoe's center
with respect to her starting position in the left
side.
So the final position of the canoe and
woman relative to the water can be found
simply by rotating the picture 180° about the
system's CM.
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 LECLAIR,A
 mechanics, Momentum

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