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Understanding Newton's Third Law and Momentum Conservation

This chapter delves into Newton's Third Law, stating that for every action, there is an equal and opposite reaction. It explains action-reaction force pairs, which are equal in size but opposite in direction, and occur on different objects. Momentum is introduced as the product of mass and velocity, with examples showing how to calculate momentum for various objects, like bowling balls and ostriches. The law of conservation of momentum is also discussed, highlighting how total momentum remains constant in an isolated system, illustrated with examples of collisions and rocket propulsion.

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Understanding Newton's Third Law and Momentum Conservation

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  1. Chapter 11 Newton’s Third Law • for every action force, there is an equal and opposite reaction force. • Forces always occur in action-reaction pairs. Action-reaction force pairs are • equal in size • opposite in direction.

  2. Chapter 11 Action and Reaction Forces • Force pairs do not act on the same object. • Equal forces don’t always have equal effects. Example: actionforce of Earth pulling on an object and causing it to fall is much more obvious than the equal and opposite reaction force of the falling object pulling on Earth.

  3. Chapter 11 Momentum • quantity defined as product of mass and velocity of an object. • momentum = mass  velocity • p = mv • Moving objects have momentum. • more mass ,greater momentum • Faster object moving, greater its momentum • When you force an object to change its motion, you force it to change its momentum.

  4. p m v Momentum Calculate the momentum of a 6.00 kg bowling ball moving at 10.0 m/s down the alley toward the pins. GIVEN: p = ? m= 6.00 kg v = 10 m/s WORK: p = mv p = (6.00 kg) (10 m/s) p = 60 kg x m/s down the alley

  5. p m v Momentum A 135 kg ostrich running north at 16.2 m/s. Calculate the momentum. GIVEN: p = ? m= 135 kg v = 16.2 m/s WORK: p = mv p = (135 kg) (16.2 m/s) p = 2190 kg x m/s north

  6. p m v Momentum Calculate the velocity of a 0.8 kg kitten with a momentum of 5 kg x m/s forward. GIVEN: p = 5 kg x m/s forward m= 0.8 kg v = ? WORK: v = p/m v = 5 kg x m/s / 0.8 kg v = 6.3 m/s forward

  7. Chapter 11 Law of Conservation of Momentum • total amount of momentum in an isolated system is conserved (stays same). • When moving object hits second object, some or all of momentum of first object is transferred to second object. • rocket propulsion. • Cars collided • Billiards

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