Block 7-Momentum and Collision Day 1 odd 11/18 even 11/19 Day 2 odd 11/20 even 11/30 Day 3 odd 12/1 even 12/2 Day 4 odd 12/3 even 12/4 Due Day

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1 Block 7-Momentum and Collision Day 1 odd 11/18 even 11/19 Day 2 odd 11/20 even 11/30 Day 3 odd 12/1 even 12/2 Day 4 odd 12/3 even 12/4 Due Day comets asteroids in solar system Day 1 11/18 or 11/19 Warm-up Word Bank: Related, Equal, un-same, Opposite, Weird, Smelly, Foo-sure, Dumb, Not Equal, Opposition, F#@K nugget, stentchful Complete the sentence: For every action there is a(n) and reaction.

2 Notes Momentum Momentum is inertia in motion. momentum=mass x velocity p=mv m=? v =? p=? Impulse Changes Momentum Force produces acceleration. When the force is increased the change in momentum and velocity also increases. Impulse = force x time = change in momentum I=Ft F=? t=? I= The greater the impulse, the greater the change in momentum. In order to increase momentum, apply the greatest force possible for as long as possible. In order to decrease momentum, decrease the impact time. A longer impact time reduces force. Conservation of Momentum To change momentum exert an impulse. Without a net force there cannot be a change in momentum Momentum is a vector quantity, having magnitude and direction. If no net force acts on the system then no change in momentum can take place within that system. Conservation of Energy: In the absence of external force, the momentum of the system remains unchanged. When momentum does not change it is conserved. Gun back fire

3 Collisions Elastic Collisions are when two objects collide, but do not produce heat and are not permanently altered. Inelastic collisions are when two objects collide and stay together. Both forms of collisions show the conservation of momentum: net momentum before = net momentum after Collsions ususally involve external forces, which usually have little consequence during the collision so the net momentum does not change during impact. Momentum Vectors Momentum is conserved when objects do not move in a straight path. Momentum is a vector quantity. The total momentum of the crash of the car is the momentum of car A and car B before the collision. If the momentum of car A and car B are equil the combined momentum after the collision will be in a northeast direction and is.212 times the momenton of either car A or B before the collision.

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5 Day 2 Warm-up Check your notes The equation for Momentum is The equation for Impulse is in Momentum is Impulse Conservation of Momentum 1. A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttle breaks. The astronaut is able to throw a 10.0 kg oxygen tank in a direction away from the shuttle with a speed of 12 m/s, propelling the astronaut back to the shuttle. Assuming that the astronaut starts from rest, find the final speed of the astronaut after throwing the tank. Here's what you know, m = 63.0 kg, tanks = 10.0 kg, tank's v = 12.0 m/s. Use the formula mv = mv. (10.0 kg) (12 m/s) = (63.0 kg)(v) v = 1.9 m/s. 2. An 85.0 kg fisherman jumps from a dock into a kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.30 m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat? Here's what you know, fisherman's m = 85.0 kg, boat's m = kg, fisherman's v = 4.30 m/s. Use the formula mv = mv. (85.0 kg + 0)( m/s) = (85.0 kg kg)(v) v = 1.66 m/s

6 3. Each croquet ball in a set has mass of 0.50 kg. The green ball, traveling at 12.0 m/s, strikes the blue ball, which is at rest. Assuming that the balls slide on a frictionless surface and all collisions are head-on, fine the final speed of the blue ball in each of the following situations: a. The green ball stops moving after it strikes the blue ball. b. The green ball continues moving after the collision at 2.4 m/s in the same direction. c. The green ball continues moving after the collision at 0.3 m/s in the same direction. Here's what you know m = 0.50 kg, green ball's v = 12.0 m/s a. Use the formula mv = mv. Plug in (0.50 kg)(12.0 m/s) = (0.50 kg)(v) v = 12.0 m/s b. Use the formula mv = mv + mv. Plug in (0.50 kg)(12.0 m/s) = (0.50 kg)(2.4 m/s) + (0.50 kg)(v) v = 9.6 m/s. c. Use the formula mv = mv + mv. Plug in (0.50 kg)(12.0 m/s) = (0.50 kg)(0.3 m/s) + (0.50 kg)(v). v = 11.7 m/s 4. A boy on a 2.0 kg skateboard initially at rest tosses an 8.0 kg jug of water in the forward direction. If the jug has a speed of 3.0 m/s relative to the ground and the boy and skateboard move in the opposite direction at 0.60 m/s, find the boy's mass. The total momentum before and after is zero, so if the jug of water has a momentum of (p = mv) (8.0 kg)(3.0 m/s) = 24 kgm/s forward, which means the skateboard and boy must have gained an equal but opposite momentum backward. So if p = mv, and p = 24 kgm/s backwards, and v =.60 m/s: 24 kgm/s = m(0.60 m/s) m = 40kg then the boy and his board must have a total mass of 40 kg, which means the boy has a mass of 38 kg, if the board has a mass of 2 kg.

7 Perfectly Inelastic and Elastic Collisions 1. A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collision. What is the final velocity of the two vehicle mass? Car has mass 1500 kg, and initial velocity 15.0 m/s. Truck has mass 4500 kg and initial velocity 0 m/s. Final mass is 6000 kg 1500 (15) (0) = 6000(v) = 6000 v v = 3.8 m/s 2. A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg shopping cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and bag? Rice has mass of 9.0 kg and initial velocity of 5.5 m/s Cart has a mass of 18.0 kg and an initial speed of 0 m/s Bag and cart have a mass of 27 kg 9 (5.5) ( 0 )= 27 ( v) = 27 v v = 1.83 m/s 3. A 1.50x10^4 kg railroad car moving at 7.00 m/s to the north collides with and sticks to another railroad car of the same mass that is moving in the same direction at 1.50 m/s. What is the velocity of the joined cars after the collision? First car has mass of 1.5 x 10 4 kg and initial velocity of 7.00 m/s Second car has a mass of 1.5 x x10 4 kg and an initial speed of 1.5 m/s Two cars have a mass of 3 x 10 4 kg (1.5 x 10 4 )( 7.00 )+ (1.5 x 10 4 )( 1.5 ) = (3 x 10 4 )( v ) (1.275 x 10 5 ) = (3 x 10 4 )v v = 4.0 m/s to the north 4. A dry cleaner throws a 22 kg bag of laundry onto a stationary 9.0 kg cart. The car and laundry bag begin moving at 3.0 m/s to the right. Find the velocity of the laundry bag before the collision. The laundry bag has mass of 22 kg and an unknown velocity The cart has a mass of 9.0 kg and an initial velocity of 0 m/s Together mass of 31 kg and a final velocity of 3.0 m/s 22 ( v )+ 9.0 (0) = 31 ( 3.0 ) 22 v + 0 = 93 v = 4.2 m/s, to the right

8 Warm-up Day 3 Explain the difference between Angular and Linear momentum. walter lewin angular momentum ice skater rotation spinning tube lycra demo Circ & Grav Assignment Questions 1. A 1500 kg satelite is launched into a geostationary orbit. a) Calculate the kinetic energy in this orbit. 7.0 x 10 9 = ½ mv 2 v=? b) Calculate the potential energy in this orbit x = m(9.8)h h=? c) Calculate the total energy in this orbit x 10 9 = a+b 2. Calculate the orbital speed of an asteroid orbiting 2.8 x m from thesun. 6.9 x 10 3 m/s 3. Locate the position of a spaceship on the Earth/Moon center line such that the net gravitational force would be zero Newton s. 3.5 x 10 8 m from Earth show all steps fallout floor

9 tony hawk ramp human run loop hotwheel track loop corkscrew track

10 Day 4 Work 1. A 44 kg student on in-line skates is playing with a 22 kg exercise ball. Disregarding friction, explain what happens during the following situations. a. Explain what happens when the student is holding the ball, and both are at rest. The student then throws the ball horizontally, causing the student to glide back at 3.5 m/s. b. What happens to the ball in part (a) in terms of the momentum of the student and the momentum of the ball? Help p=mv p s = m s (v s ) =? p b = -? = 22 V b Solve for V b c. The student is initially at rest. The student then catches the ball, which is initially moving to the right at 4.6 m/s. Calculate the velocity of the student and ball together. Help m s (v s ) +m b (v b )= (m s +m b )v v=? 2. A boy stands at one end of a floating raft that is stationary relative to the shore. He then walks in a straight line to the opposite end of the raft, away from the shore. a. Does the raft move? Explain. b. What is the total momentum of the boy and the raft before the boy walks across the raft? c. What is the total momentum of the boy and the raft after the boy walks across the raft? 3. High-speed stroboscopic photographs show the head of a 215 g golf club traveling at 55.0 m/s just before it strikes a 46 g golf ball at rest on a tee. After the collision, the club travels (in the same direction) at 42.0 m/s. Use the law of conservation of momentum to find the speed of the golf ball just after impact. Before momentum = After momentum plug all values in and solve for the missing v b m c v c + m b v b = m c v c + m b v b 4. Two isolated objects have a head-on collision. For each of the following questions, explain your answer. m c v c + m b v b = m c v c + m b v b a. If you know the change in momentum of one object, can you find the change in momentum of the other object? Explain b. If you know the initial and final velocity of one object and the mass of the other object, do you have enough information to find the final velocity of the second object? Explain c. If you know the masses of both objects and the final velocities of both objects, do you have enough information to find the initial velocities of both objects?

11 d. If you know the masses and initial velocities of both objects and the final velocity of one object, do you have enough information to find the final velocity of the other object? e. If you know the change in momentum of one object and the initial and final velocities of the other object, do you have enough information to find the mass of either object? 5. A 4.0 kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially at rest. The first ball stops after the collision. a. Find the velocity of the second ball after the collision. Before momentum = After momentum plug all values in and solve for the missing v 2 m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 b. Verify your answer by calculating the total kinetic energy before and after the collision. KE before = 1/2m 1 v /2m 2 v 2 = KE after = 1/2m 1 v /2m 2 v 2 =

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