23
1
22
6.4
10
kg m/s
tan
28 .
1.2
10
kg m/s
θ
−
−
−
§·
×⋅
==
°
¨¸
©¹
(c) Combining the two equations
p = mv
and
Km
v
=
1
2
2
, we obtain (with
p = p
n r
and
m = m
n r
)
()
2
22
2
19
26
1.4 10
kg m/s
1.6 10
J.
2
2 5.8 10
kg
p
K
m
−
−
−
=
×
×
100. (a) We find the momentum
G
p
nr
of the residual nucleus from momentum
conservation.
22
23
ˆˆ
0
( 1.2 10
kg m/s)i ( 6.4 10
kg m/s) j
ni
e
v
p
ppp
p
−−
=++
¡
=− ×
⋅
+− ×
⋅
+
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Momentum

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