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From Earth to Space
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  • Target check, surprise
  • Momentum measures the quantity of motion, in simple terms how much stuff is moving (mass) and how fast it is moving (velocity) Look at M1 book, according to Newton it can be considered as a measure of stoppability. So from the clear formula we can deduce that momentum is the product of the mass and the velocity of an object.
  • In any collision, the total momentum before the collision is equal to the total momentum after the collision, provided that there is no external force acting
  • To see whether you have understood so far what has been taught, we have a little surprise for you:
  • The correct terminology for the particles sticking together is the term: coalesce
  • In this case, the particles will coalesce.
  • If the collision is elastic, we can also use conservation of energy.
  • By the law of conservation of momentum , m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Before firing, both the gun and the bullet have zero momentum i.e. m 1 u 1 + m 2 u 2 = 0 After firing, the bullet moves forward, and have a forward momentum So, the gun must have a backward momentum


  • 1. PHYSICSFurther Mechanics Andres Igea Friday 19th October 2012
  • 2. Further MechanicsLinear MomentumConservation of Linear MomentumTarget checkElastic and Inelastic Collisions in one dimensionTarget checkTwo-dimensional collisionsApplicationsResources
  • 3. Linear MomentumThe linear momentum of a particle of mass m and velocity v is defined as:   p = m⋅vThe 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 checkA 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 collisionsTwo 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. ApplicationsConservation of momentum: Rocket being launched into space.Rocket gainsmomentum inthe upwardsdirectionThe hot gasesgainmomentum inthedownwardsdirection
  • 12. ResourcesAQA Mechanics 1 bookEDEXCEL Physics A2 Textbookwww.physicsclassroom.comwww.nasa.gov
  • 13. Thanks a lot forwatching Andres