homework 27 – PAPAGEORGE, MATT – Due: Apr 3 2008, 4:00 am
1
Question 1, chap 10, sect 99.
part 1 of 2
10 points
Two particles of masses
m
1
= 6
.
6 kg and
m
2
= 21
.
2 kg are moving toward each other
along the
x
axis with equal speeds 3
.
85 m
/
s.
Specifically,
v
1
x
= +3
.
85 m
/
s (particle 1
moves to the right) and
v
2
x
=

3
.
85 m
/
s
(particle 2 moves to the left).
The particles collide elastically.
After the
collision, the first particle moves at
θ
1
= 90
◦
to its original direction while the second par
ticle is
deflected
through
a smaller
angle
θ
2
<
90
◦
.
x
y
vectorv
1
vectorv
2
vectorv
′
1
vectorv
′
2
θ
1
= 90
◦
θ
2
Find the final speed

v
′
1

of the first particle.
Correct answer: 5
.
51284 m
/
s (tolerance
±
1
%).
Explanation:
In elastic collision, the net momentum
vector
P
and the net kinetic energy
K
are both con
served. In components,
P
x
=
m
1
v
1

m
2
v
2
= 0

m
2
v
′
2
cos
θ
2
,
(1)
P
y
= 0
=
m
1
v
′
1

m
2
v
′
2
sin
θ
2
,
(2)
K
=
m
1
v
2
1
2
+
m
2
v
2
2
2
=
m
1
v
′
2
1
2
+
m
2
v
′
2
2
2
.
(3)
Given
v
1
=
v
0
,
v
2
=

v
0
, eq. (1) gives
v
′
2
cos
θ
2
=
parenleftbigg
1

m
1
m
2
parenrightbigg
v
0
(4)
while eq. (2) implies
v
′
2
sin
θ
2
=
m
1
m
2
v
′
1
;
(5)
therefore
v
′
2
2
= (
v
′
2
cos
θ
2
)
2
+ (
v
′
2
sin
θ
2
)
2
=
parenleftbigg
1

m
1
m
2
parenrightbigg
2
v
2
0
+
m
2
1
m
2
2
v
′
2
1
.
(6)
Substituting this formula into the energy con
servation eq. (3), we arrive at
m
1
+
m
2
2
v
2
0
=
m
1
2
v
′
2
1
+
m
2
2
bracketleftBigg
parenleftbigg
1

m
1
m
2
parenrightbigg
2
v
2
0
+
m
2
1
m
2
2
v
′
2
1
bracketrightBigg
=
m
1
(
m
2
+
m
1
)
2
m
2
v
′
2
1
+
(
m
2

m
1
)
2
2
m
2
v
2
0
(7)
and hence (after a bit of algebra)
v
′
2
1
v
2
0
=
3
m
2

m
1
m
2
+
m
1
.
(8)
Consequently,
v
′
1
=
v
0
×
radicalbigg
3
m
2

m
1
m
2
+
m
1
= 5
.
51284 m
/
s
.
(9)
Question 2, chap 10, sect 99.
part 2 of 2
10 points
Find the deflection angle
θ
2
of the second
particle.
Correct answer: 32
.
915
◦
(tolerance
±
1 %).
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 Spring '09
 KLEINMAN
 Physics, Mass, Work

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