Solution for Longanswer Homework 8 Energy problems
Solution to Longanswer Homework Problem 8.1()
Problem:
A car with inertia
1200kg
is backing out of a parking space at
5
m
s
. An unobservant driver of a
1800kg
pickup truck is coasting through the parking lot at a speed of
3
m
s
, and runs straight into the rear bumper of the
car.
(a)What is the change in internal energy of the system, if the speed of the pickup is
1
.
5
m
s
backwards
after they collide?
(b)What is the coefficient of restitution for this collision?
Solution to Part (a)
Conserving momentum:
m
c
v
c,i
+
m
t
v
t,i
=
m
c
v
c,f
+
m
t
v
t,f
v
c,f
=
m
c
v
c,i
+
m
t
(
v
t,i

v
t,f
)
m
c
v
c,f
=
(1200kg)
(

5
m
s
)
+ (1800kg)
(
3
m
s

(

1
.
5
m
s
))
1200kg
v
c,f
= 1
.
75
m
s
Energy is conserved, so
Δ
E
+ Δ
K
= 0
Δ
E
=

Δ
K
=
K
i

K
f
=
1
2
m
c
(
v
2
c,i

v
2
c,f
)
+
1
2
m
t
(
v
2
t,i

v
2
t,f
)
Δ
E
=
1
2
(1200kg)
parenleftbigg
parenleftBig

5
m
s
parenrightBig
2

parenleftBig
1
.
75
m
s
parenrightBig
2
parenrightbigg
+
1
2
(1800kg)
parenleftbigg
parenleftBig
3
m
s
parenrightBig
2

parenleftBig

1
.
5
m
s
parenrightBig
2
parenrightbigg
Δ
E
= 19237
.
5J
Notice this is positive, which means the internal energy of the system increases during the collision.
Grading Key: Part (a) 3 Points
Solution to Part (b)
The coefficient of restitution is given by:
e
=

v
rel,f
v
rel,i
So
e
=

v
c,f

v
t,f
v
c,i

v
t,i
=

1
.
75
m
s

(

1
.
5
m
s
)

5
m
s

3
m
s
e
= 0
.
40625
Grading Key: Part (b) 2 Points
Total Points for Problem: 5 Points
1
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 Spring '08
 Stewart
 Physics, Energy, Inertia, Kinetic Energy, Mass, Work, Special Relativity, Longanswer Homework Problem

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