in almost all of the problems we will consider, the total momentum will be conserved the total...
TRANSCRIPT
In almost all of the problems we will consider, the total momentum will be conserved
The total mechanical energy may or may not be conserved
Two kinds of collisions:1. Elastic – energy conserved (special
case) 2. Inelastic – energy not conserved (general)
Example – Problem 7.38
A ball is dropped from rest at the top of a 6.10-m tall building, falls straight downward, collides inelastically with the ground, and bounces back. The ball loses 10.0% of its kinetic energy every
time it collides with the ground. How many bounces can the ball make and still reach a window sill that is 2.44 m above the ground?
Solution:
Method: since the ball bounces on the ground, there is an external force. Therefore, we can not use conservation of linear momentum.
An inelastic collision means the total energy is not conserved, but we know by how much is not conserved. On every bounce 10% of the KE is lost:
Given: h0 = 6.10 m, hf = 2.44 m
bounces ofnumber theis ,...,3,2,1,1.0 nniKEQ ii
o
ooo
EKEEmghPEE
11
o
1 2
3
4 5
6Since energy is conserved from point 0 to point 1.
However, between point 1 and 2, energy is lost
432
2
121
122
12
9.01.01.0
1.0
EEEEEEEEEKE
EEQEEQW
KEQ
o
oo
initialfinalNC
Total energy after one bounce
By the same reasoning
o
o
EEEEEE
25
245
9.0)9.0(9.09.09.0
Total energy after two bouncesThe total energy after n bounces is then
)/log()9.0log()/log()9.0log(
/9.09.0
9.09.0
of
ofn
ofn
on
f
on
f
on
f
hhnhh
hhhhmghmgh
EE
7.8)9.0log(
)10.6/44.2log()9.0log(
)/log(
n
n
hhn of
Answer is 8 bounces
Center of Mass The average location for the total mass of a collection of particles (system)
Consider two particles located along a straight line
If m1=m2=m
x
x2
x1m1
m2
cm
xcm
21
2211
mm
xmxmxcm
22
)( 212121 xx
m
xxm
mm
mxmxxcm
N
ii
N
iii
N
ii
N
iii
cm
N
NNcm
mMM
xm
m
xmx
mmmm
xmxmxmxmx
1
1
1
1
321
332211
,
...
...
Also can define the center of mass for multiple particles - N number of particles
Where M is the total mass of the system
Can also define the center of mass along other coordinate axes - ycm
cm
N
iii
N
iii
cmcm
MP
Pm
M
m
mm
mm
v
momentum totalvbut
vor v ,
vvv
21
2211
Can also define the velocity of the center of mass
If linear momentum is conserved, then
cmfcmo
cmfcmofo MMPP
,,
,,
vvvv
The velocity of the center of mass is constant. It is the same before and after a collision.
Return to earlier Example
m1 = 45 kg, m2 = 12 kg, vo1 = +5.1 m/s, vo2 = 0
We found vf1 = +2.95 m/s and vf2 = +8.05 m/s
Now, compute the velocity of the center of mass
xx
mm
xmxmx
mm
mmmm
mm
cm
ffcmf
oocmo
21.01245
)(12)0(450.4
1245
)05.8(12)95.2(45vvv
0.41245
)0(12)1.5(45vvv
21
2211
sm
21
2211,
sm
21
2211,
Taking x1=0 and x=x2-x1=x2