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Physics Topic 6 Notes 3

Physics Topic 6 Notes 3. Learning outcomes. When is momentum conserved?. What are the two types of collisions?. When is kinetic energy conserved in a collision?. When is the kinetic energy NOT conserved in a collision?. How do you measure or calculate momentum before and after collisions?.

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Physics Topic 6 Notes 3

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  1. Physics Topic 6 Notes 3

  2. Learning outcomes When is momentum conserved? What are the two types of collisions? When is kinetic energy conserved in a collision? When is the kinetic energy NOT conserved in a collision? How do you measure or calculate momentum before and after collisions?

  3. Momentum and collisions

  4. What is conservation of momentum? If two objects collide or interact, the forces acting on each one will be the same size but in opposite directions. The same is true for the change in momentum of each object. This means that the momentum lost by one of the objects will be gained by the other object. Therefore, whenever two objects collide or interact, momentum is conserved.

  5. Using conservation of momentum

  6. Elastic Collisions • objects bounce apart after collision • momentum conserved • KE conserved • Perfectly elastic collisions never actually happen – some energy is lost to heat, sound or light

  7. Inelastic Collisions • objects stick together • total mass • momentum conserved • KE is NOT conserved

  8. Collision Problem Strategy Make a chart and fill in known numbers Calculate what you can right away – then figure out what you need to answer the question Inelastic collisions will only have one combined object after the collision

  9. Conservation of momentum question Two trolleys collide and stick together. From the data below, calculate the velocity of the trolleys after the collision. (General equation: m1v1 + m2v2 =mTvf) Before After trolley A trolley B Pf = mTvf mass = 3kg velocity = 8m/s pf = 4kgm/s mass = 5kg velocity = -4m/s piA = (3kg)(8 m/s) mT= 8kg (add masses because they stick together: 3kg + 5 kg) piA = 24kgm/s piB = (5 kg)(-4m/s) piB = -20kgm/s vf = pf / mT = 4kgm/s / 8 kg = 0.5 m/s pi = (24kgm/s + -20kgm/s) pi= 4kgm/s

  10. Investigating momentum

  11. Calculation: A 13 kg bowling ball traveling at 21 m/s slams into a 5 kg bowling pin that is at rest. If the pin moves at 8 m/s after the collision what is the speed of the bowling ball? (General equation: m1v1i +m2v2i = m1v1f + m2v2f) mball = mpin= 5 kg 13 kg After collision: Before collision: vfpin = 8 m/s viball = 21 m/s pball + ppin (We don’t add masses together, they bounce) pf = vipin = 0 m/s pi = pf = mballvfball + mpinvfpin pi = pball + ppin vfball = pi - mpinvfpin mball mballviball + 0 = = (13 kg)(21 m/s) = 273 kgm/s - (5 kg)(8 m/s) 13kg = 273 kgm/s = 17.9 m/s

  12. Quiz time  Judy, at rest on ice, catches her leaping dog, Atti. Atti jumps at 3.1 m/s initially. What is the final velocity of Judy and Atti if Atti has a mass of 15 kg and Judy’s mass is 55 kg? pi = pf (General equation: m1v1 + m2v2 =mTvf) mJ = 55 kg mA = 15 kg mJ+A = 70 kg pf = pJ+A vJ+A = pf mJ+A = 46.5 kgm/s 70 kg = mJ+AvJ+A vJi = 0 m/s 3.1 m/s vAi = pi = mJvJi + mAVAi = 0 + mAVAi = (15kg)(3.1 m/s) = 46.5 kgm/s = .66 m/s

  13. Quiz time  - Chart strategy Judy, at rest on ice, catches her leaping dog, Atti. Atti jumps at 3.1 m/s initially. What is the final velocity of Judy and Atti if Atti has a mass of 15 kg and Judy’s mass is 55 kg?

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