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CHAPTER-9
Center of Mass and Linear Momentum
Ch 9-2 The Center of Mass
 Center of mass (com) of a
system of particles is
required to describe the
position and motion of the
system
 The com of a system of
particles is the point that
moves as if the entire mass
of the system were
concentrated there and all
external forces were
applied there
Ch 9-2 The Center of Mass
com is defined with
reference to origin of an
axis
For Fig. b
Xcom=(m1x1+m2x2)/(m1+m2)
=(m1x1+m2x2)/M
where M= m1+m2 =mi
xcom =mixi / M= mixi/mi
ycom=miyi /M;
zcom=miyi /M
rcom= xcom i +ycom j +zcom k
= miri / M
Ch 9-2 The Center of Mass
(solid bodies)
For a solid body having continuous distribution of
matter, the particle becomes differential mass
element dm
xcom=(1/M) x dm ; ycom=(1/M) y dm ;
zcom=(1/M) z dm
M is mass of the object and its density  are
related to its volume through  = M/V =dm/dV
Then xcom=(1/V) x dV ; ycom=(1/V) y dV ;
zcom=(1/V) z dV
Ch-9 Check Point 1
The figure shows a uniform square plate from
which four identical squares at the corners
will be removed
a) where is com of plate originally?
b) where it is after removal of square 1
c) after removal of square 1 and 2
d) after removal of square 1 and 3
e) after removal of square 1, 2 and 3
f) All four square
Answers in term of Quadrant, axes or points
(a) origin;
(b) fourth quadrant;
(c) on y axis below
origin;
(d) origin;
(e) third quadrant;
(f) origin
Ch 9-3 Newton’s Second Law for a
System of Particles
For a system of particles with com defined by:
rcom = miri / M , Newton’s Second law : Fnet=M acom
Mrcom = miri ;
differentiating w.r.t time
Mvcom= mivi ;
differentiating w.r.t time
Macom= miai = Fi= Fnet
Fnet-x= Macom-x; Fnet-y= Macom-y; Fnet-z= Macom-z
Ch 9-4,5 Linear Momentum
 Linear momentum of a particle p- a vector quantity p = mv
(linear momentum of a particle)
 Newton’s second law of motion:
Time rate change of the momentum of a particle is equal to
the net force acting on the particle and is in the direction of
force
Fnet = d/dt (p) = d/dt (mv) = m dv/dt = ma
 Linear momentum P of a system of particles is
vector sum of individuals particle’s linear momenta p
P = pi = mivi =Mvcom (System of particles)
Fnet = d/dt (P)= Macom (System of particles)
Ch-9 Check Point 3
The figure gives the magnitude of
the linear momentum versus
time t for a particle moving
along an axis. A force
directed along the axis acts
on the particle.
(a) Rank the four regions
indicated according to the
magnitude of the force,
greatest first
b) In which region particle is
slowing
F= dp/dt
Region 1:
2:
3:
4:
Ans:
(a) 1, 3,
(zero
(b) 3
largest slope
Zero slope
Negative slope
Zero slopes
then 2 and 4 tie
force);
Ch 9-6 Collision and Impulse
 Momentum p of any point- like
object can be changed by
application of an external force
 Single collision of a moving
particle-like object (projectile)
with another body ( target)
 Ball (Projectile-R) – bat (target-L)
system
 Change in momentum of ball in time
dt , dp=F(t) dt.
 Net change  dp=  F(t) dt.
 Impulse J=  F(t) dt. = Favg t
 Change in momentum P =Pf-Pi =J
Ch 9-7 Conservation of Linear Momentum
 If Fnet-external = 0 then J=  F(t) dt =0
 P =Pf-Pi =J=0; then Pf = Pi
 Law of conservation of linear momentum:
 If no net external force acts on a system of
particles, the total linear momentum of the system
of particles cannot change.
 Pfx = Pix ; Pfy = Piy
Ch-9 Check Point 4
A paratrooper whose chute
fails to open lands in snow,
he slightly hurt. Had he
landed on bare ground, the
stoppong time would be 10
times shorter and the
collision lethal. Does the
presence of snow increases,
decreases or leaves
unchanged the values of
(a) the paratrooper change in
momentum
(b) the impulse stopping the
paratrooper
C) the force stopping the
paratrooper
Answer:
(a)p =m(vf-vi)
unchanged;
(b) J= p ; unchanged;
(c) J=F.dt ; t increase,
F decrease
Ch-9 Check Point 5
The figure shows an overhead view
of a ball bouncing from a
vertical wall without any change
in its speed. Consider the
change p in the balls’ linear
momentum
a) Is px positive, negative, or
zero
b) Is px positive, negative, or
zero
b) What is direction of p?
q
q
x
px=pxf-pxi
= 0
py=pyf-pyi=pyf-(-pyi)
=positive
Direction of P towards y-axis
Ch-9 Check Point 6
An initially stationary device
lying on a frictionless floor
explodes into two pieces,
which then slides across the
floor. One piece slides in
the positive direction of an
x axis.
a) What is the sum of the
momenta of the two pieces
after the explosion?
b) Can the second piece move
at an angle to the x-axis?
c) What is the direction of
the momentum of the
second piece?
a) Pi= Pf =0
b) No because
Pf = 0=P1fx+P2
Then P2 =-P1fx
c) Negative x-axis
Ch 9-8 Momentum and Kinetic Energy in
collisions
 Elastic collision: Momentum and kinetic energy of
the system is conserved
 then Pf = Pi and Kf = Ki
 Inelastic collision: Momentum of the system is
conserved but kinetic energy of the system is not
conserved
 then Pf = Pi and Kf  Ki
 Completely inelastic collision: Momentum of the
system is conserved but kinetic energy of the
system is not conserved. After the collision the
colliding bodies stick together and moves as a one
body.
Ch-9-9 Inelastic Collision in one
Dimension
 One-dimensional inelastic collision
 For a two body system
Total momentum Pi before collision =
Total momentum Pf after collision
p1i+p2i=p1f+p2f
m1v1i+m2v2i=m1v1f+m2v2f
 One-dimensional completely inelastic
collision (v2i=0)
m1v1i=(m1+m2)V
m1v1i/(m1+m2)
Hence Vv1i [m1  (m1+m2)]
Ch-9-9: Velocity of the Center of Mass
 For One-dimensional completely
inelastic collision of a two body system
 P=(m1+m2)vcom
 Pi=Pf and m1v1i=(m1+m2)V
Pi=(m1+m2)vcom= m1v1i and Pf=(m1+m2)vcom
=(m1+m2)V
 vcom = m1v1i /(m1+m2)= V
 Vcom has a constant speed
Sample-Problem-9-8
Ballistic Pedulum
 One dimensional
completely inelastic
collision
Conservation of linear
momentum
mv =(m+M)V; V=mv/(m+M)
 Conservation of
mechanical energy
(K+PE)initial= (K+PE)final
(m+M)V2/2= (m+M)gh
V= (2gh); V=mv/(m+M)
v= [(m+M)(2gh)]/m
Ch-9-10: Elastic Collision in One Dimensions
 Elastic collision: Momentum
and kinetic energy of the
system is conserved
then Pf = Pi and Kf = Ki
Two classes:
 Stationary target (v2i=0)
m1v1i=m1v1f+m2v2f
m1v1i2/2=m1v1f2/2+m2v2f2/2
 v1f=v1i(m1-m2)/(m1+m2)
 v2f=2v1im1/(m1+m2)
Ch-9-10: Elastic Collision in One
Dimensions
 Stationary target (v2i=0)
 v1f=v1i(m1-m2)/(m1+m2)
 v2f=2v1im1/(m1+m2)
 Three cases:
 Equal masses: m1=m2

v1f=0 and v2f=v1i
 Massive target : m2>>m1

v1f=-v1i and v2f 
(2m1/m2)v1i
 Massive projectile : m1>>m2

v1f  v1i and v2f 2v1i
Ch-9-10: Elastic Collision in One
Dimensions
Moving target
m1v1i+ m2v2i = m1v1f+m2v2f
m1v1i2/2+ m2v2i2/2 =
m1v1f2/2+m2v2f2/2
v1f=[v1i (m1-m2)/(m1+m2)]
+2m2v2i/(m1+m2)
v2f=2v1i m1/(m1+m2) +
[v2i (m2-m1) / (m1+m2)]
Ch-9 Check Point 10
What is the final linear momentum of
the target if the initial linear
momentum of the projectile is 6 kg.m/s
and final linear momentum of the
projectile is:
a) 2 kg.m/s
b) -2 kg.m/s
c) what is the final kinetic energy of
the target if the initial and final
kinetic energies of the projectile are ,
respectively 5 J and 2 J?
a)
b)
Pi=Pf
Pf2=Pi-Pf1
=6-2= 4 kg.m/s
=6-(-2)=8 kg.m/s
c) Ki1= Kf1+Kf2
Then
Kf2=Ki1-Kf1
= 5 – 2=3 J
Ch-9-11: Elastic Collision in Two
Dimensions
 Solve equations:
P1i + P2i = P1f + P2f
K1i + K2i = K1f + K2f
 Along x-axis
m1v1i= m1v1f cosq1+m2v2f cosq2
 Along y-axis
0= - m1v1f sinq1+m2v2f sinq2
 m1v1i2/2 = m1v1f2/2+m2v2f2/2