1. French philosopher, mathematician and physicist 1596 - 1650 Rene Descartes thought of motion as being quantifiable…but not solely based on an object’s speed. He also conceptualized that if a large object is going the same speed as a small object, the larger object has a larger “quantity of motion” Furthermore, he believed that the total quantity of motion in the universe was: determined by a Supreme Being (at the beginning of time) conserved throughout time (the total sum of momentum doesn’t change) We call the “quantity of motion” by a different name -- momentum 1 Momentum

2. Momentum is the product of an object’s mass and velocity. p = mv p = momentum (kg m/s) m = mass (kg) v = velocity (m/s) An object’s change in momentum is typically a result of a change in velocity. Δp = mΔv Δp = m(v – v 0 ) When we studied forces, we learned that changes in velocity are the result of unbalanced forces. The same is true of objects that experience a change in momentum. 2

3. Σ F = ma Σ F = m(∆v / t) Σ F = (m∆v) / t Newton’s Second Law (2 nd Version) – the net force applied to an object is equal to the rate of change of momentum of an object A net force changes an object’s momentum 3

4. A tennis ball with a mass of 0.060 kg leaves a tennis racket on the serve with a velocity of 55 m/s. If the racket is in contact with the ball for 0.004 seconds, what is the force that is exerted on the ball by the racket? +825 N 4

5. Water leaves a hose at a rate of 1.5 kg/s with a velocity of 20 m/s. If the water is aimed at a car window which stops it, what is the force exerted by the car on the water? -30 N What would be the force of the water on the car? +30 N 5

6. When you catch a water balloon, you are very careful about the act of catching. You tend to cradle the balloon, taking care to not catch it too fast. Athletes call this having “soft hands”. Having soft hands ensures catching a ball as opposed to having it bounce off your hands You prolong the process of changing the ball’s momentum. This allows you to apply the slowing force over a larger period of time. Impulse 6

7. The larger the time, the smaller the force that is needed to change the momentum. Notice that this is just another way to rewrite Newton’s 2 nd Law. It just groups the force and the time together as a product called impulse. This is the phenomenon that explains why airbags are effective at reducing car injuries. By prolonging the time to decrease a driver’s momentum, it allows the force required to be smaller. Δp = change in momentum (kg x m/s) J = Impulse (kg x m/s or Ns) 7 A change in momentum experienced by an object is caused by an impulse on that object. Impulse – the product of a force exerted on an object and the time over which the force acts

8. 8 The average force by a football helmet on the head of a player is 1770 N and endured for 7.78 milliseconds. Using a head mass of 5.20 kg and presuming the head to be a free body, determine the velocity change experienced in such an impact. Putting more padding in a helmet increase the impact time to 9.53 ms. Determine how much the impact force changes. -2.65 m/s 1446 N

9. 9 A trampolinist uses three impulses from the mat to slow herself down. Just prior to this series of impulses, her 48.5-kg body is moving downward at 8.20 m/s. On the first impulse, she experiences an average upward force of 1070 N for 0.65 seconds. The second impulse of 500 Ns lasts for 0.55 seconds. The last impulse involves an average upward force of 883 N which causes a 265 kgm/s momentum change. What vertical velocity does she have after these three impulses? 1.29 m/s

10. When two separate objects interact with each other, they can be considered to be a system of objects. These are examples of systems of objects: Law of Conservation of Momentum 10

11. When two objects interact, the total momemtum of the system remains constant. In a system of interacting objects, the individual momentums of the objects may change but they cancel each other out the interaction forces are internal and cancel out (Newton’s 3 rd Law) the sum of their momentums stays constant m A v A + m B v B = m A v A ' + m B v B ' p A + p B = p A ' + p B ‘ Σp = Σp' 11

12. This is called the Law of Conservation of Momentum. It applies only to the system of objects that are interacting with each other. The individual objects, however, each experience an individual momentum change… 12 If we limit our system to just object A, it experiences a dramatic change in momentum because of the external force exerted by B If we limit our system to just object B, it experiences a momentum change because of the external force exerted by A Both are examples of Newton’s 2 nd Law in the context of momentum.

13. Explosions Initially, an system has a singular momentum. Some event causes the system to break into several objects that travel in various directions. All of the resultant individual momentums have to add up to the momentum of the initial system. Σp = Σp‘ (m A + m B )v = m A v A + m B v B 13

14. What is the recoil velocity of a stationary 5.0-kg rifle after shooting a 0.020-kg bullet at a speed of 620 m/s? -2.48 m/s 14

15. Collisions In collisions, a system starts out with individual objects that collide and then move independently as separate objects afterward. There are two basic types: 1.Elastic collisions – collisions where kinetic energy is conserved in addition to momentum 2.Inelastic collisions – collisions where kinetic energy is NOT conserved in the system even though momentum is conserved Σp = Σp‘ m A v A + m B v B = m A v A ' + m B v B ' 15

16. To determine whether or not a collision is elastic, you have to compare the total kinetic energies before the collision to the total kinetic energies after the collision. 16

17. There is one special type of inelastic collision that is obvious. A completely inelastic collision is one where two separate objects in a system collide and stick together, moving with a singular velocity after the collision. 17 Σp = Σp‘ m A v A + m B v B = (m A + m B )v

18. A ball of mass 0.440 kg (Ball A) moving east with a speed of 3.30 m/s collides head-on with a 0.220-kg ball (Ball B) at rest. After the collision, the speed of Ball A is 1.30 m/s and the speed of ball B is unknown. Both balls are moving east. Determine the speed of ball B and if the collision is elastic or inelastic. v B = 4.00 m/s Inelastic because E k is not conserved 18

19. A bullet is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m/s right before it hits, at what speed will the block/bullet system rise after the collision and how high will the block subsequently travel? 10.35 m/s 5.47 m 19