101. The mass of each ball is
m
, and the initial speed of one of the balls is
1
2.2m s.
i
v
=
We apply the conservation of linear momentum to the
x
and
y
axes respectively.
11
1
2
2
2 2
cos
0
sin
if
f
ff
mv
θθ
=+
=−
The mass
m
cancels out of these equations, and we are left with two unknowns and two
equations, which is sufficient to solve.
(a) The
y
momentum equation can be rewritten as, using
2
60
θ
=°
and
2
1.1 m/s
f
v
=
,
sin
(1.1 m/s)sin 60
0.95 m/s.
f
v
=
and the
x
momentum equation yields
cos
(2.2 m/s) (1.1 m/s)cos60
1.65 m/s.
f
v
°
=
Dividing these two equations, we find tan
1
= 0.576 which yields
1
= 30°. We plug the
value into either equation and find
1
f
v
≈
1.9 m/s.
(b) From the above, we have
1
= 30°, measured
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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, Mass, Momentum

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