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 1c) after removal of square 1 and 2d) after removal of square 1 and 3e) after removal of square 1, 2 and 3f) All four squareAnswers 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 timeMvcom= mivi ;

differentiating w.r.t timeMacom= 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 3The 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/dtRegion 1: largest slope 2: Zero slope 3: Negative slope 4: Zero slopesAns:(a) 1, 3, then 2 and 4 tie

(zero force); (b) 3

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 5The 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?

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 becausePf = 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 systemTotal 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 Vv1i [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-9-8Ballistic 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)ghV= (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=m1v1f

2/2+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 targetm1v1i+ m2v2i = m1v1f+m2v2f

m1v1i2/2+ m2v2i

2/2 = m1v1f

2/2+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/sc) 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?

Pi=Pf

a) Pf2=Pi-Pf1

=6-2= 4 kg.m/sb) =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 cos1+m2v2f

cos2

Along y-axis 0= - m1v1f sin1+m2v2f sin2

m1v1i2/2 = m1v1f

2/2+m2v2f2/2