physics 1 chapter 4: linear momentum and collisions … · · 2017-11-07checkpoint 3 (p. 210) (a)...
TRANSCRIPT
4.1. The Center of Mass, Newton’s Second
Law for a System of Particles
4.2. Linear Momentum and Its Conservation
4.3. Collision and Impulse
4.4. Momentum and Kinetic Energy in Collisions
Chapter 4: Linear Momentum and Collisions
Physics 1
4.1. The Center of Mass. Newton’s Second Law for a System of Particles
4.1.1. The center of mass
•Consider a system of 2 particles of masses m1
and m2 separated by distance d:
dmm
mxcom
21
2
•If m1 at x1 and m2 at x2:
M
xmxm
mm
xmxmxcom
2211
21
2211
where M is the total mass of the system
•If the system has n particles that are strung out along the x axis:
n
i
iinn
com xmMM
xmxmxmx
1
2211 1...
a. Systems of Particles
•If the n particles are distributed in three dimensions:
n
i
iicom
n
i
iicom
n
i
iicom zmM
zymM
yxmM
x111
1,
1,
1
kzjyixr iiiiˆˆˆ
•If the position of particle i is given by a vector:
kzjyixr comxomcomcomˆˆˆ
•The center of mass of the system is determined by:
n
i
iicom rmM
r1
1
b. Solid Bodies
zdmM
zydmM
yxdmM
x comcomcom
1,
1,
1
where M is the mass of the object
•For uniform objects, their density are:
V
M
dV
dm
dVV
Mdm
zdVV
zydVV
yxdVV
x comcomcom
1,
1,
1
Sample Problem (p. 204)
Determine the center of mass of the plate
0
PS
PPSSPS
mm
xmxmx
P
SSP
m
mxx
3
1
)2( 22
2
RR
R
area
area
thickness
thickness
m
m
P
S
P
S
P
S
P
S
RxS
RxP3
1
4.1.2. Newton’s Second Law for a System of Particles
)1(comnet aMF
forces external all of forcenet the:netF
system theof mass ofcenter theofon accelerati the:coma
system theof mass total the:M
zcomznetycomynetxcomxnet MaFMaFMaF ,,,,,,
Proof of Equation (1):
n
i
iicom rmrM1
net
n
i
i
n
i
iicom
n
i
iicom FFamaMvmvM
111
;
4.2. Linear Momentum and Its Conservation
The linear momentum of a particle is a vector quantity defined as:p
vmp
ly.respective particle, theof velocity theand mass theare and where vm
Newton’s second law is expressed in terms of momentum:
dt
pdFnet
particle. on the force externalnet theis where netF
Checkpoint 3 (p. 210)(a) rank the magnitude of forces(b) in which region is the particleslowing?
1, 3, 2, 4; 3
a. Linear Momentum
(Unit: kg m/s)
•For a system of particles:
nnn vmvmvmpppP
...... 221121
The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass.
comcom aM
dt
vdM
dt
Pd
dt
PdFnet
comvMP
Question: Why do we need momentum?
Because momentum provides us a tool for studying collisionof 2 or more objects.
b. Conservation of Linear Momentum:If the net external force acting on a system of particles is zero,
0netF
constantP
:z)or y, x,X(0 If , XnetFconstantXP
Homework: 2, 5, 13, 14, 22 (pages 230-233)
4.3. Collision and Impulse• Consider a collision between a bat and a ball:
The change in the ball’s momentum is:
dttFpd )(
from a time ti to a time tf:
f
i
f
i
t
t
t
tdttFpd )(
• The impulse of the collision is defined by:
f
i
t
tdttFJ )(
Jp
the change in the object’s momentum
the impulse ofthe object
(Unit: kg m/s)
If we do not know the F(t) function, we can use:
tFJ avg
Examples:
1. A 0.70 kg ball is moving horizontally with a speed of 5.0 m/s whenit strikes a vertical wall. The ball rebounds with a speed of 2.0 m/s.What is the magnitude of the change in linear momentum of the ball?
vmpvmp
;Since the ball is moving horizontally, therefore, this is one dimensional motion:
xx vmp
)( ifxx vvmvmp
m/s) (kg 4.95)-(-20.7 :m/s 5 m/s; 2 xif pvv
ivmp ˆkg.m/s)9.4(
x
iv
fv
2. A 1500-kg car travelling at a speed of 5.0 m/s makes a 900 turn in a time of 3.0 s and emerges from this turn with a speed of 3.0 m/s: (a) What is the magnitude of the impulse that acts on the car during this turn? Draw the impulse vector. (b) What is the magnitude of the average force on the car during this turn? (Final exam, June 2014)
iv
fv
ip
fppJ
(a)
if pppJ
)m/s kg(874622 ifppJ
(b)
(N)29153
8746avg
t
JF
4.4. Momentum and Kinetic Energy in Collisions
Three types of collisions: We consider a system of 2 bodies
1. Inelastic collision:
constantP :momentum total
ffii pppp 2121
constantKESome energy (KE) is transferred to other forms, e.g. heat, sound.
constant21
21
21
21
21
mm
pp
mm
pp
mm
Pv
ffiicom
2. Elastic collision: conserved. are and KEp
ffii pppp 2121
ffii KKKK 2121
• In one dimension: ffii vmvmvmvm 22112211
2
222
112
222
112
1
2
1
2
1
2
1ffii
vmvmvmvm
Special cases:
:02 iv
if
if
vmm
mv
vmm
mmv
1
21
12
1
21
211
2
:21 mm iff vvv 121 ;0
:12 mm ifif v
m
mvvv 1
2
1211
2;
:21 mm ifif vvvv 1211 2;
3. Perfectly inelastic collision: two bodies stick together after collision:
KE.not but conserved p
:21 fff vvv fii vmmvmvm )( 212211
Case 3
3.1. In one dimension:
3.2. In two dimensions:
fii vmmvmvm
)( 212211
Example: (Perfectly inelastic collision)
A 1000-kg car travelling east at 80.0 km/h collides with a 3000 kg car traveling south at 50.0 km/h. The two cars stick together after the collision. What is the speed of the cars after the collision?(Final exam, June 2014)
fii vmmvmvm
)( 212211
80 km/h
50 km/hfii ppp
21
ip1
ip2
fp
)km/h kg(170000
0.5030000.801000 2222
21
f
iif
p
ppp
m/s 11.8or )km/h(5.42)( 21
mm
pv
ff
Homework: 25, 38, 49, 56, 67, 60, 64, 74 (p. 233-237)