Section 7-4: Conservation of Energy & Momentum 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!

mAvA + mBvB = mA(vA) + mB(vB)

HOLDS for ALL collisions!

• Ideal Very Special Case: 2 very hard objects (like

billiard balls) collide. An “Elastic Collision”• To analyze Elastic Collisions:

Rule # 1 Still holds!

mAvA + mBvB = mAvA + mBvB

Rule # 2 For Elastic Collisions ONLY (!!)

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

(½)mA(vA)2 + (½) mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)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!)

Special case: Head-on Elastic CollisionsCan analyze in 1 dimension

Types of head-on collisions

2 masses colliding elastically We know the masses & the initial speeds. Both momentum & kinetic energy are conserved, so we have 2 equations. Doing algebra, we can solve for 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!

Sect. 7-5: Elastic Collisions in 1 Dimension• Special case: Head-on Elastic Collisions.

– Momentum is conserved (ALWAYS!)

Pbefore = Pafter

mAvA + mBvB = mAvA + mBvB

vA, vB, vA, vB are one dimensional vectors!

– Kinetic Energy is conserved (ELASTIC!)

(KE)before = (KE)after

(½)mA(vA)2 + (½)mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2

– 2 equations, 6 quantities: vA,vB,vA, vB, mA, mB

Clearly, we must be given 4 out of 6 to solve problems!

Solve with CAREFUL algebra!!

mAvA + mBvB = mAvA + mBvB (1)

(½)mA(vA)2 + (½)mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2 (2)

• Now, some algebra with (1) & (2), the results of

which will help to simplify problem solving:

– Rewrite (1) as: mA(vA - vA) = mB(vB - vB) (a)

– Rewrite (2) as:

mA[(vA)2 - (vA)2] = mB[(vB)2 - (vB)2] (b)

– Divide (b) by (a):

vA + vA = vB + vB or

vA - vB = vB - vA = - (vA - vB) (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:

mAvA + mBvB = mAvA + mBvB (1)

along with:

vA - vB = vB - vA = - (vA - vB) (3)• (1) & (3) are equivalent to momentum conservation +

Kinetic Energy conservation, since (3) was derived from these conservation laws!

use these!

Example 7-7: Pool (Billiards)

mA = mB = m, vA = v, vB = 0, vA = ?, vB = ?

Momentum Conservation: mv +m(0)=mvA + mvB

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

v - 0 = vB - vA (II)

Solve (I) & (II) simultaneously for vA & vB :

vA = 0, vB = v

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

Before:Ball A Ball B

v v = 0

Before:Ball 1

vBall 2

v = 0

Example: Unequal Masses, Target at Rest

A very common practical situation is for a moving object (mA) to strike a second object (mB, the “target”) at rest (vB = 0). Assume the objects have unequal masses, and that the collision is elastic and occurs along a line (head-on). (a) Derive equations for vB and vA in terms of the initial velocity vA of mass mA and the masses mA and mB. (b) Determine the final velocities if the moving object is much more massive than the target (mA >> mB). (c) Determine the final velocities if the moving object is much less massive than the target (mA << mB).

Example 7-8: Nuclear Collision

A proton (p) of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/s has an elastic head-on collision with a helium (He) nucleus (mHe = 4.00 u) initially at rest. What are the velocities of the proton and helium nucleus after the collision? Assume the collision takes place in nearly empty space.

Section 7-6: 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 7-9: Railroad cars againSame rail cars as Ex. 7-3. Car A, mass mA = 10,000 kg, traveling at speed vA = 24 m/s strikes car B (same mass), initially at rest (vB = 0). Cars lock together after collision. Ex. 7-3: Find speed v after collision.

Ex. 7-3 Solution: vA = 0, (vA) = (vB) = v Use Momentum

Conservation: mAvA+mBvB = (mA + m2B)v v = [(mAvA)/(mA + mB)] = 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)mA(vA)2 = 2.88 106 J

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

Example 7-10: 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.

Ex. 7-10 & Probs. 32 & 33 (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

Problem 71

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

7-7 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