conservation of momentum & energy in collisions. given some information, & using...

Conservation of Momentum & Energy in Collisions

• Given some information, & using conservation laws, we can determine a LOT about collisions without knowing the collision forces! To analyze ALL collisions:

Rule #1

Momentum is ALWAYS (!!!)conserved in a collision!

m1v1i + m2v2i = m1v1f + m2v2f

HOLDS for ALL collisions!

• An Ideal Very Special Case: 2 very hard objects (like billiard balls) collide. An “Elastic Collision”

• To analyze Elastic Collisions:Rule # 1 Still holds!


Rule # 2 • For Elastic Collisions ONLY (!!)

Total Kinetic Energy (KE) is conserved!!(KE)before = (KE)after

(½)m1(v1i)2 + (½) m2(v2i)2 = (½)m1(v1f)2 + (½)m2(v2f)2

Note!!

• Total Kinetic energy (KE) is conserved for ELASTIC COLLISIONS ONLY!!

• Inelastic Collisions Collisions which are AREN’T elastic.

• Is KE conserved for Inelastic Collisions?

NO!!!!!!• Is momentum conserved for Inelastic Collisions?

YES!!By Rule # 1: Momentum is ALWAYS

conserved in a collision!

Types of CollisionsMomentum is conserved in all collisions• Inelastic collisions: rubber ball and hard ball

– Kinetic energy is not conserved– Perfectly inelastic collisions occur when the

objects stick together• Elastic collisions: billiard ball

– both momentum and kinetic energy are conserved• Actual collisions

– Most collisions fall between elastic and perfectly inelastic collisions

Collisions Summary• In an elastic collision, both momentum and kinetic

energy are conserved• In a non-perfect inelastic collision, momentum is

conserved but kinetic energy is not. Moreover, the objects do not stick together

• In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same

• Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types

• Momentum is conserved in all collisions

Special case: Head-on Elastic CollisionsThese can be analyzed in 1dimension. Some Types ofhead on collisions:For 2 masses collidingelastically: Often we knowthe masses & the initialvelocities. Both momentum &kinetic energy are conserved,so we have 2 equations.Doing algebra, we can solvefor the 2 unknown final speeds.

• Special case: Head-on Elastic Collisions.1 dimensional collisions: Some possible types:

before collision

or

aftercollision or

vA, vB, (vA), (vB), are 1 dimensional vectors!

Elastic Collisions in 1 Dimension• Special case: Head-on Elastic Collisions.• Momentum is conserved (ALWAYS!)

Pbefore = Pafter


v1i, v2i, v1f, v2f are one dimensional vectors!• Kinetic Energy is conserved (ELASTIC!)

(KE)before = (KE)after

(½)m1(v1i)2 + (½) m2(v2i)2 = (½)m1(v1f)2 + (½)m2(v2f)2 • 2 equations, 6 quantities: v1i,v2i,v1f, v2f, m1, m2

Clearly, we must be given 4 out of 6 to solve problems! Solve with CAREFUL algebra!!

m1v1i + m2v2i = m1v1f + m2v2f (1)

(½)m1(v1i)2 + (½) m2(v2i)2 = (½)m1(v1f)2 + (½)m2(v2f)2 (2)

• Now, some algebra with (1) & (2), the results of which will help to simplify problem solving:

• Rewrite (1) as: m1(v1i – v1f) = m2(v2f - v2i) (a)

• Rewrite (2) as: m1[(v1i)2 - (v1f)2] = mB[(v2f)2 - (v2i)2 (b)

• Divide (b) by (a):

v1i + v1f = v2i + v2f or

v1i – v2i = v2f – v1f = - (v1f – v2f) (3)

Relative velocity before= - Relative velocity after

Elastic head-on (1d) collisions only!!

• Summary: 1d Elastic collisions: Rather than directly use momentum conservation + KE conservation, often convenient to use:

Momentum conservation:m1v1i + m2v2i = m1v1f + m2v2f (1)

Along with:v1i – v2i = v2f – v1f = - (v1f – v2f) (3)

• (1) & (3) are equivalent to momentum conservation + Kinetic Energy conservation, since (3) was derived from these conservation laws!

Example: Pool (Billiards)

m1 = m2 = m, v1i = v, v2i = 0, v1f = ?, v2f = ?

Momentum Conservation: mv +m(0)=mv1f + mv2f

Masses cancel v = v1f + v2f (I)• Relative velocity results for elastic head on collision:

v - 0 = v2f - v1f (II)

Solve (I) & (II) simultaneously for v1f & v2f :

v1f = 0, v2f = v

Ball 1: to rest. Ball 2 moves with original velocity of ball 1

Before:Ball 1 Ball 2

v v = 0

Before:Ball 1

vBall 2

v = 0

Example: Unequal Masses, Target at RestA very common practical situation is for a moving object (m1) to strike a second object (m2, the “target”) at rest (v2 = 0). Assume the objects have unequal masses, and that the collision is elastic and occurs along a line (head-on).

(a) Derive equations for v2f & v1f in terms of the initial velocity v1i of mass m1 & the masses m1 & m2.

(b) Determine the final velocities if the moving object is much more massive than the target (m1 >> m2).

(c) Determine the final velocities if the moving object is much less massive than the target (m1 << m2).

Inelastic Collisions

Inelastic Collisions Collisions which

Do NOT Conserve Kinetic Energy!

Some initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained in explosions (there is addition of chemical or nuclear energy).

A Completely Inelastic Collision is one in which the objects stick together afterward, so there is only one final velocity.

• Total Kinetic energy (KE) is conserved for ELASTIC COLLISIONS ONLY!!

• Inelastic Collisions Collisions which are NOT elastic.

• Is KE conserved for Inelastic Collisions? NO!!!!• Is momentum conserved for Inelastic Collisions?

YES!! (Rule # 1: Momentum is ALWAYS conserved in a collision!).

• Special Case: Completely Inelastic Collisions Inelastic collisions in which the 2 objects collide & stick together.

• KE IS NOT CONSERVED FOR THESE!!

Example: Railroad cars againSame rail cars as Ex. 7-3. Car A, mass m1 = 10,000 kg, traveling at speed v1i = 24 m/s strikes car B (same mass), initially at rest (v2i = 0). Cars lock together after collision. Ex. 7-3: Find speed v after collision.

Ex. 7-3 Solution: v2i = 0, v1f = v2f = v Use Momentum Conservation:

m1v1i+m2v2i = (m1 + m2)v v = [(m1v1i)/(m1 + m2)] = 12 m/s

Before Collision

After Collision

Ex. 7-9: Cars lock together after collision. Find amount of initial KE transformed to thermal or other energy forms:

Initially: KEi = (½)m1(v1i)2 = 2.88 106 J

Finally: KEf = (½)(m1+ m2)(v)2 = 1.44 106 J ! (50% loss!)

More about Perfectly Inelastic Collisions

• When two objects stick together after the collision, they have undergone a perfectly inelastic collision

• Conservation of momentum

• Kinetic energy is NOT conserved

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An SUV Versus a Compact An SUV with mass 1.80103 kg is travelling

eastbound at +15.0 m/s, while a compact car with mass 9.00102 kg is travelling westbound at -15.0 m/s. The cars collide head-on, becoming entangled.

(a) Find the speed of the entangled cars after the collision.

(b) Find the change in the velocity of each car.

(c) Find the change in the kinetic energy of the system consisting of both cars.

(a) Find the speed of the entangled cars after the collision.


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More About Elastic Collisions• Both momentum and kinetic energy are

conserved

• Typically have two unknowns• Momentum is a vector quantity

– Direction is important– Be sure to have the correct signs

• Solve the equations simultaneously

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Elastic Collisions

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Summary of Types of Collisions

• In an elastic collision, both momentum and kinetic energy are conserved

• In an inelastic collision, momentum is conserved but kinetic energy is not

• In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same

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Example: Ballistic pendulum

The ballistic pendulum is a device used to measure speeds of projectiles, such as a bullet.

A projectile, mass m, is fired into a large block, mass M, which is suspended like a pendulum. After the collision, pendulum & projectile swing up to a maximum height h.

Find the relation between the initial horizontal speed of the projectile, v & the maximum height h.

Example: Inelastic Collisions

Before After

Momentum Conservation mv = (m + M)v´

Mechanical Energy (½)(m +M)(v´)2 = (m + M)ghConservation

v = [1 +(M/m)](2gh)½

a

v = 0aaa

aa

a

ℓ - h

ℓ ℓ

Example

A bullet, m = 0.025 kg hits

& is embedded in a block,

M = 1.35 kg. Friction

coefficient between block &

surface: μk = 0.25. Moves d = 9.5 m before stopping. Find v

of the bullet it before hits the block. Multi-step problem!

1. Find V using Work-Energy Principle with friction.

2. Find v using momentum conservation. But, to find V, first we need to

3. Find the frictional force!

Ffr = μkFN = μk(M+m)g

1. Friction force:

Ffr = μkFN = μk(M+m)g

2. The Work- Energy

Principle: Wfr = -Ffrd

= KE = 0 – (½)(M+m)V2 OR: -Ffrd = - (½)(M+m)V2

μk(M+m)gd = (½)(M+m)V2 (masses cancel!)

Stops in distance d = 9.5 m

V = 6.82 m/s

3. Momentum conservation:

mv + 0 = (M+m) V

v = (M+m)V/m = 375 m/s (bullet speed)

Summary: Collisions• Basic Physical Principles:• Conservation of Momentum: Rule # 1:

Momentum is ALWAYS conserved in a collision!

• Conservation of Kinetic Energy:

Rule # 2: KE is conserved for elastic collisions

ONLY !! – Combine Rules #1 & #2 & get relative velocity

before = - relative velocity after.• As intermediate step, might use Conservation of

Mechanical Energy (KE + PE)!!

vA – vB = vB – vA

Collisions in Two or Three DimensionsConservation of energy & momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy.

Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.

Elastic Collisions in 2Dqualitative here, quantitative in the text

Physical Principles: The same as in 1D1. Conservation of VECTOR momentum:

PAx + PBx = PAx + PBxPAy + PBy = PAy + PBy

2. Conservation of KE (½)mA(vA)2 + (½)mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2

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