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 MOMENTUM:  Inertia in motion  Linear momentum of an object equals the product of its mass and velocity  Moving objects have momentum  Vector quantity.

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Presentation on theme: " MOMENTUM:  Inertia in motion  Linear momentum of an object equals the product of its mass and velocity  Moving objects have momentum  Vector quantity."— Presentation transcript:

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3  MOMENTUM:  Inertia in motion  Linear momentum of an object equals the product of its mass and velocity  Moving objects have momentum  Vector quantity The momentum vector points in the same direction as the velocity vector Proportional to mass and velocity p = mv p = momentum (kg * m/s) m = mass (kg) v = velocity (m/s)

4  Collisions:  Momentum- Useful concept when applied to collisions  In a collision, two or more objects exert forces on each other for a brief instant of time, and these forces are significantly greater than any other forces they may experience during the collision

5 What is the taxi cab’s momentum? * Mass of the taxi = 0.14 kg * Velocity of the taxi = 1.2 m/s Answer:p = mv p = (0.14 kg)(1.2 m/s) p = 0.17 kg * m/s to the left v = 1.2 m/s p = 0.17 kg * m/s

6  Newton’s 2 nd Law & the definition of acceleration  ΣF = ma = m(Δv/Δt)  Momentum  p = mv ** Here, you see a car accelerating − or you could say you see a car whose momentum is changing over time. Either statement leads to the conclusion that a net force is acting on the car. In fact, what we call Newton’s second law was expressed by him in terms of a change in momentum, not in terms of acceleration. ΣF = Δp/Δt

7 Definition: Newton’s 2 nd Law: F = ma or, since M=V so, t If you substitute m = p v F= p x v so, v t The result: The change in momentum over time equals the force!

8  ΣF = Δp/Δt **Net force equals the change in momentum per unit time  Rearranging this equation  Δp = ΣFΔt  Impulse (J)  The change in momentum is called the impulse of the force (Impulse- momentum theorem)  Vector quantity  Units: kg * m/s  J = Δp = F avg Δt p = Momentum J = Impulse F avg = Average force Δt = Elapsed time  The greater the net force, or the longer the interval of time it is applied, the more the object’s momentum changes  the same as saying the impulse increases

9  Changing Momentum: Scenario 1  if you want to decrease a large momentum, you can have the force applied for a longer time. If the change in momentum occurs over a long time, the force of impact is small.  Examples: Air bags in cars. Crash test video FtFt

10  Changing Momentum: Scenario 2  If the change in momentum occurs over a short time, the force of impact is large.  Karate link Karate link  Boxing video Boxing video FtFt

11  Baseball player swings a bat and hits the ball, the duration of the collision can be as short as 1/1000 th of a second and the force averages in the thousands of newtons  The brief but large force the bat exerts on the ball = Impulsive force

12  A long jumper's speed just before landing is 7.8 m/s. What is the impulse of her landing? (mass = 68 kg)  J = p f – p i  J = mv f – mv i  J = 0 – (68kg)97.8m/s)  J = -530 kg * m/s *Negative sign indicates that the direction of the impulse is opposite to her direction of motion

13  View Kinetic books section 8.4- Physics at play: Hitting a baseball BASEBALL EXAMPLE The ball arrives at 40 m/s and leaves at 49 m/s in the opposite direction. The contact time is 5.0×10 −4 s. What is the average force on the ball? J = F avg Δt J = Δp = mΔv F avg Δt = mΔv F avg = mΔv/Δt F avg = (0.14kg)(49 – (-40)m/s)/5.0×10 −4 s F avg = 2.5×10 4 N

14  Momentum  p = mv  Newton’s 2 nd Law & the definition of acceleration  ΣF = ma = m(Δv/Δt)  Different way to express Newton’s 2 nd Law  Net force is equal to the change in momentum over time  ΣF = Δp/Δt

15  Impulse = Change in momentum  J = Δp = F avg Δt  Change in momentum  Δp = mΔv ** In conclusion, there are 2 different equations for impulse J = F avg Δt J = Δp = mΔv F avg Δt = mΔv  F avg = mΔv/Δt

16  Conservation of momentum:  The total momentum of an isolated system is constant  No net external force acting on the system  Momentum before = Momentum after  Playing pool example: Kinetic books 8.6

17  Momentum  p = mv  Conservation of momentum  Momentum before = Momentum after  p i1 + p i2 +…+ p in = p f1 + p f2 +…+ p fn p i1, p i2, …, p in = initial momenta p f1, p f2, …, p fn = final momenta  m 1 v i1 + m 2 v i2 = m 1 v f1 + m 2 v f2 m 1, m 2 = masses of objects v i1, v i2 = initial velocities v f1, v f2 = final velocities

18 The law of conservation of momentum can be derived from Newton’s 2 nd and 3 rd laws Newton’s 2 nd law  F = ma Newton’s 3 rd law  Forces are equal but opposite * Refer to Kinetic Books- 8.7 For step-by-step derivation

19 A 55.0 kg astronaut is stationary in the spaceship’s reference frame. She wants to move at m/s to the left. She is holding a 4.00 kg bag of dehydrated astronaut chow. At what velocity must she throw the bag to achieve her desired velocity? (Assume the positive direction is to the right.)

20  VARIABLES:  Mass of astronaut m a = 55 kg  Mass of bag m b = 4 kg  Initial velocity of astronaut v ia = 0 m/s  Initial velocity of bag v ib =0 m/s  Final velocity of astronautv fa = -0.5 m/s  Final velocity of bag v fb = ?  EQUATION:  m 1 v i1 + m 2 v i2 = m 1 v f1 + m 2 v f2  m a v ia + m b v ib = m a v fa + m b v fb  0 = m a v fa + m b v fb  V fb = - (m a v fa / m b )  V fb = - ((55kg)(-0.5m/s))/(4kg) = m/s

21  Elastic collision  Kinetic energy is conserved KE before = KE after KE = 1/2mv 2  Inelastic collision  Kinetic energy is NOT conserved KE before ≠ KE after  Momentum is conserved in any collision  Elastic or inelastic

22  Momentum is always conserved in a collision  Collision video Collision video  Classification of collisions:  Newton’s cradle: Demonstration  ELASTIC Both energy & momentum are conserved  INELASTIC Momentum conserved, not energy Perfectly inelastic -> objects stick Lost energy goes to heat

23 Catching a baseball: VideoVideo Football tackle Cars colliding and sticking Bat eating an insect Examples of Perfectly Elastic Collisions Superball bouncing Electron scattering Bouncing ball video Examples of Perfectly Inelastic Collisions

24 Mv(initial) = Mv (final) An astronaut of mass 80 kg pushes away from a space station by throwing a kg wrench which moves with a velocity of 24 m/s relative to the original frame of the astronaut. What is the astronaut’s recoil speed? m/s

25 Final velocities are the same

26 A 5879-lb (2665 kg) Cadillac Escalade going 35 mph =smashes into a 2342-lb (1061 kg) Honda Civic also moving at 35 mph=15.64 m/s in the opposite direction.The cars collide and stick. a) What is the final velocity of the two vehicles? b) What are the equivalent “brick-wall” speeds for each vehicle? a) 6.73 m/s = 15.1 mph b) 19.9 mph for Cadillac, 50.1 mph for Civic

27 1.Conservation of Energy: 2.Conservation of Momentum: Rearrange both equations and divide:

28 A proton (m p =1.67x kg) elastically collides with a target proton which then moves straight forward. If the initial velocity of the projectile proton is 3.0x10 6 m/s, and the target proton bounces forward, what are a) the final velocity of the projectile proton? b) the final velocity of the target proton? x10 6 m/s

29 Final equations for head-on elastic collision: Relative velocity changes sign Equivalent to Conservation of Energy

30 An proton (m p =1.67x kg) elastically collides with a target deuteron (m D =2m p ) which then moves straight forward. If the initial velocity of the projectile proton is 3.0x10 6 m/s, and the target deuteron bounces forward, what are a) the final velocity of the projectile proton? b) the final velocity of the target deuteron? v p =-1.0x10 6 m/s v d = 2.0x10 6 m/s Head-on collisions with heavier objects always lead to reflections

31  Video Video  Balancing Activity: video demovideo demo


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