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Homework7
Chapt. 10 & 11
1/ the last part of Introduction of Collision
Most real collisions are somewhere between elastic and perfectly inelastic. This is indicated by
the
elasticity
E of the collision, which measures the difference in the velocities of the particles
after the collision compared with the difference in velocities before the collision. For instance, in
a perfectly inelastic collision, the two particles stick together after colliding. The elasticity of
such a collision is E = 0, because the difference in velocities between the particles is 0 after they
collide.
Technically, the elasticity is defined by the relation
, where
and
are
the initial and final velocities of particle 1, and
and
are the initial and final velocities of
particle 2. In this problem, the formula is simplified by our definition of
and the
hypothesis
. So, using
for the final velocity of particle 1 and
for the final velocity of
particle 2, we obtain the simpler formula
.
This final form will be most useful to you in solving Part F.
Part F
If the two particles with equal masses m collide with elasticity E = 0.650 , what are the final
velocities of the particles? Assume that particle 1 has initial velocity
and particle 2 is initially at
rest. Look at the applet to be sure that your answer is reasonable.
Give the velocity
of particle 1 followed by the velocity
of particle 2, separated by a comma.
Express the velocities in terms of
.
1/ read the problem, and understand what is elasticity E
2/
Since from definition:
2
1
2
1
(
)
f
f
i
i
V
V
E V
V

= 

, in this problem, V(2i) = 0, E = 0.650, so the
equation goes to
2
1
1
0.65
)
f
f
i
V
V
V

=
and from the momentum conservation law
1
1
2
i
f
f
mV
mV
mV
=
+
ie.
1
1
2
i
f
f
V
V
V
=
+
, solve these two equations,
you
will get V
1f
=0.175v, and V
2f
=0.825v.
10.5
10.5
The figure shows a ball’s beforeandafter pictorial representation for the three situations in parts (a), (b) and (c).
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View Full Document Solve:
The quantity
K
+
U
g
is the same during free fall:
f
gf
i
gi
K
U
K
U
+
=
+
. We have
(a)
( )
2
2
1
1
0
0
2
2
2
2
2
1
0
1
1
1
2
2
2
[(10 m/s)
(0 m/s) ]/(2 9.8 m/s )
5.10 m
mv
mgy
mv
mgy
y
v
v
g
+
=
+
⇒
=

=

×
=
5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground.
(b)
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This note was uploaded on 04/18/2010 for the course PHYS ? taught by Professor Zhou during the Spring '10 term at Georgia State University, Atlanta.
 Spring '10
 ZHOU

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