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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  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 = FΔt p = Momentum J = Impulse F = 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

7  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

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

9  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

10  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

11  Impulse = Change in momentum  J = Δp = F avg Δt  Change in momentum  Δp = mΔv ** In conclusion, there are different equations for impulse J = F Δt J = Δp = mΔv = mv f – mv i F Δt = mΔv

12  Conservation of momentum:  The total momentum of an isolated system is constant  No net external force acting on the system  Momentum before = Momentum after

13  Momentum  p = mv  Conservation of momentum  Momentum before = Momentum after  p i1 + p i2 +…+ p in = p f1 + p f2 +…+ p fn  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

14  Elastic collision  Objects start apart and end apart  Inelastic collision  Objects start apart and end together  Objects start together and end apart  Momentum is conserved in any collision  Elastic or inelastic

15 A 55.0 kg astronaut is stationary in the spaceship’s reference frame. She wants to move at 0.500 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.)

16  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) = 6.875 m/s

17  Momentum is always conserved in a collision  Classification of collisions:  ELASTIC Both energy & momentum are conserved  INELASTIC Momentum conserved, not energy Perfectly inelastic -> objects stick Lost energy goes to heat

18 Catching a baseball Football tackle Cars colliding and sticking Bat eating an insect Examples of Perfectly Elastic Collisions Superball bouncing Electron scattering Examples of Perfectly Inelastic Collisions


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