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- 1. PHYSICSFurther Mechanics Andres Igea Friday 19th October 2012
- 2. Further MechanicsLinear MomentumConservation of Linear MomentumTarget checkElastic and Inelastic Collisions in one dimensionTarget checkTwo-dimensional collisionsApplicationsResources
- 3. Linear MomentumThe linear momentum of a particle of mass m and velocity v is defined as: p = m⋅vThe linear momentum is a vector quantity. Its direction is along v.
- 4. Conservation of Linear Momentum p = ∑ p = constant or: ∑ pi =∑ p f p1,i + p2,i = p1, f + p2, f
- 5. Target checkA 2kg marble travels to the right at 0.4 m/s, on a smooth, level surface. It collides head-on with a 6kg marble moving to the left at 0.2 m/s. After the collision, the 2 kg marble rebounds at 0.1 m/s.Task: Find the velocity of the 6kg marble after the collision. Why are the marbles so heavy ?
- 6. Elastic and Inelastic Collisions inone dimension Momentum is conserved in any collision, elastic and inelastic. Mechanical Energy is only conserved in elastic collisions.Perfectly inelastic collision: After colliding, particles coalesce(stick together). There is a loss of energy.Elastic collision: Particles bounce off each other without lossof energy.Inelastic collision: Particles collide with some loss of energy(deformation), but don’t coalesce.
- 7. Perfectly inelastic collision of two particles pi = p f Notice that p and v are are vector m1v1i + quantities = (m1 + m2 )v f m2 v2i and, thus have a direction (+/-). Ki − Eloss = K f1 2 1 2 1 2 m1v1i + m2v2i = ( m1 + m2 )v f + Eloss2 2 2 There is a loss in energy Eloss
- 8. Elastic collision of two particlesMomentum is conserved m1v1i + m2 v2i = m1v1 f + m2 v2 fEnergy is conserved1 2 1 2 1 2 1 2 m1v1i + m2 v2i = m1v1 f + m2 v2 f2 2 2 2
- 9. Target check A bullet (m = 0.01kg) is fired into a block (0.1 kg) sitting at the edge of a table. The block (with the embedded bullet) flies off the table (h = 1.2 m) and lands on the floor 2 m away from the edge of the table.a.) What was the speed of the bullet?b.) What was the energy loss in the bullet-block collision? vb = ? h = 1.2 m x=2m
- 10. Two-dimensional collisionsTwo particles:Conservation of momentum: pi = p f m1v1i + m2 v2i = m1v1 f + m2 v2 fSplit into components: p x ,i = p x , fm1v1ix + 2 v2 ix =m1v1 fx + 2 v2 fx m m p y ,i = p y , fm1v1iy + 2 v2 iy =m1v1 fy + 2 v2 fy m m
- 11. ApplicationsConservation of momentum: Rocket being launched into space.Rocket gainsmomentum inthe upwardsdirectionThe hot gasesgainmomentum inthedownwardsdirection
- 12. ResourcesAQA Mechanics 1 bookEDEXCEL Physics A2 Textbookwww.physicsclassroom.comwww.nasa.gov
- 13. Thanks a lot forwatching Andres

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