Chapter 7. Linear Momentum

Transcription

1 Chapter 7 Linear Momentum

2 The Impulse-Momentum Theorem Example: A Rain Storm Rain comes down with a velocity of -15 m/s and hits the roof of a car. The mass of rain per second that strikes the roof of the car is kg/s. Assuming that rain comes to rest upon striking the car, find the average force exerted by the rain on the roof. F ( ) Δt = m v f m v o

3 The Impulse-Momentum Theorem Neglecting the weight of the raindrops, the net force on a raindrop is simply the force on the raindrop due to the roof. FΔt = mv f mv o & F= $ % m Δt #! v " o F = ( kg s)( 15m s) = N

4 The Impulse-Momentum Theorem Conceptual Example: Hailstones Versus Raindrops Instead of rain, suppose hail is falling. Unlike rain, hail usually bounces off the roof of the car. If hail fell instead of rain, would the force be smaller than, equal to, or greater than that calculated in the previous Example?

5 The Principle of Conservation of Linear Momentum WORK-ENERGY THEOREM óconservation OF ENERGY IMPULSE-MOMENTUM THEOREM ó??? Apply the impulse-momentum theorem to the midair collision between two objects..

6 The Principle of Conservation of Linear Momentum The midair collision between two objects. Internal forces Forces that objects within the system exert on each other. External forces Forces exerted on objects by agents external to the system.

7 The Principle of Conservation of Linear Momentum F ( ) Δt = m v f m v o W 1 + F 1 OBJECT 1 ( ) Δt = m 1 v f1 m 1vo1 External forces (gravity) W + F 1 Internal forces OBJECT ( ) Δt = m v f m vo

8 The Principle of Conservation of Linear Momentum W 1 + F 1 ( ) Δt = m 1 W + F 1 + ( ) Δt = m v f1 m 1vo1 v f m vo W 1 + W + F 1 + F 1 ( ) Δt = m 1 v f1 + m vf ( ) ( m 1vo1 + m vo ) F 1 F 1 P = f Internal forces cancel from Newton s 3rd law Total final momentum P o Total initial momentum

9 The Principle of Conservation of Linear Momentum The internal forces cancel out. W 1 + W ( )Δt = P f P o More generally, ( sum of average external forces)δt = P f P o F ( ) Δt = P EXT f P 0

10 The Principle of Conservation of Linear Momentum F ( ) Δt = P EXT f P 0 If the sum of the external forces is zero, then = P f P o = 0 P f Po PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The total linear momentum of an isolated system is constant (conserved). An isolated system is one for which the sum of the average external forces acting on the system is zero.

11 The Principle of Conservation of Linear Momentum Conceptual Example: Is the Total Momentum Conserved? Imagine two balls colliding on a billiard table that is friction-free. Use the momentum conservation principle in answering the following questions. (a) Is the total momentum of the two-ball system the same before and after the collision? (b) Answer part (a) for a system that contains only one of the two colliding balls.

12 The Principle of Conservation of Linear Momentum PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The total linear momentum of an isolated system is constant (conserved). An isolated system is one for which the sum of the average external forces acting on the system is zero. In the top picture the net external force on the system is zero. In the bottom picture the net external force on the system is not zero.

13 The Principle of Conservation of Linear Momentum Applying the Principle of Conservation of Linear Momentum 1. Decide which objects are included in the system.. Relative to the system, identify the internal and external forces. 3. Verify that the system is isolated. 4. Set the final momentum of the system equal to its initial momentum. Remember that momentum is a vector.

14 The Principle of Conservation of Linear Momentum Example: Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54 kg woman and one is a 88 kg man. The woman moves away with a velocity of +.5 m/s. Find the recoil velocity of the man.

15 The Principle of Conservation of Linear Momentum P f = P o m v 1 f 1 + mv f = 0 v f = m v 1 m f 1 ( 54 kg)( +.5m s) v f = = 88 kg 1.5m s

16 Collisions in One Dimension The total linear momentum is conserved when two objects collide, provided they constitute an isolated system. Elastic collision -- One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. Inelastic collision -- One in which the total kinetic energy of the system after the collision is not equal to the total kinetic energy before the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.

17 Elastic collision of an isolated system à both momentum and kinetic energy are conserved: Initial Final m 1 m m 1 m v 01 v0 v f 1 v f head-on collision m 1 v01 + m v0 = m 1 v f 1 + m v f m 1 v 01 + m v 0 = m 1 v f 1 + m v f 1 m 1 v m v 0 = 1 m 1v f m v f à can be shown by combining these equations for head-on collisions (see book): v 01 v 0 = (v f 1 v f )

18 Example: Head-on elastic collision of an isolated system. A softball of mass 0.00 kg that is moving with a speed 8.3 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3. m/s. a) Calculate the velocity of the target ball after the collision, and b) calculate the mass of the target ball. a) v 01 v 0 = (v f 1 v f ) v 01 = 8.3 m/s v 0 = 0 v f 1 = 3. m/s v f =? v f = v f 1 + v 01 v 0 = = 5.1 m/s b) m 1 v 01 + m v 0 = m 1 v f 1 + m v f m 1 = 0.00 kg m =? # & % $ v f v ( 0 ' v m ( v f v 0 ) = m 1 ( v 01 v f 1 ) m = m 01 v f 1 1 % # m = 0.00% $ 8.3 ( 3.) & ( = 0.45 kg '

19 Collisions in One Dimension Example: Inelastic collision -- Ballistic Pendulum The mass of the block of wood is.50 kg and the mass of the bullet is kg. The block swings to a maximum height of m above the initial position. Find the initial speed of the bullet.

20 Collisions in One Dimension Apply conservation of momentum to the collision: m v + m v = m v + m 1 f 1 f 1 o1 o v ( m 1 + m )v f = m 1 v o1 v o1 = m 1 + m ( )v f m 1

21 Collisions in One Dimension Applying conservation of energy to the swinging motion: mgh = 1 mv ( m 1 + m )gh f = 1 m ( 1 + m )v f gh = 1 f v f v f = gh f

22 Collisions in One Dimension v f = gh f ( v o1 = m + m 1 )v f m 1 ( v o1 = m + m 1 ) m 1 gh f! v o1 = # " kg +.50 kg kg $ & 9.80m s % ( ) m ( ) = +896m s

23 Center of Mass The center of mass is a point that represents the average location for the total mass of a system. m x x 1 1 = cm m m m x

24 Center of Mass The motion of the center-of-mass is related to conservation of momentum. Consider the change in the positions of the masses in some short time Δt: Δ m Δx 1 1 x cm = m1 + + m m Δx Divide by Δt m v v 1 1 = cm m mv m

25 Center of Mass P v cm = m 1v 1 + m v m 1 + m = P M In vector form, M v cm = m 1v 1 + m v = m 1 + m P M In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.

26 F ( ) Δt = P EXT f P 0 = ΔP Divide both sides by Δt F = Δ P EXT Δt P = M v CM ΔP Δt = Δ M Δt v CM ( ) = M Δ v CM Δt = M a CM F EXT = Ma CM Newton s nd Law for a system of particles which act like a single particle of mass M concentrated at the center of mass

27 The Principle of Conservation of Linear Momentum Example: Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54 kg woman and one is a 88 kg man. The woman moves away with a velocity of +.5 m/s. Find the recoil velocity of the man.

28 The Principle of Conservation of Linear Momentum P f = P o m v 1 f 1 + mv f = 0 v f = m v 1 m f 1 ( 54 kg)( +.5m s) v f = = 88 kg 1.5m s

29 Center of Mass Consider the Ice Skater example which was an isolated system: BEFORE v cm = m v m 1 1 = 1 + m v + m 0 AFTER v cm = (54 kg)(+.5 m/s) + (88 kg)(-1.5 m/s) ( 88 kg)( 1.5m s) + ( 54 kg)( +.5m s) 54 88kg kg kg =