Collisions and Conservation of Momentum

1. In collisions, momentum of SYSTEM is conserved. • What does conservation look like? • How do we make conservation calculations? • Are all collisions the same? How do we classify them? Collisions and Conservationof Momentum

2. Conservation of Momentum Animation

3. Conservation Example 1 Classic two-car collision. One is moving and one is stationary. System is BOTH cars. What is the velocity of car 2 after collision? Conservation says: p car 1 + pcar 2 before collision = p car 1 + pcar 2 after collision One way to write this: p1i + p2i = p1f + p2f Because only the velocities change, not masses, this is m1v1i + m2v2i = m1v1f + m2v2f Givens: m1 = 2 kg v1i = +5 m/s v1f = +1 m/s m2 = 2 kg v2i = 0 m/s v2f = ? Substitute: 2·5 + 2·0 = 2·1 + 2· v2f Solve: 10 = 2 + 2 v2f v2f = +4 m/s

4. Types of Collisions What happens to momentum during collisions? To Kinetic Energy? To the objects themselves?

5. Perfectly Elastic

6. Perfectly Inelastic

7. Inelastic

8. Recoil For the system, pibefore explosion = pfafter explosion pi = 0 pf = 100 kg · -vcannon+ 5kg · +vcannonball Is KE conserved? KEinitial = 0 KEfinal= very not zero!

9. Example 2 A 1000 kg car going north at 30 m/s collided head-on with a 5000 kg truck going south at 5 m/s. The two vehicles stuck together forming one tangled mass of metal and moved off together, but at what speed and direction? a) What velocity did the tangled mass have? c for car t for truck mc = 1000 kg mt = 5000 kg vci = +30 m/s vti = -5m/s vcf=? vtf = ? Since car and truck stuck together, there is only 1 vf mcvci + mtvti = mcvcf + mtvtf Since they form one object, this simplifies to: mcvci + mtvti = (mc + mt)vf (1000)(30) + (5000)(-5) = (1000 + 5000)(vf) (30000 – 25000)/6000 = vfvf = + .83 m/s, or .83 m/s north

10. Example 2, part B A 1000 kg car going north at 30 m/s collided head-on with a 5000 kg truck going south at 5 m/s. The two vehicles stuck together forming one tangled mass of metal and moved off together, but at what speed and direction? b) What percentage of the initial KE was conserved? c for car t for truck mc = 1000 kg mt = 5000 kg vci = +30 m/s vti = -5m/s vcf= .83 m/s vtf= .83 m/s Since car and truck stuck together, there is only 1 vf ΣKEi = KEci + KEti ΣKEi= ½(1000)(302) + ½(5000)(52) = 512,500 J KEf = ½(1000 + 5000)(.832) = 2067 J   % conserved = 2067/512,500 x 100 = .40 % of initial KE.What kind of collision? Perfectly Inelastic

11. Example 3 A 0.005 kg bullet moving 670 m/s strikes a 0.7 kg wooden block at rest on a frictionless surface. The bullet passes clear through the block and emerges with a speed of 430 m/s in the same direction. Of course, the block begins to move, but with what velocity? The system is bullet and block. Before collision: Bullet approaches block. After collision: Bullet has left block & both bullet & block are moving. b for bullet w for wooden block mb = 0.005 kg mw = 0.7 kg vbi = +670 m/s vwi = 0 vbf = +430 m/s vwf = ? Momentum is conserved: mbvbi + mwvwi = mbvbf + mwvwf (0.005)(670) + (0.7)(0) = (0.005)(430) + (0.7)( vwf) Vwf = +1.7 m/s

12. Example 3 continued Characterize the collision. The objects don’t stick together, so it’s not perfectly inelastic. But is it inelastic or perfectly elastic? That depends. Is the KE conserved? KEi = ½ (0.005)(6702) = 1122 J KEf = ½ (0.005)(4302) + ½(0.7)(1.72) = 463 J  So about 60% of the KE was lost, making this an inelastic collision.

13. Tips & Tricks Draw a sketch and label the pieces. Decide when the collision occurs. What is before and what is after? Write a list of givens. Watch the vector signs. Organize! Write the equations. Don’t shortcut! Do a reality check. Is this reasonable?