Presentation on theme: "1 Ch7. Impulse and Momentum There are situations in which force acting on an object is not constant, but varies with time. Two new ideas: Impulse of the."— Presentation transcript:
1 Ch7. Impulse and Momentum There are situations in which force acting on an object is not constant, but varies with time. Two new ideas: Impulse of the force and Linear momentum of an object.
2 Impulse-Momentum Theorem
3 Definition of Impulse The impulse J of a force is the product of the average force and the time interval during which the force acts: J = Impulse is a vector quantity and has the same direction as the average force. SI Unit of Impulse: newton. second (N. s)
4 Definition of Linear Momentum: The linear momentum p of an object is the product of the object’s mass m and velocity v: p=mv Linear momentum is a vector quantity that points in the same direction as the velocity. SI Unit of Linear Momentum: kilogram. meter/second(kg. m/s)
6 Impulse-Momentum Theorem When a net force acts on an object, the impulse of this force is equal to the change in momentum of the object: Impulse Final momentum Initial momentum Impulse=Change in momentum
7 Example 1. A Well-Hit Ball A baseball (m=0.14kg) has initial velocity of v 0 =-38m/s as it approaches a bat. The bat applies an average force that is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of v f =+58m. (a)Determine the impulse applied to the ball by the bat. (b) Assuming time of contact is =1.6 * s, find the average force exerted on the ball by the bat.
8 = kg.m/s (a) (b)
9 Example 2. A Rain Storm Rain comes straight down with velocity of v 0 =-15m/s and hits the roof of a car perpendicularly. Mass of rain per second that strikes the car roof is 0.06kg/s. Assuming the rain comes to rest upon striking the car (v f =0m/s), find the average force exerted by the raindrop.
10 = -(0.06kg/s)(-15m/s)=0.9 N According to action-reaction law, the force exerted on the roof also has a magnitude of 0.9 N points downward: -0.9N
11 Conceptual Example 3. Hailstones Versus Raindrops Suppose hail is falling. The hail comes straight down at a mass rate of m/ =0.06kg/s and an initial velocity of v 0 =15m/s and strikes the roof perpendicularly. Hailstones bounces off the roof. Would the force on the roof be smaller than, equal to, or greater than that in example 2? Greater.
12 Check your understanding 1 Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, determine which one is correct. A In bringing you to a halt, the sand exerts a greater impulse on you than does the water. B In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force. C In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force. B
13 The Principle of Conservation of Linear Momentum
14 Two types of forces act on the system: 1.Internal forces: Forces that the objects within the system exert on each other. 2.External forces: Forces exerted on the objects by agents external to the system. External force Internal force
15 = p f - p 0 (Sum of average external forces) = p f - p 0 Then 0 = p f - p 0 or p f = p 0 m 1 v f1 +m 2 v f2 = m 1 v 01 +m 2 v 02 pfpf p0p0 sum of average external forces sum of average internal forces () + internal forces cancel F 12 = -F 21 If sum of external forces is zero (an isolated system)
16 Principle of Conservation of Linear Momentum: The total linear momentum of an isolated system remains constant(is conserved). An isolated system is one for which the vector sum of the average external forces acting on the system is zero.
17 Conceptual Example 4. Is the Total Momentum Conserved? Two balls collide on the billiard table that is free of friction. (a)Is the total momentum of the two ball system the same before and after the collision? (b)Answer (a) for a system that contains only one ball. (a)The total momentum is conserved. (b) The total momentum of one ball system is not conserved.
18 Example 5. Assembling a Freight Train Car 1 has a mass of m 1 =65 * 10 3 kg and moves at a velocity of v 01 =+0.8m/s. Car 2 has a mass of m 2 =92 * 10 3 kg and a velocity of v 02 =+1.3m/s. Neglecting friction, find the common velocity v f of the cars after they become coupled.
19 (m 1 +m 2 ) v f = m 1 v 01 + m 2 v 02 After collisionBefore collision =+1.1 m/s
20 Example 6. Ice Skaters Starting from rest, two skaters push off against each other on smooth level ice (friction is negligible). One is a woman (m 1 =54kg), and one is a man(m 2 =88kg). The woman moves away with a velocity of v f1 =2.5m/s. Find the recoil velocity v f2 of the man.
21 For the two skater system, what are the forces? In horizontal direction F 12 F 21 internal forces system taken together 0 No external forces. isolated system conservation of momentum
22 m 1 v f1 + m 2 v f2 = 0 after pushingbefore pushing It is important to realize that the total linear momentum may be conserved even when the kinetic energies of the individual parts of a system change.
23 Check your understanding 2 A canoe with two people aboard is coasting with an initial momentum of 110kg.m/s. Then person 1 dives off the back of the canoe. During this time, the net average external force acting on the system is zero. The table lists four possibilities for the final momentum of person 1 and final momentum of person 2 plus the canoe, immediately after person 1 leaves the canoe. Which possibility is correct? Person 1 Person 2 & canoe a-60kg.m/s+170kg.m/s b-30kg.m/s+110kg.m/s c-40kg.m/s-70kg.m/s d+80kg.m/s-30kg.m/s a
24 Collisions in One Dimension 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 is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.
25 Example 7. A Collision in One Dimension A ball of mass m 1 =0.25kg and velocity v 01 =5m/s collides head-on with a ball of mass m 2 =0.8kg that is initially at rest(v 02 =0m/s). No external forces act on the balls. If the collision in elastic, what are the velocities of the balls after the collision?
26 Total momentum after collision Total momentum before collision Total kinetic energy after collision Total kinetic energy before collision (1) (3) (2)
27 Substitute in (2)
29 Substitute in (1)
31 m 1 =0.25, m 2 =0.8 v 01 =5 m/s, v 02 = 0
32 Types of Collisions
33 Example 8. A Ballistic Pendulum The ballistic pendulum consists of a block of wood(mass m 2 =2.5kg)suspended by a wire of negligible mass. A bullet(mass m 1 =0.01kg)is fired with a speed v 01. After collision, the block has a speed v f and then swings to a maximum height of 0.65m above the initial position. Find the speed v 01 of the bullet, assuming air resistance is negligible.
34 Is conservation of energy valid? No (completely inelastic) Is conservation of momentum valid? Yes (no external forces ) Just before collision Just after collision m 1 +m 2 vfvf
35 Total momentum after collision Total momentum before collision
36 Applying conservation of energy Total mechanical energy at top of swing, all potential Total mechanical energy at bottom of swing, all kinetic vfvf h f =0.65 m
38 Check your understanding 3 Two balls collide in a one-dimensional, elastic collision. The two balls constitute a system, and the net external force acting on them is zero. The table shows four possible sets of values for the initial and final momenta of the two balls as well as their initial and final kinetic energies. Which one is correct?
41 Collisions in Two Dimensions v f2 =0.7 m/s Ball 1 m 1 =0.15 kg Ball 2 m 2 =0.26 kg beforeafterbeforeafter v 01 sin50=v f1 cos 0v 02 v f2 cos35 -v 01 cos50=v f1 sin 00-v f2 sin35 x component y component
42 x component y component P fx P 0x P fy P 0y
43 Use momentum conservation to determine the magnitude and direction of the final velocity of ball 1 after the collision. x component Ball 1 after Ball 2 after Ball 1 before Ball 2 before
44 y component Ball 1 afterBall 2 after Ball 1 beforeBall 2 before
46 Center of Mass
47 Suppose m 1 =5kg, m 2 =12kg x 1 =2m, x 2 =6m
48 During a time displacements of the particles, displacement of cm
49 Before collision After collision m 1 =0.25 kg, m 2 =0.8 kg v 01 =5 m/s, v 02 =0 m/s
50 Check your understanding 4 Water, dripping at a constant rate from a faucet, falls to the ground. At any instant, there are many drops in the air between the faucet and the ground. Where does the center of mass of the drops lie relative to the halfway point between the faucet and the ground: above it, below it, or exactly at the halfway point? Above the halfway point (consider gravity)
51 Concepts & Calculations Example 10. A Scalar and a Vector
52 Jim and Tom are both running at a speed of 4m/s. Jim has a mass of 90kg, and Tom has a mass of 55kg. Find the kinetic energy and momentum of the two-jogger system when (a)Jim and Tom are both running due north. (b)Jim is running due north and Tom is running due south.
55 Concepts & Calculations Example 11. Momentum and Kinetic Energy MassSpeed Object A 2.0 kg6.0 m/s Object B 6.0 kg2.0 m/s Find the magnitude of the momentum and the kinetic energy for each object.
57 Problem 6 REASONING The impulse that the roof of the car applies to the hailstones can be found from the impulse- momentum theorem, Equation 7.4. Two forces act on the hailstones, the average force exerted by the roof, and the weight of the hailstones. Since it is assumed that is much greater than the weight of the hailstones, the net average force is equal to. SOLUTION From Equation 7.4, the impulse that the roof applies to the hailstones is:
58 Solving for (with up taken to be the positive direction) gives This is the average force exerted on the hailstones by the roof of the car. The positive sign indicates that this force points upward. From Newton's third law, the average force exerted by the hailstones on the roof is equal in magnitude and opposite in direction to this force. Therefore, Force on roof = -1.8 N The negative sign indicates that this force points downward.
59 Problem 20 REASONING During the time that the skaters are pushing against each other, the sum of the external forces acting on the two-skater system is zero, because the weight of each skater is balanced by a corresponding normal force and friction is negligible. The skaters constitute an isolated system, so the principle of conservation of linear momentum applies. We will use this principle to find an expression for the ratio of the skater’s masses in terms of their recoil velocities. We will then obtain expressions for the recoil velocities by noting that each skater, after pushing off, comes to rest in a certain distance. The recoil velocity, acceleration, and distance are related by Equation 2.9 of the equations of kinematics.
60 Ignore kinetic friction Skater 1 glides twice as far as skater 2 Conservation of momentum Initial momentum = final momentum
61 v f1 and v f2 are just after pushing each other. That will be the initial velocity for the motion of the skaters and will come to rest. v f1 v f2 x2x2 x1x1 V end =0 stop a1a1 a2a2 For skater 1
62 For skater 2 Magnitude of the acceleration is the same, i.e.,
63 If x 1 is positive, x 2 is negative, ( 1 glides twice as far as 2 ) (opposite direction)
64 Problem 34 REASONING AND SOLUTION Momentum is conserved in the horizontal direction during the "collision." Let the coal be object 1 and the car be object 2. Then The direction of the final velocity is to the right. vfvf
65 Problem 39 REASONING The two balls constitute the system. The tension in the wire is the only non-conservative force that acts on the ball. The tension does no work since it is perpendicular to the displacement of the ball. Since W nc =0 J, the principle of conservation of mechanical energy holds and can be used to find the speed of the 1.50-kg ball just before the collision. Momentum is conserved during the collision, so the principle of conservation of momentum can be used to find the velocities of both balls just after the collision. Once the collision has occurred, energy conservation can be used to determine how high each ball rises.
66 SOLUTION a. Applying the principle of energy conservation to the 1.50-kg ball, we have If we measure the heights from the lowest point in the swing, h f =0 m, and the expression above simplifies to Solving for v f, we have h0=h0= m1=m1= m2=m2= v0=v0= h f =0, v f =?
67 Just before Just after m1m1 m2m2 v 01 v 02 v f1 v f2 Conservation of momentum
68 Conservation of energy
69 b. If we assume that the collision is elastic, then the velocities of both balls just after the collision can be obtained from Equations 7.8a and 7.8b: and Since v 01 corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the quantity v f calculated in part (a). With the given values of and,and the value of obtained in part (a), Equations 7.8a and 7.8b yield the following values: The minus sign in v f1 indicates that the first ball reverses its direction as a result of the collision. v f1 = m/s
70 c. If we apply the conservation of mechanical energy to either ball after the collision we have where v 0 is the speed of the ball just after the collision, and h f is the final height to which the ball rises. For either ball, h 0 = 0 m, and when either ball has reached its maximum height, v f = 0 m/s. Therefore, the expression of energy conservation reduces to
71 Thus, the heights to which each ball rises after the collision are
72 Problem 41 REASONING AND SOLUTION The location of the center of mass for a two-body system is given by Equation 7.10:
73 where the subscripts "1" and "2" refer to the earth and the moon, respectively. For convenience, we will let the center of the earth be coincident with the origin so that and, the center-to- center distance between the earth and the moon. Direct calculation then gives
74 Problem 50 REASONING AND SOLUTION a. According to Equation 7.4, the impulse-momentum theorem,. Since the only horizontal force exerted on the puck is the force exerted by the goalie. Since the goalie catches the puck,. Solving for the average force exerted on the puck, we have -2.2 * 10 3 N
75 By Newton’s third law, the force exerted on the goalie by the puck is equal in magnitude and opposite in direction to the force exerted on the puck by the goalie. Thus, the average force exerted on the goalie is 2.2 * 10 3 N b. If, instead of catching the puck, the goalie slaps it with his stick and returns the puck straight back to the player with a velocity of –65 m/s, then the average force exerted on the puck by the goalie is = -4.4 * 10 3 N -3
76 The average force exerted on the goalie by the puck is thus +4.4 * 10 3 N. The answer in part (b) is twice that in part (a). This is consistent with the conclusion of Conceptual Example 3. The change in the momentum of the puck is greater when the puck rebounds from the stick. Thus, the puck exerts a greater impulse, and hence a greater force, on the goalie.