TTU PHYS 1408 - Conservation of Energy and Momentum in Collisions

Unformatted text preview:

Section 9-4: Conservation of Energy & Momentum in CollisionsSlide 2Slide 3Slide 4Slide 5Slide 6Sect. 9-5: Elastic Collisions in 1 DimensionSlide 8Slide 9Example 9-7: Pool (Billiards)Slide 11Example 9-9: Nuclear CollisionSlide 13Slide 14Slide 15Slide 16Slide 17Ex. 9-11 & Probs. 42 & 43 (Inelastic Collisions)ProblemSlide 20Summary: CollisionsElastic Collisions in 2D qualitative here, quantitative in the textSection 9-4: 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 use: Rule # 1Momentum is ALWAYS (!!!) conserved in a collision! mAvA + mBvB = mA(vA) + mB(vB)HOLDS for ALL collisions!A VERY Special (idealized!) Case: 2 very hard objects (~billiard balls) collide ( “Elastic Collision”)•To analyze Elastic Collisions:•Rule # 1 Still holds, of course! mAvA + mBvB = mA(vA) + mB(vB) •Rule # 2 Holds For Elastic Collisions ONLY (!!)The total kinetic energy (K) is conserved!!(K)before = (K)after (½)mA(vA)2 + (½) mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2 Note!!!•Total Kinetic energy (K) is conserved forELASTIC COLLISIONS ONLY!!•Inelastic Collisions Collisions which are NOT elastic.•Is Kinetic Energy conserved in Inelastic Collisions? NO!!!!•Is momentum conserved in Inelastic Collisions? YES!!!!! by Rule # 1: Momentum is ALWAYS conserved in a collision!Special case: Head-on Elastic CollisionsCan analyze in 1 dimensionPossible 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oraftercollisionorNote that vA, vB, (vA), (vB) are 1 dimensional vectors!Sect. 9-5: Elastic Collisions in 1 Dimension•Special Case: Head-on Elastic Collisions.–Momentum is conserved (ALWAYS!)Pbefore = Pafter or mAvA + mBvB = mAvA + mBvBvA, vB, vA, vB are one dimensional vectors! –Kinetic Energy is conserved (ELASTIC!)(K)before = (K)after (½)mA(vA)2 + (½)mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)22 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 afterElastic head-on (1d) collisions only!!•Summary: 1d Elastic Collisions:Rather than directly use momentum conservation + K conservation, it is often convenient to use:Momentum conservation mAvA + mBvB = mAvA + mBvB (1)along withvA - 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 9-7: 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 collisions: v - 0 = vB - vA (II)Solve (I) & (II) simultaneously for vA & vB : vA = 0, vB = v Ball A comes to rest. Ball B moves with original velocity of ball A.Before:Ball ABall B vv = 0 After:Ball A vv = 0 Ball BExample 9-8: Unequal masses, target at restA common practical situation is for a moving object (mA) to strike a second object (mB, the “target”) at rest (vB = 0). Assume that mA & mB, & are unequal, that the collision is elastic & that it occurs along a line (head-on). a. Derive equations for vB & vA in terms of the initial velocity vA of mass mA & the masses mA and mB.b. Calculate vB & vA if mA is very large compared to mB (mA >> mB). c. Calculate vB & vA if mA is very small compared to mB (mA << mB).Example 9-9: Nuclear Collision A proton (p) of mass mp = 1.01 u (atomic mass units), traveling with speed vp = 3.60  104 m/s has an elastic head-on collision with a helium (He) nucleus (mHe = 4.00 u) initially at rest. Calculate the velocities of the proton & He nucleus after the collision.Section 9-6: Inelastic CollisionsInelastic 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 Collisionis one in which the objectsstick together afterwardso there is only one final velocity.•Given some information, using conservation laws, we can determine a LOT about collisions without knowing forces of collision.•To analyze ALL collisions: Rule # 1: Momentum is ALWAYS (!!!) conserved in a collision! m1v1 + m2v2 = m1v1 + m2v2HOLDS for ALL collisions!•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 9-10: Railroad cars againSame rail cars as Ex. 9-3. 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. 9-3: Find speed v after collision.Ex. 9-3 Solution: vA = 0, (vA) = (vB) = v Use Momentum Conservation: mAvA+mBvB = (mA + m2B)v  v = [(mAvA)/(mA + mB)] = 12 m/sBefore CollisionAfter CollisionEx. 9-10: Cars lock together after collision. Find amount of initial KE transformed to thermal or other energy forms:Initially: Ki = (½)mA(vA)2 = 2.88  106 JFinally: Kf = (½)(mA+ mB)(v)2 = 1.44  106 J!


View Full Document

TTU PHYS 1408 - Conservation of Energy and Momentum in Collisions

Download Conservation of Energy and Momentum in Collisions
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Conservation of Energy and Momentum in Collisions and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Conservation of Energy and Momentum in Collisions 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?