introductory physics-i part 3particle.korea.ac.kr/class/2011/introphy/introphysics-iii.pdf · 1 kg...
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Introductory Physics-IPart 3
Eunil WonDepartment of Physics
Korea University
Fundamentals of Physics by Eunil Won, Korea University 2
Energy and WorkEnergy?
Work? : human labor (wikipedia)
Fundamentals of Physics by Eunil Won, Korea University 3
Energy and Work
Energy: a scalar quantity that is associated with a state (I know this is too vague)
Kinetic energy: K =1
2mv
2
SI unit of kinetic energy (and every other type of energy): Joule (J)
1 Joule = 1 J = 1 kg m2/s2
Work is energy transferred to or from an object by means of a force acting on the object
Fundamentals of Physics by Eunil Won, Korea University 4
Work and Kinetic EnergyConsider a bead that can slide along a frictionless wire, which is stretched along a horizontal x axis
A constant force, directed at an angle to the wire, accelerates the bead
!F!
Fx = max
!d
!v0
: displacement of the bead
: initial velocity
We know from the basic constant-acceleration equations v2
= v2
0 + 2axd
1
2mv
2!
1
2mv
2
0 = Fxd
The work W done on the bead by the force (the energy transfer due to the force) is
W = Fxd W = Fd cos ! W =!F ·
!d
Fundamentals of Physics by Eunil Won, Korea University 5
Work-Kinetic energy TheoremA force does positive (negative) work if it has a vector component in the same (opposite) direction as the displacement
Work-Kinetic Energy Theorem:
SI unit for work: Joule (same as kinetic energy unit)
!K = Kf ! Ki = W!
change in the kineticenergy of a particle
"
=
!
net work done onthe particle
"
!
kinetic energy afterthe net work is done
"
=
!
kinetic energybefore the net work
"
+
!
the network done
"
Kf = Ki + W
Fundamentals of Physics by Eunil Won, Korea University 6
How “large” is one Joule?
1 kg of a rock, from 1 m high, falling on your foot:
W = mgh = 1 kg · 9.8 m/s2 · 1 m ≈ 10 J
So you will fill as much as 10 Joule of pain on your foot!
1 m
1 kg
g = 9.8 m/s2
Fundamentals of Physics by Eunil Won, Korea University 6
How “large” is one Joule?
1 kg of a rock, from 1 m high, falling on your foot:
W = mgh = 1 kg · 9.8 m/s2 · 1 m ≈ 10 J
So you will fill as much as 10 Joule of pain on your foot!
1 m
1 kg
g = 9.8 m/s2
Fundamentals of Physics by Eunil Won, Korea University 7
Sample ProblemIn 1896 in Waco, Texas, William Crush parked two locomotives at opposite ends of a 6.4 km-long track, fired them up, tied their throttles open, and then allowed them to crash head-on at full speed in front of 30,000 spectators. Hundreds of people were hurt by flying debris: several were killed. Assuming each locomotive weighted 1.2x106 N and its acceleration was a constant 0.26 m/s2, what was the total kinetic energy of the two locomotives just before the collision?
v2 = v20 + 2a(x− x0)
v0 = 0 and (x− x0) = 3.2× 103 m
v2 = 0 + 2(0.26 m/s2)(3.2× 103 m)
or
v ∼= 40.8 m/s
mass?
speed?
m =1.2× 106 N
9.8 m/s2= 1.22× 105 kg
K = 2× 1
2mv2 = 2× 1
2(1.22× 105 kg)(40.8 m/s)2
= 2.0× 108 J
Fundamentals of Physics by Eunil Won, Korea University 8
Work Done by a Gravitational ForceThe work done by the gravitational force:
Wg = mgd cos !
In the case of the figure left, the object moves upward:
Wg = mgd cos 1800
= !mgdThe negative sign indicates that the gravitational force transfers energy (mgd) from the kinetic energy of the object
If the object moves downward:
Wg = mgd cos 00
= +mgdThe positive sign indicates that the gravitational force transfers energy (mgd) to the kinetic energy of the object
Fundamentals of Physics by Eunil Won, Korea University 9
Work Done by a Gravitational ForceWork Done in Lifting and Lowering an Object
Suppose we lift an object by applying a vertical force, our applied force does positive work Wa on the object
!K = Kf ! Ki = Wa + Wg
!KThe change in the kinetic energy of the object
If object is stationary before and after the lift, Wa = !Wg
And it becomes Wa = !mgd cos !
Fundamentals of Physics by Eunil Won, Korea University 10
Work Done by a Spring ForceThe Spring Force
Relaxed state
!F = !k!d (Hooke’s law)
k : spring constant (force constant)
The negative sign indicates the spring force is always opposite to the displacement
Note: a spring force is a variable force
!F = !F (x)
(This is the first time to see this type of force)
Fundamentals of Physics by Eunil Won, Korea University 11
Work Done by a Spring Force
Since the spring force is a variable force, we cannot assume
The angle = 0 and the work done within each segment (j) can be written as
The net work Ws done
by spring, from xi to xf is
the sum of all works:
W = Fd cos !
! Fj!x
Ws =!
Fj!xIn the limit as each segment goes to zero,
Ws =
!xf
xi
Fdx
Ws =1
2kx
2
i !1
2kx
2
f Ws = !
1
2kx
2
Using Hooke’s law (F=-kx)
We get
If xi=0 and xf=x,
(work by a spring force)
The work done by a spring force
Ws =
! xf
xi
(!kx)dx = !k
! xf
xi
x dx
= (!1
2k)[x2]
xfxi = (!
1
2k)(x2
f ! x2
i )
Fundamentals of Physics by Eunil Won, Korea University 12
Work Done by a Spring ForceThe work done by an applied force
Now suppose that we displace the block along the x axis while continuing to apply a force to it. Then!Fa
!K = Kf ! Ki = Wa + Ws
If the block is stationary before and after the displacement,
Wa = !Ws
Fundamentals of Physics by Eunil Won, Korea University 13
Work Done by a General Variable ForceConsider one-dimensional variable force (see figure)
Fj,avg
!Wj = Fj,avg!x
: average value of F(x) within the j-th interval (constant)
work done by the force in the j-th interval:
The total work done from xi to xf is then
W =!
!Wj =!
Fj,avg!x
W = lim!x!0
!Fj,avg!x
W =
!xf
xi
F (x)dx
We let Δx approach zero:
then we get the integral of the function F(x) (The usual trick in calculus)
(This becomes the area between the F(x) curve and the x axis)
Fundamentals of Physics by Eunil Won, Korea University 14
Work Done by a General Variable ForceThree dimensional analysis:
consider now a particle in a three-dimensional force !F = Fxi + Fy j + Fz k
Now let the particle move through an incremental displacement
d!r = dxi + dyj + dzk
dW = !F · d!r = Fxdx + Fydy + FzdzThen the increment of work dW is
Since we know how to calculate this component by component, we get
W =
! rf
ri
dW =
! xf
xi
Fxdx +
! yf
yi
Fydy +
! zf
zi
Fzdz
Fundamentals of Physics by Eunil Won, Korea University 15
Work Done by a General Variable ForceWork-Kinetic Energy Theorem with a Variable Force
consider now a particle of mass m, moving along the x axis and acted on a net force F(x) that is directed along that axis
The work done on the particle by this force is: W =
!xf
xi
F (x)dx =
!xf
xi
ma dx
ma dx = mdv
dtdx
= mdv
dx
dx
dtdx
= mdv
dxvdx
= mv dv
Since (chain rule here)
We get W =
! vf
vi
mv dv = m
! vf
vi
v dv
=1
2mv
2
f !
1
2mv
2
ior W = Kf - Ki = ΔK
Fundamentals of Physics by Eunil Won, Korea University 16
PowerPower: the time rate at which work is done by a force
Average power:
Instantaneous power:
Pavg =W
!t
P =
dW
dt
SI unit of power : watt (W) 1 W = 1 J/s
P =
dW
dt=
F cos !dx
dt= F cos !
!
dx
dt
"
= Fv cos !
For a particle that is moving along a straight line and is acted on by a constant force directed at some angle φ to that line:
P =!F · !vIn general case we write:
Fundamentals of Physics by Eunil Won, Korea University 17
Sample ProblemA block of mass m = 0.40 kg slides across horizontal frictionless spring with speed v=0.50 m/s. It then runs into and compresses a spring of spring constant k=750 N/m.
When the block is momentarily stopped by the spring, by what distance d is the spring compressed?
Kf −Ki = −12kd2
d = v
�m
k= 0.50 m/s
�0.40 kg750N/m
= 1.2× 10−2 m = 1.2 cm
Fundamentals of Physics by Eunil Won, Korea University 18
Potential Energy
Potential energy is the energy that can be associated with the configuration
ex) gravitational potential energy elastic potential energy
These two states has two different potential energy
Work and Potential Energy
The change ΔU in gravitational potential energy is defined to equal the negative of the work done on the tomato by gravitational force
!U = !W
Fundamentals of Physics by Eunil Won, Korea University 19
Conservative and Nonconservative Forces
Conservative force:
Nonconservative force:
Work done by a force to change configuration is same but negative to the work done to restore the configuration
ex) gravitational force
A force that is not conservative
ex) kinetic frictional energy (thermal energy cannot be transferred back to kinetic energy)
Fundamentals of Physics by Eunil Won, Korea University 20
Path Independence of Conservative Forces
The net work done by a conservative force on a particle moving around every closed path is zero
1
2mv
2
0
1
2mv
2
0
The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle
Wab,1 + Wba,2 = 0
Wab,1 = !Wba,2 = Wab,2
Fundamentals of Physics by Eunil Won, Korea University 21
Potential Energy ValuesConsider a particle with a conservative force
Gravitational Potential Energy:
W =
!xf
xi
F (x)dx !U = !
!xf
xi
F (x)dxor
!U = !
! yf
yi
(!mg)dy = mg
! yf
yi
dy = mg(yf ! yi)
U(y) = mgy (if we take Ui = yi = 0)
Elastic Potential Energy:
U(x) =1
2kx
2 (if we take Ui = xi = 0)
!U = !
! xf
xi
(!kx)dx = k
! xf
xi
x dx =1
2kx
2
f !
1
2kx
2
i
Fundamentals of Physics by Eunil Won, Korea University 22
Conservation of Mechanical EnergyThe mechanical Energy Emec Emec = K + U
When a conservative force does work W on an object, it transfers energy between K and U
!K = W = !!U
K2 ! K1 = !(U2 ! U1)
K2 + U2 = K1 + U1
!Emec = !K + !U = 0
from the top, we find that
and becomes
Principle of conservation of mechanical energy:
Fundamentals of Physics by Eunil Won, Korea University 23
Conservation of Mechanical Energy
K + U is constant all the time
ex) a pendulum
Fundamentals of Physics by Eunil Won, Korea University 24
Potential Energy Curve
We had this before: !U = !
!xf
xi
F (x)dx
and it becomes at the differential limit: F (x) = !
dU(x)
dx(obtaining the force from the potential)
ex) elastic potential U(x) =1
2kx
2
U(y) = mgy F = !mg
F (x) = !kx
gravitational potential
Fundamentals of Physics by Eunil Won, Korea University 25
Potential Energy CurvePotential Energy Curve
unstable equilibrium(U = 3J)
neutral equilibrium(U = 4J)
stable equilibrium(U = 1J)
Fundamentals of Physics by Eunil Won, Korea University 26
Work Done on a System by an External Force
Work is energy transferred to or from a system by means of an external force acting on that system
Definition of work with external force:
No friction involved:
W = !Emec
Work done on the system is equal to the change in the mechanical energy
Fundamentals of Physics by Eunil Won, Korea University 27
Work Done on a System by an External ForceFriction involved: A constant force F pulls a block, increasing
the block’s velocity from v0 to v
From Newton’s 2nd law, we write
acceleration is constant as forces are constant, so we can use the following equation
v2
= v2
0 + 2ad
F ! fk = ma
a =1
2d(v2
! v2
0)
Fd =1
2mv2
!
1
2mv2
0 + fkd
and,
Fd = !K + fkd
Fd = !Emec + fkd
it becomes
(if we include the vertical motion as well)
!Eth(increase in thermal energy by sliding)
Fundamentals of Physics by Eunil Won, Korea University 28
Conservation of EnergyThe total energy E of a system can change only by amounts of energy that are transferred to or from the system
Isolated system: there can be no energy transfers to or from the isolated system
The total energy E of an isolated system cannot change
W = !E = !Emec + !Eth + !Eint
!Emec + !Eth + !Eint = 0
(internal Energy)
Fundamentals of Physics by Eunil Won, Korea University 28
Conservation of EnergyThe total energy E of a system can change only by amounts of energy that are transferred to or from the system
Isolated system: there can be no energy transfers to or from the isolated system
The total energy E of an isolated system cannot change
W = !E = !Emec + !Eth + !Eint
!Emec + !Eth + !Eint = 0
(internal Energy)
Fundamentals of Physics by Eunil Won, Korea University 29
The Center of Mass
The motion of the baseball bat looks complicated? Yes, but there is a special point moves like a particle (parabolic path) called the center of mass
Fundamentals of Physics by Eunil Won, Korea University 29
The Center of Mass
The motion of the baseball bat looks complicated? Yes, but there is a special point moves like a particle (parabolic path) called the center of mass
For a system of two particles shown in the figure left, the center of mass is defined as
xcom =m1x1 + m2x2
m1 + m2
Fundamentals of Physics by Eunil Won, Korea University 30
The Center of Mass
Fundamentals of Physics by Eunil Won, Korea University 30
The Center of Mass
For a system of n particles along the x axis:(M: total mass of the system)
M = m1 + m2 + ... + mn
Fundamentals of Physics by Eunil Won, Korea University 30
The Center of Mass
xcom =m1x1 + m2x2 + ... + mnxn
M
=1
M
n!
i=1
mixi
For a system of n particles along the x axis:(M: total mass of the system)
M = m1 + m2 + ... + mn
Fundamentals of Physics by Eunil Won, Korea University 30
The Center of Mass
xcom =m1x1 + m2x2 + ... + mnxn
M
=1
M
n!
i=1
mixi
For a system of n particles along the x axis:(M: total mass of the system)
M = m1 + m2 + ... + mn
If the particles are distributed in three dimensions, the center of mass must be identified by three coordinates
Fundamentals of Physics by Eunil Won, Korea University 30
The Center of Mass
xcom =m1x1 + m2x2 + ... + mnxn
M
=1
M
n!
i=1
mixi
For a system of n particles along the x axis:(M: total mass of the system)
M = m1 + m2 + ... + mn
xcom =1
M
n!
i=1
mixi, ycom =1
M
n!
i=1
miyi, zcom =1
M
n!
i=1
mizi,
If the particles are distributed in three dimensions, the center of mass must be identified by three coordinates
Fundamentals of Physics by Eunil Won, Korea University 30
The Center of Mass
xcom =m1x1 + m2x2 + ... + mnxn
M
=1
M
n!
i=1
mixi
For a system of n particles along the x axis:(M: total mass of the system)
M = m1 + m2 + ... + mn
xcom =1
M
n!
i=1
mixi, ycom =1
M
n!
i=1
miyi, zcom =1
M
n!
i=1
mizi,
If the particles are distributed in three dimensions, the center of mass must be identified by three coordinates
!rcom =1
M
n!
i=1
mi!riEquivalent single vector equation:
Fundamentals of Physics by Eunil Won, Korea University 31
The Center of Mass
Fundamentals of Physics by Eunil Won, Korea University 31
The Center of MassSolid Bodies
An ordinary object contains many particles (atoms) that we treat it as a continuous distribution of matter. The coordinates of the center of mass are defined as
Fundamentals of Physics by Eunil Won, Korea University 31
The Center of MassSolid Bodies
An ordinary object contains many particles (atoms) that we treat it as a continuous distribution of matter. The coordinates of the center of mass are defined as
xcom =1
M
!x dm ycom =
1
M
!y dm zcom =
1
M
!z dm
Fundamentals of Physics by Eunil Won, Korea University 31
The Center of MassSolid Bodies
An ordinary object contains many particles (atoms) that we treat it as a continuous distribution of matter. The coordinates of the center of mass are defined as
xcom =1
M
!x dm ycom =
1
M
!y dm zcom =
1
M
!z dm
! =
dm
dV=
M
V
If the object has the uniform density (ρ = constant)
Fundamentals of Physics by Eunil Won, Korea University 31
The Center of MassSolid Bodies
An ordinary object contains many particles (atoms) that we treat it as a continuous distribution of matter. The coordinates of the center of mass are defined as
xcom =1
M
!x dm ycom =
1
M
!y dm zcom =
1
M
!z dm
! =
dm
dV=
M
V
If the object has the uniform density (ρ = constant)
xcom =1
V
!x dV ycom =
1
V
!y dV zcom =
1
V
!z dV
The coordinates of the center of mass become:
Fundamentals of Physics by Eunil Won, Korea University 32
Newton’s 2nd Law for a System of ParticlesFor the system of particles, the vector equation that governs the motion of the center of mass of such a system of particle is:
The center of mass of the fragments would continue to follow the original parabolic path
Fundamentals of Physics by Eunil Won, Korea University 32
Newton’s 2nd Law for a System of ParticlesFor the system of particles, the vector equation that governs the motion of the center of mass of such a system of particle is:
The center of mass of the fragments would continue to follow the original parabolic path
!Fnet = M!acom
Fundamentals of Physics by Eunil Won, Korea University 32
Newton’s 2nd Law for a System of ParticlesFor the system of particles, the vector equation that governs the motion of the center of mass of such a system of particle is:
The center of mass of the fragments would continue to follow the original parabolic path
!Fnet = M!acom
Proof) From the definition of rcom, we write
M!rcom = m1!r1 + m2!r2 + ... + mn!rn !rcom =1
M
n!
i=1
mi!ri
Fundamentals of Physics by Eunil Won, Korea University 32
Newton’s 2nd Law for a System of ParticlesFor the system of particles, the vector equation that governs the motion of the center of mass of such a system of particle is:
The center of mass of the fragments would continue to follow the original parabolic path
!Fnet = M!acom
M!acom = m1!a1 + m2!a2 + ... + mn!an
Differentiating the above with respect to time twice leads to
Proof) From the definition of rcom, we write
M!rcom = m1!r1 + m2!r2 + ... + mn!rn !rcom =1
M
n!
i=1
mi!ri
Fundamentals of Physics by Eunil Won, Korea University 32
Newton’s 2nd Law for a System of ParticlesFor the system of particles, the vector equation that governs the motion of the center of mass of such a system of particle is:
The center of mass of the fragments would continue to follow the original parabolic path
!Fnet = M!acom
M!acom = m1!a1 + m2!a2 + ... + mn!an
Differentiating the above with respect to time twice leads to
M!acom = !F1 + !F2 + ... + !Fn
= !Fnet
From Newton’s 2nd law we know that thusmi!ai =!Fi
Proof) From the definition of rcom, we write
M!rcom = m1!r1 + m2!r2 + ... + mn!rn !rcom =1
M
n!
i=1
mi!ri
Fundamentals of Physics by Eunil Won, Korea University 33
Linear Momentum
Fundamentals of Physics by Eunil Won, Korea University 33
Linear Momentum
!p = m!vLinear Momentum of a particle:
(We will discuss angular momentum later)
Fundamentals of Physics by Eunil Won, Korea University 33
Linear Momentum
!p = m!vLinear Momentum of a particle:
(We will discuss angular momentum later)
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 direction of that force
!Fnet =
d!p
dt
Fundamentals of Physics by Eunil Won, Korea University 33
Linear Momentum
!p = m!vLinear Momentum of a particle:
(We will discuss angular momentum later)
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 direction of that force
!Fnet =
d!p
dt
Linear Momentum of a System of Particles:
!P = !p1 + !p2 + ... + !pn
= m!v1 + m!v2 + ... + m!vn
Fundamentals of Physics by Eunil Won, Korea University 33
Linear Momentum
!p = m!vLinear Momentum of a particle:
(We will discuss angular momentum later)
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 direction of that force
!Fnet =
d!p
dt
Linear Momentum of a System of Particles:
!P = !p1 + !p2 + ... + !pn
= m!v1 + m!v2 + ... + m!vn
!P = M!vcom
From the argument in the previous slide (differentiating with respect to time only once)
!Fnet =
d!P
dt
or (system of particles)
Fundamentals of Physics by Eunil Won, Korea University 34
Conservation of Linear MomentumSuppose that the net force acting on an isolated system of particles is zero
Fundamentals of Physics by Eunil Won, Korea University 34
Conservation of Linear MomentumSuppose that the net force acting on an isolated system of particles is zero
!Fnet = 0d!P
dt= 0 !P = constant
Fundamentals of Physics by Eunil Won, Korea University 34
Conservation of Linear MomentumSuppose that the net force acting on an isolated system of particles is zero
!Fnet = 0d!P
dt= 0 !P = constant
In words, if no net force acts on an isolated system of particles, the total linear momentum of the system cannot change
Law of conservation of linear momentum:
It can also be written as:
!Pi =!Pf
!
total linear momentum
at some initial time ti
"
=
!
total linear momentum
at some later time tf
"
or
Fundamentals of Physics by Eunil Won, Korea University 35
Momentum and Kinetic Energy in CollisionsIf the total kinetic energy of the system of two colliding bodies is unchanged by the collision, it is called an elastic collision
ex) collision of billiard balls
If the kinetic energy of the system is not conserved, it is called an inelastic collision
In a closed, isolated system containing a collision, the total linear momentum cannot change, whether the collision is elastic or inelastic
(another statement of the law of conservation of linear momentum)
ex) collision of two cars
Fundamentals of Physics by Eunil Won, Korea University 36
Inelastic Collisions in One Dimension
We assume the two bodies form our system, which is closed and isolated
!
total momentum !Pi
before the collision
"
=
!
total momentum !Pf
after the collision
"
!p1i + !p2i = !p1f + !p2f
m1v1i + m2v2i = m1v1f + m2v2f
or
completely inelastic collision (two particles stick together):
m1v1i = (m1 + m2)V
V =m1
m1 + m2
v1ior
(V is less than v1i)
Fundamentals of Physics by Eunil Won, Korea University 37
Inelastic Collisions in One Dimension
In a closed, isolated system, the velocity of the center of mass cannot be changed by a collision (no net external force)
after the collision
!P = M!vcom = (m1 + m2)!vcom
!P = !p1i + !p2i
!vcom =
!P
m1 + m2
=!p1i + !p2i
m1 + m2
before the collision
from the above:
(constant)
Fundamentals of Physics by Eunil Won, Korea University 38
Elastic Collisions in One Dimension: total kinetic energy of the system does not change after a collision
Stationary targetm1v1i = m1v1f + m2v2f
1
2m1v
2
1i =1
2m1v
2
1f +1
2m2v
2
2f
(linear momentum)
(kinetic energy)
m1(v1i ! v1f ) = m2v2f
m1(v1i ! v1f )(v1i + v1f ) = m2v2
2f
above two equations can be rewritten as
dividing the latter from the former we get
(v1i + v1f ) = v2f
eliminating v2f from the above
m1v1i ! m1v1f = m2v1i + m2v1f v1f =m1 ! m2
m1 + m2
v1i
v2f =2m1
m1 + m2
v1isimilarly we get
Fundamentals of Physics by Eunil Won, Korea University 39
Elastic Collisions in One Dimension
v1f =m1 ! m2
m1 + m2
v1i
v2f =2m1
m1 + m2
v1i
we obtained
Equal mass: m1 = m2 v1f = 0 and v2f = v1ithen
ex)
A massive target: m2 ! m1 then v1f ! "v1i and v2f !
2m1
m2
v1i
A massive projectile: m1 ! m2 v1f ! v1i and v2f ! 2v1ithengolf ball
bowling ball
Fundamentals of Physics by Eunil Won, Korea University 40
Elastic Collisions in One Dimension
Moving Target: m1v1i + m2v2i = m1v1f + m2v2f
1
2m1v
2
1i +1
2m2v
2
2i =1
2m1v
2
1f +1
2m2v
2
2f
m1(v1i ! v1f ) = !m2(v2i ! v2f )
m1(v1i ! v1f )(v1i + v1f ) = !m2(v2i ! v2f )(v2i + v2f )
(linear momentum)
(kinetic energy)
rewriting above equations we get
dividing the latter from the former we get (v1i + v1f ) = (v2i + v2f )
m1(v1i ! v1f ) + m2v2i = m2v2f
v1f =m1 ! m2
m1 + m2
v1i +2m2
m1 + m2
v2i v2f =2m1
m1 + m2
v1i +m2 ! m1
m1 + m2
v2i
since
similar way
m2(v1i + v1f ) = m2v2i + m1(v1i ! v1f ) + m2v2i
Finally solving for v1f:
Fundamentals of Physics by Eunil Won, Korea University 41
System with Varying Mass: A RocketWe discuss now a system with varying mass. An example would be a rocket: the most confusing problem in my class
In a) M: mass of the rocket v: speed of the rocket at an arbitrary time t
In b) M + dM: mass of the rocket (dM is negative quantity) v + dv : speed of the rocket at an arbitrary time t+dt -dM : mass of the exhaust products U : velocity of the exhaust products
Linear momentum of the system must be conserved during dt: (assume no gravitational force)
Pi = Pf
Mv = !dM U + (M + dM)(v + dv)
(we observe the rocket from the ground)
Fundamentals of Physics by Eunil Won, Korea University 42
System with Varying Mass: A Rocket
vrel: relative speed between the rocket and the exhaust products
In symbols, (v + dv) = vrel + U or
U = v + dv ! vrel
Mv = !dM U + (M + dM)(v + dv)
Substituting this result for U into the below equation (that we derived just before)
Mv = !dM(v + dv ! vrel) + (M + dM)(v + dv)
= !dMv ! dMdv + dMvrel + Mv + Mdv + dMv + dMdv
!dMvrel = Mdv
!
dM
dtvrel = M
dv
dt
or
!
velocity of rocketrelative to us
"
=
!
velocity of rocketrelative to products
"
+
!
velocity of productsrelative to us
"
Fundamentals of Physics by Eunil Won, Korea University 43
System with Varying Mass: A Rocket
We replace dM/dt (the rate at which the rocket loses mass) by -R (R: positive mass rate of fuel consumption)
!
dM
dtvrel = M
dv
dt
Then, the previous equation becomes
Rvrel = Ma
thrust
Velocity?
dv = !vrel
dM
M! vf
vi
dv = !vrel
! Mf
Mi
dM
M
vf ! vi = vrel lnMi
Mf