bhakta (vsb255) – HW 10 – coker – (58245)
1
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24
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001
10.0 points
A 2 kg steel ball strikes a wall with a speed
of 14
.
9 m
/
s at an angle of 49
.
5
◦
with the
normal to the wall.
It bounces off with the
same speed and angle, as shown in the figure.
x
y
14
.
9 m
/
s
2 kg
14
.
9 m
/
s
2 kg
49
.
5
◦
49
.
5
◦
If the ball is in contact with the wall for
0
.
277 s, what is the magnitude of the average
force exerted on the ball by the wall?
Correct answer: 139
.
737 N.
Explanation:
Let :
M
= 2 kg
,
v
= 14
.
9 m
/
s
,
and
θ
= 49
.
5
◦
.
The
y
component of the momentum is un
changed. The
x
component of the momentum
is changed by
Δ
P
x
=
−
2
M v
cos
θ .
Therefore, using impulse formula,
F
=
Δ
P
Δ
t
=
−
2
M v
cos
θ
Δ
t
=
−
2 (2 kg) (14
.
9 m
/
s) cos 49
.
5
◦
0
.
277 s
bardbl
vector
F
bardbl
=
139
.
737 N
.
Note:
The direction of the force is in negative
x
direction, as indicated by the minus sign.
002
10.0 points
A 0.35 kg soccer ball approaches a player
horizontally with a velocity of 16 m
/
s to the
north. The player strikes the ball and causes
it to move in the opposite direction with a
velocity of 24 m
/
s.
What impulse was delivered to the ball by
the player?
Correct answer:
−
14 kg
·
m
/
s.
Explanation:
Let north be positive:
Let :
m
= 0
.
35 kg
,
v
i
= 16 m
/
s
,
and
v
f
=
−
24 m
/
s
.
Δ
vectorp
=
mvectorv
f
−
mvectorv
i
−
(0
.
35 kg)(16 m
/
s)
=
−
14 kg
·
m
/
s
,
which is directed to the south.
003
(part 1 of 2) 10.0 points
A railroad car of mass 42700 kg moving at
3
.
9 m
/
s collides and couples with two coupled
railroad cars, each of the same mass as the
single car and moving in the same direction
at 1
.
89 m
/
s.
What is the speed of the three coupled cars
after the collision?
Correct answer: 2
.
56 m
/
s.
Explanation:
Given :
m
= 42700 kg
,
v
1
= 3
.
9 m
/
s
,
and
v
2
= 1
.
89 m
/
s
.
By conservation of momentum,
m v
1
+ (2
m
)
v
2
= (3
m
)
v
f
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bhakta (vsb255) – HW 10 – coker – (58245)
2
v
f
=
m v
1
+ 2
m v
2
3
m
=
(42700 kg) (3
.
9 m
/
s)
3 (42700 kg)
+
2 (42700 kg) (1
.
89 m
/
s)
3 (42700 kg)
=
2
.
56 m
/
s
.
004
(part 2 of 2) 10.0 points
How much kinetic energy is lost in the colli
sion?
Correct answer: 57504
.
1 J.
Explanation:
The change in kinetic energy is
Δ
KE
=
KE
f
−
KE
i
=
1
2
(3
m
)
v
2
f
−
bracketleftbigg
1
2
m v
2
1
+
1
2
(2
m
)
v
2
2
bracketrightbigg
=
3
2
(42700 kg) (2
.
56 m
/
s)
2
−
1
2
(42700 kg) (3
.
9 m
/
s)
2
−
(42700 kg) (1
.
89 m
/
s)
2
=
−
57504
.
1 J
,
which represents a loss of
2
.
56 m
/
s
.
005
(part 1 of 2) 10.0 points
Given two masses,
m
1
= 4 kg and
m
2
= 6 kg,
m
1
is moving with a velocity
v
1
colliding with
m
2
, which is suspended by a string of length
8 m.
The two masses are stuck together as the
result of the collision. The compound system
swings to the right, passes point
B
as shown
in the figure, and stops at the horizontal level.
The acceleration of gravity is 9
.
8 m
/
s
2
.
v
1
m
1
B
m
3
8 m
A
m
2
θ
m
3
=
m
1
+
m
2
Find the kinetic energy
K
of the compound
system immediately after the collision.
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 Spring '08
 Turner
 Kinetic Energy, Momentum, m/s

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