6) Two cars collide at an icy intersection and stick together afterward. The first car
has a mass of 1200 kg and is approaching at 8.00 m/s due south. The second car
has
mass of 850 kg and is approaching at 17.0 m/s due west. (a) Calculate the
final velocity (magnitude and direction) of the cars. (b) How much kinetic
energy is lost in the collision? (This energy goes into deformation of the cars.)
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M ssong
mis 1200 kg
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Answers
Answer:
Given data
The mass of the first car is.
m
1
=
1200
k
g
.
The initial velocity of first car is
v
1
=
8
m
/
s
.
The mass of second car is
m
2
=
850
k
g
.
The initial velocity of second car is
v
2
=
17.0
m
/
s
.
(a)
The final velocity of the cars in west direction can be calculated as,
m
1
v
1
+
m
2
v
2
=
(
m
1
+
m
2
)
v
x
(
1200
k
g
×
8
m
/
s
)
+
(
850
k
g
×
0
)
=
(
1200
k
g
+
850
k
g
)
×
v
x
v
x
=
9600
k
g
m
/
s
2050
k
g
v
x
=
4.41
m
/
s
The final velocity of the cars in south direction can be calculated as,
m
1
v
1
+
m
2
v
2
=
(
m
1
+
m
2
)
v
y
(
1200
k
g
×
0
)
+
(
850
k
g
×
17.0
m
/
s
)
=
(
1200
k
g
+
850
k
g
)
×
v
y
v
y
=
14450
k
g
m
/
s
2050
k
g
v
y
=
7.04
m
/
s
The final velocity of both the cars can be calculated as,
v
′
=
√
v
2
x
+
v
2
y
v
′
=
√
(
4.41
m
/
s
)
2
+
(
7.04
m
/
s
)
2
v
′
=
8.30
m
/
s
The direction can be calculated as,
tan
θ
=
v
y
v
x
tan
θ
=
7.04
m
/
s
4.41
m
/
s
θ
=
tan
−
1
1.59
θ
=
57.83
∘
Thus, the magnitude and direction of the final velocity is
v
′
=
8.30
m
/
s
and
θ
=
57.83
∘
.
(b)
The kinetic energy lost in collision can be calculated as,
Δ
K
E
=
K
E
′
−
(
K
E
1
+
K
E
2
)
Δ
K
E
=
1
2
(
m
1
+
m
2
)
v
′
2
−
1
2
(
m
1
v
2
1
+
m
2
v
2
2
)
Δ
K
E
=
1
2
(
1200
k
g
+
850
k
g)(8.3m/s)2−12(
(1200kg)
(8.00m/s)2+
(850kg)(17.0m/s)
2)ΔKE=−90612.75J
Thus, the kinetic energy lost in collision is Δ
KE=−90612.75J.