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Applied Science

Applied Science. Yr 12. Force x time = change in momentum. What does it mean?. E k = ½ mv 2. How do we calculate kinetic energy E k = ½ mv 2 What are the units? So work me out the kinetic energy of a 1250kg car riding at 25 m/s.

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Applied Science

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  1. Applied Science Yr 12

  2. Force x time = change in momentum

  3. What does it mean? • Ek = ½ mv2

  4. How do we calculate kinetic energy • Ek = ½ mv2 • What are the units? • So work me out the kinetic energy of a 1250kg car riding at 25 m/s. • What is Ek of a horse weighing 170 riding eastwards at 15 m/s.

  5. Remember the formula for impulse

  6. If you drop a tennis ball and it returns to the EXACT height that you dropped it from, it has the same energy. This is known as an ELASTIC COLLISION. This is known as E=0 If energy is lost during a collision it is known as a non-elastic collision.

  7. Conservation of Momentum An important principle: The total momentum of a system remains constant provided that no external forces act on the system.   This has important implications in the study of collisions.  In simple terms, we can say that the total momentum before =  total momentum after

  8. Collision Formula You will need to write to write this down! Conservation of momentum: Total momentum before = total momentum after Mu1 + mu2 = Mv1 + mv2

  9. The diagram shows two cars at a fairground, before and after bumping into each other. One car and driver has a total mass of 500 kg, while the other car and driver has a total mass of 400 kg.

  10. Total energy before the collision • Kinetic Energy = 1/2 mv2 • Kinetic Energy of blue car = 1/2 x 500 kg x (5 m/s)2 = J • Kinetic Energy of yellow car = 1/2 x 400 kg x (2 m/s)2 =  J • Total energy = 1st cars kinetic energy+ 2nd cars kinetic energy.

  11. Work out what the momentum would be if the first car was travelling at double the momentum. • Work out what the Ek would be if the first car had twice the amount of mass.

  12. Kinetic Energy = 1/2 mv2 • Kinetic Energy of blue car = 1/2 x 500 kg x (3 m/s)2 = …………. • Kinetic Energy of yellow car = 1/2 x 400 kg x (4.5 m/s)2 = ………. • Total energy = J • Total loss = J

  13. Is this reaction elastic?

  14. What is the speed of the 500 kg car after the collision?

  15. Momentum before = momentum after  • Momentum before = m1u1 + m2u2 • Momentum before = (500 kg x 5 m/s) + (400 kg x 2 m/s) • Momentum before = 2500 kg m/s + 800 kg m/s = 3300 kg m/s

  16. Momentum after = m1v1 + m2v2 Momentum after = (500 kg x v m/s) + (400 kg x 4 m/s) Momentum after = 500 v kg m/s + 1600 kg m/s Therefore: 3300 kg m/s = 500 v kg m/s + 1600 kg m/s 500 v = 3300 kg m/s - 1600 kg m/s = 1700 kg m/s v = 1700 ÷ 500 = 3.4 m/s

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