Chapter 9

Systems of Particles

Section 9.2:Center of Mass in a Two Particle

System Center of Mass is the point at which all forces are assumed to act. For a two body system, this force is calculated using the equation:

)(1

221121

xmxmmm

xcm

Where m represents the mass of each body and x represents its position.

Equation 1

Velocity and Acceleration of the Center of Mass

The Velocity and Acceleration of the center of mass of a system is calculated using a similar equation:

)(1

221121

vmvmmm

vc

)(1

221121

amammm

ac

Equation 2

Equation 3

Section 9.2:Many Particle Systems

Expanding equations 1,2,& 3 from a two particle system to a many particle system is simply a matter of expanding the terms to as many particles as the system holds, thus the general form of these equations are:

)...(..

12211

21nn

ncm xmxmxm

mmmx

)..(..

12211

21nn

nc vmvmvm

mmmv

)..(..

12211

21nn

nc amamam

mmma

Analyzing Multi-particle Systems

Since we can calculate the center of mass of a system and its velocity and acceleration, we can then analyze the system using the following rule: The overall translational motion of a system of particles can be analyzed using Newton’s Laws as if all the mass were concentrated at the center of mass and the total external force were applied at that point.

A corollary which follows is: If the net external force on a system of particles is zero, then the center of mass of the system move with a constant velocity.

The Center of Mass of Solid Objects

If an object has spherical symmetry – the center of mass is at the geometric center

If an object is not symmetrical then you calculate the center of mass along each axis and this determines the center of mass of that object for translational motion in that direction.

Section 9.3: Newton’s Second Law for a System of

Particles Fnet = Macm where:

Fnet – the sum of all external forces acting on the system.

M – total mass of the system acm – the acceleration of the center of mass of the

system. This is equivalent to the three equations:

Fnet,x = Macm,x Fnet,y = Macm,y Fnet,z = Macm,z

Section 9.4Linear Momentum of a Particle

Linear Momentum is the product of the mass of an object and its linear velocity.

ρ = mv where ρ (roe) is the momentum, m is the mass in kilograms, and v is velocity in m/s.

F

Section 9.5:Linear Momentum in System of

Particles In a system of particles: So

dt

dFnet

cmcm Madt

dvM

dt

d

Section 9:6Change in Momentum: Impulse

The rate of change of momentum of a body (impulse) is equal to the resultant force acting on the body and is the direction of that force. Since ρ = mv and F = ma and a = Δv/t Δρ =mΔv = FΔt

Hence change in momentum is a product of the force acting on an object and the time during which it acts.

Section 9.6Conservation of Linear

Momentum Like energy – momentum in a system is also conserved.

For example, prior to firing a shotgun, neither the gun, nor the shell inside are moving, hence the momentum of the system is zero.

mgvg + msvs = 0 After the gun is fired, the shell moves forward, and

by conservation of momentum, the gun moves backward

mgvg = -(msvs)

Sample Problem

A 3 kg shotgun contains a .140 kg shell. When the shotgun is fired, the shell leaves the barrel with a velocity of 400 m/s. What is the recoil velocity of the shotgun?

If the Shotgun impacts the hunter’s shoulder for 0.03 seconds, what force does it exert on her shoulder?

Solution

ssggsg vmvmVmm )(By Conservation of Momentum:

Hence

(3 kg + 0.140 kg)(0 m/s) = (3 kg)(Vg) + (0.140 kg)(400 m/s)

-(3 kg)(Vg) = 56 kg•m/s

smkg

smkgVg /67.18

3

/56

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Section 9.8:Momentum and Kinetic Energy in

CollisionsTwo type of Collisions Elastic

Two objects collide and there are two objects move away In a perfectly elastic collision kinetic energy is conserved.

Inelastic collision Two objects collide and stick together- hence they move off as

one object after the collision The converse of an inelastic collision is when one object breaks

into two pieces and both move off after the breaking (often an explosion)

While total energy is conserved in the collision, often it is changed into heat energy or energy of deformation so kinetic energy is not conserved.

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Section 9.9Inelastic Collisions

Inelastic Collisions can be described mathematically by the equation:

m1v1 + m2v2 = (m1 + m2 )v’

since kinetic energy is not conserved this is the only equation for an inelastic collision

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Sample Problem

A car with a mass of 1000 kg moving at a velocity of 25 m/s, strikes a second car with a mass of 1500 kg that is at rest. The bumpers lock and the cars travel together after the collision. What is the velocity of the two car system immediately after the collision?

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Sample ProblemSolution

m1v1 + m2v2 = (m1 + m2)v’

(1000 kg)(25 m/s) + (1500 kg)(0 m/s) = (2500 kg)(v’)

25000 kg m/s = (2500 kg)(v’)

V’ = 10 m/s

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Section 9.10Elastic Collisions

Elastic Collisions can be described mathematically by the following: Conservation of Momentum

m1v1 + m2v2 = m1v1’ + m2v2’ Conservation of kinetic energy

v1 - v2 = v2’ - v1’

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Sample Problem

A ball with a mass of 0.5 kg rolling at a velocity of 10 m/s, strikes a second ball with a mass of 0.75 kg that is at rest in an elastic collision. What is the velocity of each ball after the collision?

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Sample ProblemSolution

m1v1 + m2v2 = m1v1’ + m2v2’(0.5 kg)(10 m/s) + (0.75 kg)(0 m/s) = (0.5 kg)(v1’)

+ (0.75 kg)(v2’)v1 - v2 = v2’ - v1’

10 m/s - 0 m/s = v2’ - v1’V2’ = 10m/s + v1’

5 kg m/s = (0.5 kg)(v1’) + 7.5 kg m/s + 0.75v1’-2.5 kg m/s = 1.25 kg(v1’)

V1’ = -2 m/sV2’ = 10 m/s + -2 m/s = 8 m/s

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Section 9.11Two Dimensional Collisions

When analyzing a collision in two dimensions, the key is to remember that momentum is conserved in each direction.

ρix = ρfx and ρiy = ρfy

Remembering that since ρ = mv and v is a vector.

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Sample Problem

A gas molecule having a speed of 322 m/s collides elastically with another molecule having the same mass which is initially at rest. After the collision, the first molecule move at an angle of 30° with respect to its initial direction. Find the velocity of each molecule after the collision and the angle made with the incident direction by the recoiling target (second) molecule.

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Solution

m1v1x + m2v2x = m1v’1x + m2v’2x

and

m1v1y + m2v2y = m1v’1y + m2v’2y

Since m1=m2 mass cancels so you have

322 m/s + 0 m/s = v’1x + v’2x

0 m/s + 0 m/s = v’1y + v’2y

θ

β

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Solution (Continued)

From previous work with vectors we remember:

V’1x = V’1cos θ and V’1y = V’1sin θ

V’2x = V’2cos β and V’2y = V’2sin β So the equations become:

V1x = V’1cos θ + V’2cos β

-V’1sin θ = V’2sin β (negative sign denotes opposite direction)

Rearranging the first equation we get:

V1x - V’1cos θ = V’2cos β

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Solution (Continued)

Squaring both equations and adding them together you get:

V1x2 -2V1xV’1cos θ + V’12 sin2θ + V’12 cos2 θ = V’22

cos2β + V’22 sin2β Using a well know trig identity this becomes:

V1x2 -2V1xV’1cos θ + V’12 = V’22

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Solution (Continued)

Remember that this is an elastic collision where energy is conserved so:

V12 = V’12 + V’22

Combined withV1x

2 -2V1xV’1cos θ + V’12 = V’22

And the fact that

V1x =v1

(since there was no velocity in the Y direction initially)

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Solution (Continued)

V’22 +V’12 -2V1V’1cos θ + V’12 = V’22

2V’12 = 2V1V’1cos θ

V’1 = V1cos θ = (322 m/s)(cos 30°)= 279 m/s

Remember: V12 = V’12 + V’22 so:

(322 m/s)2 = (279 m/s)2 + V’22

V’2 = 161 m/s

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Solution (continued)

Finally, since:

V’1sin θ = V’2sin β

Sin β = V’1sin θ/V’2

Sin β = 279 m/s sin 30°/161 m/s

Β = 60°

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Momentum and Energy Combined

Momentum is often combined in problems with energy. One of the most common problems used to illustrate this is the ballistic pendulum problem. Ballistics is the study of projectiles and a ballistic pendulum is often used to measure the speed of a bullet. This is done by firing the bullet into a wooden block suspended on a string (hence the pendulum) and using the change in the pendulum’s height to back calculate the bullet’s speed.

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Sample Problem

A bullet with a mass of 14 grams is fired into a suspended wooden block with a mass of 2 kg. If the block travels to a maximum height of 0.75 meters above its rest position, what was the velocity of the bullet when it was fired?

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Sample ProblemSolution

In order to solve this problem you work backward. At the top of the arc the pendulum-bullet combination has only potential

energy. At the bottom of the arc, immediately after the bullet is embedded in the block the combination has only kinetic

energy. So:½ (m1 + m2)v’2 = (m1 + m2)gh

v’ = SQRT (2gh) = SQRT(2 x 9.8 m/s2 x 0.75 m)= 3.83 m/s

Since the bullet embedding itself into the block is an inelastic collision the equation:

m1v1 + m2v2 = (m1 + m2)v’applies, hence

(0.014 kg)(v1) + (2 kg)(0 m/s) = (2.014 kg)(3.83 m/s)V1 = 551.5 m/s

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Problem Types

Linear momentum of a single object Impulse Collisions

Elastic Conservation of Momentum and Kinetic Energy

Inelastic Conservation of Momentum but not Kinetic Energy

Locked Bumpers, explosions

Two dimensional Collisions Energy Momentum Combinations