Linear momentum and Linear momentum and Collisions Collisions Chapter 9 Chapter 9
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Linear momentum and CollisionsLinear momentum and Collisions
Chapter 9Chapter 9
Center of mass and linear momentumCenter of mass and linear momentum
I.I. The center of massThe center of mass - System of particles / - Solid body- System of particles / - Solid body
II.II. Newton’s Second law for a system of particlesNewton’s Second law for a system of particles
III.III. Linear Momentum Linear Momentum - System of particles / - Conservation- System of particles / - ConservationIV. Collision and impulseIV. Collision and impulse
- Single collision / - Series of - Single collision / - Series of collisionscollisions
V. Momentum and kinetic energy in collisionsV. Momentum and kinetic energy in collisions
VI. Inelastic collisions in 1DVI. Inelastic collisions in 1D -Completely inelastic collision/ -Completely inelastic collision/
Velocity of COMVelocity of COM
VII. Elastic collisions in 1DVII. Elastic collisions in 1DVIII. Collisions in 2DVIII. Collisions in 2D
IX. Systems with varying massIX. Systems with varying mass
X. External forces and internal energy X. External forces and internal energy changeschanges
I. Center of massI. Center of mass
The center of mass of a body or a system of The center of mass of a body or a system of bodies is the point that moves as though all the bodies is the point that moves as though all the mass were concentrated there and all external mass were concentrated there and all external forces were applied there.forces were applied there.
System of particles:System of particles:
Origin of reference system Origin of reference system coincides with coincides with mm11
dmm
mxcom
21
2
Two particles of masses Two particles of masses mm11 and and mm22 separated by a distance separated by a distance dd
System of particles:System of particles:
M
xmxm
mm
xmxmxcom
2211
21
2211
General:General:
The center of mass lies somewhere between the two particles.The center of mass lies somewhere between the two particles.
Choice of the reference Choice of the reference origin is arbitrary origin is arbitrary Shift Shift of the coordinate system of the coordinate system but center of mass is still but center of mass is still at the same distance at the same distance from each particle. from each particle.
MM = total mass of the system = total mass of the system
3D:3D:
n
iiicom
n
iiicom
n
iiicom zm
Mzym
Myxm
Mx
111
111
System of particles:System of particles:
We can extend this equation to a general situation for We can extend this equation to a general situation for nn particles particles that strung along that strung along xx-axis. The total mass of the system -axis. The total mass of the system M=mM=m11+m+m22+m+m33+……+m+……+mnn The location of center of the mass: The location of center of the mass:
n
iii
nncom xm
MM
xmxmxmxmx
1
332211 1........
3D: 3D: The vector formThe vector form
Position of the Position of the particle:particle:
n
iiicom rm
Mr
1
1 MM = mass of the = mass of the objectobject
System of particles:System of particles:
kzjyixr iiiiˆˆˆ
Position COMPosition COM kzjyixr comcomcomcomˆˆˆ
Solid bodies:Solid bodies:
Continuous distribution of matter. Particles = Continuous distribution of matter. Particles = dmdm (differential mass elements).(differential mass elements).
3D:3D: dmzM
zdmyM
ydmxM
x comcomcom111
MM = mass of the object = mass of the object
Assumption:Assumption:V
M
dV
dm
dVzV
zdVyV
ydVxV
x comcomcom111
Uniform objects Uniform objects uniform density uniform density
The center of mass of an object with a point, line or The center of mass of an object with a point, line or plane of symmetry lies on that point, line or plane.plane of symmetry lies on that point, line or plane.
The center of mass of an object does not need to lie The center of mass of an object does not need to lie within the object (Examples: doughnut, horseshoe )within the object (Examples: doughnut, horseshoe )
Problem solving tactics:Problem solving tactics:
(1)(1) Use object’s symmetry.Use object’s symmetry.
(2) If possible, divide object in several parts. Treat each of (2) If possible, divide object in several parts. Treat each of
these parts as a particle located at its own center of mass.these parts as a particle located at its own center of mass.
(3)(3) Chose your axes. Use one particle of the system as origin Chose your axes. Use one particle of the system as origin
of your reference system or let the symmetry lines be your of your reference system or let the symmetry lines be your
axis.axis.
II. Newton’s second law for a system of particlesII. Newton’s second law for a system of particles
Center of the mass of the system moves as a particle Center of the mass of the system moves as a particle whose mass is equal to the total mass of the system.whose mass is equal to the total mass of the system.
Motion of the center of mass:Motion of the center of mass:
comnet aMF
FFnetnet is the is the net of all external forcesnet of all external forces that act on the that act on the
system. Internal forces (from one part of the system to system. Internal forces (from one part of the system to another are not included).another are not included).
The system is closed: no mass enters or leaves the The system is closed: no mass enters or leaves the system during the movement. (system during the movement. (MM=total system mass).=total system mass). aacomcom is the acceleration of the system’s center of mass.is the acceleration of the system’s center of mass.
zcomznetycomynetxcomxnet MaFMaFMaF ,,,,,,
Prove:
(*)......
...
......
...
321332211
332211
22
11
332211
nnncom
nncom
nn
com
nncom
FFFFamamamamaM
vmvmvmvmvMdt
rdm
dt
rdm
dt
rdm
dt
rdM
rmrmrmrmrM
(*) includes forces that the particles of the system exert on (*) includes forces that the particles of the system exert on each other (internal forces) and forces exerted on the each other (internal forces) and forces exerted on the particles from outside the system (external).particles from outside the system (external).
Newton’s third lawNewton’s third law internal forces from third-law force internal forces from third-law force pairs cancel out in the sum (*) pairs cancel out in the sum (*) Only external forces.Only external forces.
III. Linear momentumIII. Linear momentum
vmp
The linear momentum of a particle is a vector The linear momentum of a particle is a vector p p defined as:defined as:
Momentum is a vector with magnitude equal Momentum is a vector with magnitude equal mvmv and have and have direction ofdirection of v v ..
SI unitSI unit of the momentum isof the momentum is kg-meter/secondkg-meter/second
Newton II lawNewton II law in terms of momentum:in terms of momentum:
amdt
vdm
dt
vmd
dt
pdFnet
)(
The time rate of change of the momentum of a particle is The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the equal to the net force acting on the particle and is in the direction of the force.direction of the force.
System of particles:System of particles:
nnn vmvmvmvmppppP
....... 332211321
The total linear moment The total linear moment PP is the vector sum of the is the vector sum of the individual particle’s linear momentum.individual particle’s linear momentum.
comvMP
The linear momentum of a system of particles is The linear momentum of a system of particles is equal to the product of the total mass M of the equal to the product of the total mass M of the system and the velocity of the center of mass.system and the velocity of the center of mass.
dt
PdFaM
dt
vdM
dt
Pdnetcom
com
Net external force acting on the system.Net external force acting on the system.
Conservation:Conservation:
If no external force acts on a closed, isolated system of If no external force acts on a closed, isolated system of particles, the total linear momentum particles, the total linear momentum P P of the system cannot of the system cannot change.change.
ifnet PPdt
systemisolatedClosedconstP
0
),(
Closed:Closed: no matter passes through the system boundary in no matter passes through the system boundary in any direction.any direction.
If the component of the net external force on a closed If the component of the net external force on a closed system is zero along an axis system is zero along an axis component of the linear component of the linear momentum along that axis cannot change.momentum along that axis cannot change.
The momentum is constant if no external forces act on a The momentum is constant if no external forces act on a closed particle system. internal forces can change the linear closed particle system. internal forces can change the linear momentum of portions of the system, but they cannot momentum of portions of the system, but they cannot change the total linear momentum of the entire system.change the total linear momentum of the entire system.
If no net external force acts on the system of particles the total If no net external force acts on the system of particles the total linear momentum linear momentum PP of the system cannot change. of the system cannot change.
Each component of the linear momentum is conserved Each component of the linear momentum is conserved separately if the corresponding component of the net external separately if the corresponding component of the net external force is zero.force is zero.
Conservation of Linear MomentumConservation of Linear Momentum
IV. Collision and impulseIV. Collision and impulse
Collision:Collision: isolated event in which two or more bodies exert isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time.relatively strong forces on each other for a relatively short time.
Single collisionSingle collision
pppdttFJ
dttFpddttFpddt
ift
t
t
t
p
p
f
i
f
i
f
i
)(
)()(
Measures the strength and duration of the collision Measures the strength and duration of the collision forceforce
Third law force pairThird law force pair
FFR R = - F= - FLL
Impulse:Impulse:
Impulse-linear momentum theorem Impulse-linear momentum theorem
The change in the linear momentum of a body in a The change in the linear momentum of a body in a collision is equal to the impulse that acts on that body.collision is equal to the impulse that acts on that body.
Jppp if
Units:Units: kg m/skg m/s
zzizfz
yyiyfy
xxixfx
Jppp
Jppp
Jppp
tFJ avg
FFavgavg such that: such that:
Area under Area under F(t)F(t) vs vs ΔΔtt curve curve = Area under= Area under F Favgavg vs vs tt
Series of collisionsSeries of collisionsTarget fixed in place Target fixed in place n-projectiles n-projectiles n n ΔΔpp = Total = Total change in linear momentum change in linear momentum (projectiles)(projectiles)
pnJ Impulse on the target:Impulse on the target:JJ and and ΔΔpp have opposite have opposite directions, directions, ppff < < ppii ΔΔpp
left left JJ to the right. to the right.
vmt
np
t
n
t
JFavg
n/n/ΔΔtt Rate at which the Rate at which the projectiles collide with the projectiles collide with the target.target.
a) Projectiles stop upon impact:a) Projectiles stop upon impact: ΔΔv = vv = vff-v-vi i = 0-v = -v= 0-v = -v
b) Projectiles bounce:b) Projectiles bounce: ΔΔv = vv = vff-v-vi i = -v-v = -2v= -v-v = -2v
vt
mFtinnmm avg
ΔΔm/m/ΔΔtt Rate at which mass Rate at which mass collides with the target.collides with the target.
V. Momentum and kinetic energy in collisionsV. Momentum and kinetic energy in collisions
Assumptions:Assumptions: Closed systemsClosed systems (no mass enters or leaves (no mass enters or leaves them) them)
Isolated systemsIsolated systems (no external forces act on (no external forces act on the bodies within the system) the bodies within the system)
Elastic collision:Elastic collision: If the total kinetic energy of the system of If the total kinetic energy of the system of two colliding bodies is unchanged two colliding bodies is unchanged (conserved) by the collision.(conserved) by the collision.
Inelastic collision:Inelastic collision: The kinetic energy of the system is not The kinetic energy of the system is not conserved conserved some goes into thermal some goes into thermal energy, sound, etc.energy, sound, etc.
Completely inelastic collision:Completely inelastic collision: After the collision the bodies lose After the collision the bodies lose energy and stick together.energy and stick together.
Velocity of the center of mass:Velocity of the center of mass:
21
21
21
21
21
2121
21 )(
mm
pp
mm
pp
mm
Pv
ppppconservedP
vmmvMP
ffiicom
ffii
comcom
In a closed, isolated In a closed, isolated system, the velocity of system, the velocity of COM of the system cannot COM of the system cannot be changed by a collision. be changed by a collision. (No net external force).(No net external force).
Completely inelastic Completely inelastic collision collision v = vv = vcomcom
VII. Elastic collisions in 1DVII. Elastic collisions in 1D)()( collisionafterenergykineticTotalcollisionbeforeenergykineticTotal
ffi vmvmvm 221111 Closed, isolated Closed, isolated system system
In an elastic collision, the In an elastic collision, the kinetic energy of each kinetic energy of each colliding colliding body may change, body may change, but the total kinetic energy of but the total kinetic energy of the system does not change.the system does not change.
Stationary target:Stationary target:
Linear momentumLinear momentum
222
211
211 2
1
2
1
2
1ffi vmvmvm Kinetic energyKinetic energy
)2())(()(
)1()(
11111222
21
211
22111
fififfi
ffi
vvvvmvmvvm
vmvvm
ifif
ifiiff
fif
fif
vmm
mvv
mm
mmv
vvvm
mvvvin
vvm
mvFrom
vvvDividing
121
121
21
211
1112
1121
112
12
112
2
)()3()1(
)()1(
)3()1/()2(
v2f >0 alwaysv1f >0 if m1>m2 forward mov.v1f <0 if m1<m2 bounce back
Stationary target:Stationary target:
Equal masses: Equal masses: mm11=m=m2 2 v v1f1f=0 and v=0 and v2f 2f = v= v1i1i In head-on In head-on
collisions bodies of equal masses simply exchange collisions bodies of equal masses simply exchange velocities.velocities.
Massive target:Massive target: mm22>>m>>m1 1 v v1f 1f ≈ -v≈ -v1i1i and and v v2f 2f ≈ ≈
(2m(2m11/m/m22)v)v1i1i Body 1 bounces back with Body 1 bounces back with
approximately same speed. Body 2 moves forward approximately same speed. Body 2 moves forward at low speed.at low speed.
Massive projectile:Massive projectile: mm11>>m>>m2 2 v v1f 1f ≈ v≈ v1i1i and and v v2f 2f ≈ ≈
2v2v1i1i Body 1 keeps on going barely lowed by the Body 1 keeps on going barely lowed by the
collision. Body 2 charges ahead at twice the initial collision. Body 2 charges ahead at twice the initial speed of the projectile.speed of the projectile.
ifif vmm
mvv
mm
mmv 1
21
121
21
211
2
VII. Elastic collisions in 1DVII. Elastic collisions in 1D
ffii vmvmvmvm 22112211 Closed, isolated system Closed, isolated system
iif
iif
fifififi
fifi
vmm
mmv
mm
mv
vmm
mv
mm
mmvDividing
vvvvmvvvvm
vvmvvm
221
121
21
12
221
21
21
211
2222211111
222111
2
2)1/()2(
)2())(())((
)1()()(
Moving target:Moving target:
Linear momentumLinear momentum
222
211
222
211 2
1
2
1
2
1
2
1ffii vmvmvmvm Kinetic energyKinetic energy
VIII. Collisions in 2DVIII. Collisions in 2D
Closed, isolated system Closed, isolated system
ffii PPPP 2121
Linear momentum conservedLinear momentum conserved
ffii KKKK 2121 Kinetic energy conservedKinetic energy conserved
Elastic collision Elastic collision
Example:Example:
222
211
211 2
1
2
1
2
1ffi vmvmvm
22211111 coscos ffi vmvmvmaxisx
222111 sinsin0 ff vmvmaxisy
If the collision is elastic If the collision is elastic
IV. Systems with varying massIV. Systems with varying mass
)()( dvvdMMUdMMv
Example:Example: most of the mass of a rocket on its launching is fuel most of the mass of a rocket on its launching is fuel that gets burned during the travel.that gets burned during the travel.
System:System:rocket + exhaust productsrocket + exhaust productsClosed and isolatedClosed and isolated mass of this mass of this system does not change as the system does not change as the rocket accelerates. rocket accelerates. PP=const =const PPii=P=Pff
Linear momentum of Linear momentum of exhaust products exhaust products released during the released during the interval interval dtdt
Linear momentumLinear momentumof rocket at the of rocket at the end of end of dtdt
dM < 0dM < 0AfterAfter dtdt
MdvdMvMvdvdMvdMMdvMvdMvdvdMvdMMv
dvvdMMvdvvdMMv
relrel
rel
))((])[(
RR=Rate at which the rocket losses mass=Rate at which the rocket losses mass= = -dM/dt-dM/dt = rate of = rate of fuel consumptionfuel consumption
MavRdt
dvMv
dt
dMrelrel
First rocket First rocket equationequation
Velocity of rocket relative to frameVelocity of rocket relative to frame = (velocity of rocket relative = (velocity of rocket relative to products)+ (velocity of products relative to frame)to products)+ (velocity of products relative to frame)
relrel vdvvUUvdvv )()(
)()( dvvdMMUdMMv
dt
dvMv
dt
dMdMvMdv relrel
f
irelif
ifrel
v
v
M
Mrelrelrel
M
Mvvv
MMvM
dMvdvv
M
dMdv
dt
dvMv
dt
dM f
i
f
i
ln
lnln
Second rocket equationSecond rocket equation
Two blocks of masses Two blocks of masses MM and 3 and 3MM are placed on a horizontal, are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the cord initially holding the blocks together is burned; after this, the block of mass 3block of mass 3MM moves to the right with a speed of 2.00 m/s. moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass (a) What is the speed of the block of mass MM? (b) Find the ? (b) Find the original elastic potential energy in the spring if original elastic potential energy in the spring if MM = 0.350 kg. = 0.350 kg.
(a) (a) For the system of two blocksFor the system of two blocks 0p
or or i fp p
Therefore,Therefore, 0 3 2.00 m smMv M
Solving gives: Solving gives:
6.00 m smv (motion toward the left).(motion toward the left).
(b) (b) 2 2 23
1 1 13 8.40 J
2 2 2M Mkx Mv M v
A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart initially at rest (Figure P9.55). The person slides on the cart’s top surface and finally initially at rest (Figure P9.55). The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.400. Friction between the cart and ground can be person and the cart is 0.400. Friction between the cart and ground can be neglected. (a) Find the final velocity of the person and cart relative to the ground. neglected. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the friction force acting on the person while he is sliding across the top (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (c) How long does the friction force act on the person? (d) Find surface of the cart. (c) How long does the friction force act on the person? (d) Find the change in momentum of the person and the change in momentum of the cart. the change in momentum of the person and the change in momentum of the cart. (e) Determine the displacement of the person relative to the ground while he is (e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (f) Determine the displacement of the cart relative to the ground sliding on the cart. (f) Determine the displacement of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (i) Explain why the answers to (g) and Find the change in kinetic energy of the cart. (i) Explain why the answers to (g) and (h) differ. (What kind of collision is this, and what accounts for the loss of (h) differ. (What kind of collision is this, and what accounts for the loss of
mechanical energy?) mechanical energy?)
(a)(a)
(b)(b)
60.0 kg 4.00 m s 120 60.0 kg fv
ˆ1.33 m sf v i
0yF 260.0 kg 9.80 m s 0n
0.400 588 N 235 Nk kf n
ˆ235 Nk fi
(c) For the person,(c) For the person, i fp I p
i fmv Ft mv
60.0 kg 4.00 m s 235 N 60.0 kg 1.33 m s
0.680 s
t
t
(d)(d) person: person:
cart: cart:
ˆ60.0 kg 1.33 4.00 m s 160 N sfim m v v i
ˆ120 kg 1.33 m s 0 160 N s i
(e)(e) 1 14.00 1.33 m s 0.680 s 1.81 m
2 2fi i fx x v v t
1 10 1.33 m s 0.680 s 0.454 m
2 2fi i fx x v v t (f)(f)
2 22 21 1 1 160.0 kg 1.33 m s 60.0 kg 4.00 m s 427 J
2 2 2 2fimv mv (g)(g)
22 21 1 1120.0 kg 1.33 m s 0 107 J
2 2 2fimv mv (h)(h)
(i)(i) The force exerted by the person on the cart must equal in magnitude and The force exerted by the person on the cart must equal in magnitude and opposite in direction to the force exerted by the cart on the person. The opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and changes in momentum of the two objects must be equal in magnitude and must add to zero. Their changes in kinetic energy are different in must add to zero. Their changes in kinetic energy are different in magnitude and do not add to zero. The following represent two ways of magnitude and do not add to zero. The following represent two ways of thinking about “way”. The distance the cart moves is different from the thinking about “way”. The distance the cart moves is different from the distance moved by the point of applicatio9n of the friction force to the cart. distance moved by the point of applicatio9n of the friction force to the cart. The total change of mechanical energy for both objects together, -320J, The total change of mechanical energy for both objects together, -320J, becomes +320J of additional internal energy in this perfectly inelastic becomes +320J of additional internal energy in this perfectly inelastic collision.collision.
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