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block 2 is
x
=
x
m
cos(
ω
t
+
φ
) where
φ
=
π
/2 which gives
x
=
x
m
cos(
π
/2 +
π
/2) =
–
x
m
.
This
means block 2 is at a turning point in its motion (and thus has zero speed right before the
impact occurs); this means, too, that the spring is stretched an amount of 1 cm = 0.01 m
at this moment.
To calculate its aftercollision speed (which will be the same as that of
block 1 right after the impact, since they stick together in the process) we use momentum
conservation and obtain
v
= (4.0 kg)(6.0 m/s)/(6.0 kg) = 4.0 m/s. Thus, at the end of the
impact itself (while block 1 is still at the same position as before the impact) the system
(consisting now of a total mass
M
= 6.0 kg) has kinetic energy
K
=
1
2
(6.0 kg)(4.0 m/s)
2
= 48 J
and potential energy
U
=
1
2
kx
2
=
1
2
(1.97 × 10
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Simple Harmonic Motion

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