Impulse and Momentum continued
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1 Chapter 7 Impulse and Momentum continued
2 7.2 The Principle of Conservation of Linear Momentum External forces Forces exerted on the objects by agents external to the system. Net force changes the velocity (and momentum) of a mass. Newton's 2 nd Law W 1 & W 2 System of two masses Internal forces Forces that objects within the system exert on each other. These forces have equal magnitudes and opposite directions. Newton's 3 rd Law External forces Forces exerted on the objects by agents external to the system. Net force changes the velocity (and momentum) of a mass. Newton's 2 nd Law W 1 & W 2 Also W 1 & W 2
3 7.2 The Principle of Conservation of Linear Momentum.c. p initial momentum of mass 1 and mass 2 p o1 = m vo1 1 Capital P P o = m vo1 1 + m vo2 2 + P f = m vf1 1 + m vf2 2 + p o2 = m vo2 2 initial momentum sum final momentum sum System of two masses Net External Force 0 ( ) Fexternal Δt = P f P o Changes the momentum sum Most Important case Net External Force = 0 0 = P f P o P f = P o No change in momentum sum Conservation of Linear Momentum Sum
4 7.2 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 picture, the net external force on the system is zero. Sum of Weight force and Normal Force on each mass is zero, and no net external force. However, If the cue of a pool player hits one of the balls, the force of the cue on the ball is an external force and in the collision the momentum sum changes from zero to non-zero.
5 7.2 The Principle of Conservation of Linear Momentum Why is adding momentum vectors of different masses together useful at all? P o = vo1 + m 2 vo2 + P f = vf1 + m 2 vf2 + initial momentum sum final momentum sum System of two masses Net External Force on system of two skaters is zero. P f = P o P o = 0 P f = 0 Before Still not so clear why adding momentum vectors is useful. But, P f = vf1 + m 2 vf2 = 0 Momentum vector of mass 2 m 2 vf2 = vf1 is opposite to Momentum vector of mass 1
6 7.2 The Principle of Conservation of Linear Momentum Example 6 Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. P o = 0 One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. Find the recoil velocity of the man. P f = P o vf1 + m 2 vf2 = 0 m 2 vf2 = vf1 v f2 = vf1 = 54 (+2.5 m/s) = 1.5 m/s m 2 88 Before After
7 Clicker Question 7.1 A 9-kg object is at rest. Suddenly, it explodes and breaks into two pieces. The mass of one piece is 6 kg and the other is a 3-kg piece. Which one of the following statements concerning these two pieces is correct? a) The speed of the 6-kg piece will be one eighth that of the 3-kg piece. b) The speed of the 3-kg piece will be one fourth that of the 6-kg piece. c) The speed of the 6-kg piece will be one forth that of the 3-kg piece. d) The speed of the 6-kg piece will be one half that of the 3-kg piece. e) The speed of the 3-kg piece will be one half that of the 6-kg piece.
8 Clicker Question 7.1 A 9-kg object is at rest. Suddenly, it explodes and breaks into two pieces. The mass of one piece is 6 kg and the other is a 3-kg piece. Which one of the following statements concerning these two pieces is correct? a) The speed of the 6-kg piece will be one eighth that of the 3-kg piece. b) The speed of the 3-kg piece will be one fourth that of the 6-kg piece. c) The speed of the 6-kg piece will be one forth that of the 3-kg piece. d) The speed of the 6-kg piece will be one half that of the 3-kg piece. e) The speed of the 3-kg piece will be one half that of the 6-kg piece. m 3 v 3 + m 6 v 6 = 0 v 3 = m 6 m 3 v 6 speeds v 3 = m 6 m 3 v 6 v 3 = 2v 6 or v 6 = 1 2 v 3
9 7.2 The Principle of Conservation of Linear Momentum Applying the Principle of Conservation of Linear Momentum 1. Decide which objects are included in the system. 2. 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.
10 7.3 Collisions in One Dimension In a collision of a puck with a wall, the speed of the puck hitting the wall is the same as the speed of the puck coming off the wall. Clicker Question 7.2 Which statement below is true about this collision? a) momentum of the puck is conserved b) momentum is conserved in a system containing the earth and puck. c) kinetic energy is not conserved in the collision d) total energy is not conserved in the collision e) the system consists of the puck before and after the collision.
11 7.3 Collisions in One Dimension In a collision of a puck with a wall, the speed of the puck hitting the wall is the same as the speed of the puck coming off the wall. Clicker Question 7.1 Which statement below is true about this collision? a) momentum of the puck is conserved b) momentum is conserved in a system containing the earth and puck. c) kinetic energy is not conserved in the collision d) total energy is not conserved in the collision e) the system consists of the puck before and after the collision. a) momentum before is mv, while after it is +mv (not conserved) b) no external forces are acting in a system containing earth and puck c) kinetic energy is the same before and after collision (conserved) d) total energy is the same before and after the collision (conserved) e) a system must be defined before the collision
12 7.3 Collisions in One Dimension The total linear momentum is conserved when two objects collide, provided they constitute an isolated system. Isolated system is the ball and the earth. Energy considerations in collisions 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.
13 7.3 Collisions in One Dimension Example 8 A Ballistic Pendulim The mass of the block of wood is 2.50-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, v o1. Before System of two masses After
14 7.3 Collisions in One Dimension In the collision of bullet and block, tension force of the string and weight of the mass are external forces, but the net external force is zero. Therefore, momentum is conserved. System of two masses But, the bullet is stopped in the block by friction, therefore, energy is not conserved in collision. After the collision, only gravity (a conservative force) does work. Therefore, total energy is conserved.
15 7.3 Collisions in One Dimension Apply conservation of momentum to the collision: v f 1 + m 2 v f 2 = vo1 + m 2 vo2 ( + m ) 2 v f = m vo1 1 System of two masses ( ) v o1 = m + 2 v f need v f To determine the speed of the bullet + block after collision use conservation of energy in the swing. collision v f = swing v 0
16 7.3 Collisions in One Dimension Apply conservation of momentum to the collision: v f 1 + m 2 v f 2 = vo1 + m 2 vo2 ( + m ) 2 v f = m vo1 1 System of two masses ( ) v o1 = m + 2 v f need v f To determine the speed of the bullet + block after collision use conservation of energy in the swing. collision v f = swing v 0 KE 0 + PE 0 = KE f + PE f 1 mv = mgh v 0 = 2gh, v 0 = + 2gh ( ) v o1 = m + 2 2gh = +896m s
17 7.4 Collisions in Two Dimensions Momentum conserved in each of the two dimensions, x and y. Use these masses below. = kg, m 2 = kg x-components: v f 1 + m 2 v f 2 = vo1 + m 2 vo2 ; v o1 = sin50 m/s, v o2 = m/s, v f 2 = cos35 m/s y-components: v f 1 + m 2v f 2 = m vo1 1 + m vo2 2 ; v o1 = 0.900cos50 m/s v o2 = 0, v f 2 = 0.700sin35 m/s final x : v 1x = m/s final y : v 1y = m/s v 1 = v 2 2 1x + v 1y = m/s; θ = tan 1 ( v 1y v ) 1x = 11
18 7.5 Center of Mass The center of mass is a point that represents the average location for the total mass of a system. The center of mass: x cm = x 1 + m 2 x 2 + m 2 The center of mass can move if the objects are moving.
19 7.5 Center of Mass If masess are moving!x cm =!x 1 + m 2!x 2 + m 2 v cm = m v m v2 2 + m 2
20 7.5 Center of Mass v cm = m v m v2 2 + m 2 In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.
21 7.5 Center of Mass BEFORE v cm = m v m v2 2 = 0 + m 2 AFTER v cm = ( 88 kg) ( 1.5m s) + ( 54 kg) ( +2.5m s) 88 kg + 54 kg =
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