Presentation on theme: "MOMENTUM! Momentum Impulse Conservation of Momentum in 1 Dimension Conservation of Momentum in 2 Dimensions Angular Momentum Torque Moment of Inertia."— Presentation transcript:
MOMENTUM! Momentum Impulse Conservation of Momentum in 1 Dimension Conservation of Momentum in 2 Dimensions Angular Momentum Torque Moment of Inertia
Momentum Facts p = m v Momentum is a vector quantity! Velocity and momentum vectors point in the same direction. SI unit for momentum: kg · m /s (no special name). Momentum is a conserved quantity (this will be proven later). A net force is required to change a body’s momentum. Momentum describes the tendency of a mass to keep going in the same direction with the same speed. Something big and slow could have the same momentum as something small and fast.
Momentum and Inertia Inertia is another property of mass that resists changes in velocity; however, inertia depends only on mass. Inertia is a scalar quantity. Momentum is a property of moving mass that resists changes in a moving object’s velocity. Momentum is a vector quantity.
Vocabulary angular momentum collision law of conservation of momentum elastic collision gyroscope impulse inelastic collision linear momentum momentum
Momentum The momentum of a ball depends on its mass and velocity. Ball B has more momentum than ball A.
Momentum Ball A is 1 kg moving 1m/sec, Ball B is 1kg at 3 m/sec. If a 1 N force is applied to deflect each ball's motion. What happens? Does the force deflect both balls equally? Ball B deflects much less than ball A when the same force is applied because ball B had a greater initial momentum.
Calculating Momentum The momentum of a moving object is its mass multiplied by its velocity. That means momentum increases with both mass and velocity. Velocity (m/sec) Mass (kg) Momentum (kg m/sec) p = m v
1.You are asked for momentum. 2.You are given masses and velocities. 3.Use: p = m v 4.Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s 5.Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s 1.The car has more momentum even though it is going much slower. Comparing momentum A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road. The mass of the car is 1,300 kg. A motorcycle passes the car at a speed of 30 m/sec (67 mph). The motorcycle (with rider) has a mass of 350 kg. Calculate and compare the momentum of the car and motorcycle.
Momenta Bus: m = 9000 kg; v = 16 m /s p = 144, 000 kg · m /s Train: m = 3.6 ·10 4 kg; v = 4 m /s p = 144,000 kg · m /s Car: m = 1800 kg; v = 80 m /s p = 144, 000 kg · m /s
Angular Momentum linear momentum. Momentum resulting from an object moving in linear motion is called linear momentum. angular momentum Momentum resulting from the rotation (or spin) of an object is called angular momentum.
Conservation of Angular Momentum Angular momentum is important because it obeys a conservation law, as does linear momentum. The total angular momentum of a closed system stays the same.
Calculating Angular Momentum Angular momentum is calculated in a similar way to linear momentum, except the mass and velocity are replaced by the moment of inertia and angular velocity. Angular velocity (rad/sec) Angular momentum (kg m/sec 2 ) L = I Moment of inertia (kg m 2 )
Calculating Angular Momentum The moment of inertia of an object is the average of mass times radius squared for the whole object. Since the radius is measured from the axis of rotation, the moment of inertia depends on the axis of rotation.
Gyroscopes Angular Momentum A gyroscope is a device that contains a spinning object with a lot of angular momentum. Gyroscopes can do amazing tricks because they conserve angular momentum. For example, a spinning gyroscope can easily balance on a pencil point.
Gyroscopes A gyroscope on the space shuttle is mounted at the center of mass, allowing a computer to measure rotation of the spacecraft in three dimensions. An on-board computer is able to accurately measure the rotation of the shuttle and maintain its orientation in space.
Comparison: Linear & Angular Momentum Linear Momentum, p Tendency for a mass to continue moving in a straight line. Parallel to v. A conserved, vector quantity. Magnitude is inertia (mass) times speed. Net force required to change it. The greater the mass, the greater the force needed to change momentum. Angular Momentum, L Tendency for a mass to continue rotating. Perpendicular to both v and r. A conserved, vector quantity. Magnitude is rotational inertia times angular speed. Net torque required to change it. The greater the moment of inertia, the greater the torque needed to change angular momentum.
Conservation of Momentum The law of conservation of momentum states when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change. If you throw a rock forward from a skateboard, you will move backward in response.
Collisions in One Dimension A collision occurs when two or more objects hit each other. During a collision, momentum is transferred from one object to another. Collisions can be elastic or inelastic.
Conservation of Momentum in 1-D Whenever two objects collide (or when they exert forces on each other without colliding, such as gravity) momentum of the system (both objects together) is conserved. This mean the total momentum of the objects is the same before and after the collision. before: p = m 1 v 1 - m 2 v 2 after: p = - m 1 v a + m 2 v b m1m1 m2m2 v1v1 v2v2 (Choosing right as the + direction, m 2 has - momentum.) m1m1 m2m2 vava vbvb m 1 v 1 - m 2 v 2 = - m 1 v a + m 2 v b
Elastic collisions Two 0.165 kg billiard balls roll toward each other and collide head-on. Initially, the 5-ball has a velocity of 0.5 m/s. The 10-ball has an initial velocity of -0.7 m/s. The collision is elastic and the 10- ball rebounds with a velocity of 0.4 m/s, reversing its direction. What is the velocity of the 5-ball after the collision?
You are asked for 10-ball’s velocity after collision. You are given mass, initial velocities, 5-ball’s final velocity. Diagram the motion, use m 1 v 1 + m 2 v 2 = m 1 v 3 + m 2 v 4 Solve for V 3 : (0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)=(0.165 kg) v 3 + (0.165 kg) (0.4 m/s) V 3 = -0.6 m/s Elastic Collisions
Directions after a collision On the last slide the boxes were drawn going in the opposite direction after colliding. This isn’t always the case. For example, when a bat hits a ball, the ball changes direction, but the bat doesn’t. It doesn’t really matter, though, which way we draw the velocity vectors in “after” picture. If we solved the conservation of momentum equation (red box) for v b and got a negative answer, it would mean that m 2 was still moving to the left after the collision. As long as we interpret our answers correctly, it matters not how the velocity vectors are drawn. m 1 v 1 - m 2 v 2 = - m 1 v a + m 2 v b m1m1 m2m2 v1v1 v2v2 m1m1 m2m2 vava vbvb
Sample Problem before after 3 kg 15 kg 10 m/s 6 m/s 3 kg 15 kg 4.5 m/s v A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the after picture. 3 (10) - 6 (15) = -3 (4.5) + 15 v v = -3.1 m/s Since v came out negative, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1 m/s to the left.
Sample Problem 1 7 kg v = 0 700 m/s A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passing through it, the bullet pushes the butter to the left, and the butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet? (The mass of the rifle matters not.) 35 g 7 kg v = ? 35 g 4 cm/s continued on next slide
Sample Problem 1 (cont.) 7 kg v = 0 700 m/s 35 g 7 kg v = ? 35 g 4 cm/s p before = 7 (0) + (0.035) (700) = 24.5 kg · m /s Let’s choose left to be the + direction & use conservation of momentum, converting all units to meters and kilograms. p after = 7 (0.04) + 0.035 v = 0.28 + 0.035 v p before = p after 24.5 = 0.28 + 0.035 v v = 692 m/s v came out positive. This means we chose the correct direction of the bullet in the “after” picture.
Inelastic Collisions A train car moving to the right at 10 m/s collides with a parked train car. They stick together and roll along the track. If the moving car has a mass of 8,000 kg and the parked car has a mass of 2,000 kg, what is their combined velocity after the collision? You are asked for the final velocity. You are given masses, and initial velocity of moving train car.
Diagram the problem Use m 1 v 1 + m 2 v 2 = (m 1 v 1 +m 2 v 2 ) v 3 Solve for v 3 = m 1 v 1 + m 2 v 2 (m 1 v 1 +m 2 v 2 ) v 3 = (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s) (8,000 + 2,000 kg) v 3 = 8 m/s The train cars moving together to right at 8 m/s. Inelastic Collisions
Sample Problem 2 7 kg v = 0 700 m/s 35 g Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact? v 7. 035 kg (0.035) (700) = 7.035 v v = 3.48 m/s Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional force on the wood in addition to that of the bullet, and the “system” would have to include the table as well.
Bouncing Alfred went on a date with Mabel. When Alfred dropped off Mabel after the date, he was anxious to play Angry Birds, so he forgot to kiss her on the cheek good night. She went up to her room, opened the window and threw a flower pot at Alfred. On of three things could happen. 1. The flower pot – head collision is elastic 2. The flower pot – head collision is inelastic 3. The flower pot bounces off his head Which will hurt more?????
Elastic Collision Alfred + Flower pot = Alfred + Flower pot m 1 v 1 + m 2 v 2 = m 1 v 3 + m 2 v 4 100kg(0m/s) + 10kg (15 m/s) = 100kg (v 3 ) + 10kg (0m/s) 150kgm/s = 100kg (v 3 ) 100kg 1.5 m/s = v 3 (elastic)
Inelastic Collision Alfred + Flower pot = (Alfred + Flower pot) m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v 3 100kg(0m/s) + 10kg(15 m/s) = (100kg + 10kg) (v 3 ) 150kgm/s = 110kg(v 3 ) 110kg 1.36 m/s = v 3 (inelastic) 1.5 m/s = v 3 (elastic)
Elastic Collision Alfred + Flower pot = Alfred + Flower pot m 1 v 1 + m 2 v 2 = m 1 v 3 + m 2 v 4 100kg(0m/s) + 10kg(15 m/s) = 100kg(v 3 ) + 10kg(-5m/s) 150kgm/s = 100kg(v 3 ) – 50kgm/s 200kgm/s = 100kg(v 3 ) 100kg 2.0 m/s = v 3 (bouncing) 1.5 m/s = v 3 (elastic) 1.36 m/s = v 3 (inelastic)