* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chris Khan 2008 Physics Chapter 9 Linear momentum is defined as
Tensor operator wikipedia , lookup
Quantum vacuum thruster wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Monte Carlo methods for electron transport wikipedia , lookup
Faster-than-light wikipedia , lookup
Atomic theory wikipedia , lookup
Routhian mechanics wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Classical mechanics wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Accretion disk wikipedia , lookup
Photon polarization wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Mass in special relativity wikipedia , lookup
Angular momentum wikipedia , lookup
Matter wave wikipedia , lookup
Centripetal force wikipedia , lookup
Angular momentum operator wikipedia , lookup
Equations of motion wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Work (physics) wikipedia , lookup
Mass versus weight wikipedia , lookup
Classical central-force problem wikipedia , lookup
Center of mass wikipedia , lookup
Specific impulse wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Chris Khan 2008 Physics Chapter 9 Linear momentum is defined as the product of mass and velocity of an object. o P = mv When dropping an item, the momentum before hitting the floor is –mv since the velocity is in the negative y-direction. If the item lands on the floor, final momentum is 0. If the item bounces, the final momentum is mv. o When dropping a beanbag, ∆P = Pf – Pi = 0 – m(-v) = mv. o When dropping a rubber ball, ∆P = P f – Pi = mv – m(-v) = 2mv. Since momentum is a vector quantity, the total momentum of a system is the vector sum of the momenta of all the objects. o If two 4-kg ducks and a 9-kg goose swim towards bread, swimming at 1.1 m/s and 1.3 m/s, what is the magnitude and direction of the total momentum? Since the ducks are coming from the top and the left and the goose is coming from the bottom, we have to find x and y vectors for momentum. The x-direction is equal to mv = (1.10 m/s)(4 kg) = 4.40 kg m/s (towards the right). The y-direction is equal to mvg – mvd = (1.3 m/s)(9 kg) – (1.1 m/s)(4 kg) = 11.7 – 4.40 = 7.30 kg m/s. Therefore, Newton’s Second Law can be written as F = P/t. Therefore, ∑F = m∆v/∆t. In addition, a = ∆v/∆t. Impulse is found by doing I = Fav∆t. Since Fav = ∆P/∆t, Fav∆t = ∆P. Therefore, impulse is given as the change in momentum I = ∆P. o If you bunt a 0.144-kg baseball, initially moving towards home plate with a speed of 43 m/s, with an average force of 6.50 x 103 N for 1.30 ms, what is the final speed of the ball? First, relate change in momentum to impulse: ∆P = Pf – Pi = I = Fav∆t. Therefore, Next, calculate the impulse: With these results, we can find Pf: Use these with P = mv to find vf = 2.26/0.144 = 15.7 m/s. Jumping for joy, a 72-kg balla jumps upward with a speed of 2.1 m/s. What is the impulse experienced by him? Before he jumps, the floor exerts a force of mg on the contestant. What additional average upward force does the floor exert if the contestant pushes down on it for 0.36s during the jump? First, I = ∆P = P f – Pi = mvf (since v1 = 0). I = mvf = (72)(2.1)y = (150 kg m/s)y. To find average force in terms of the impulse I and the time interval ∆t, Fav = I/∆t = (150 kg m/s)y / 0.36 s = (420 N)y. The Law of Conservation of Momentum says that if the net force acting on an object is zero, its momentum is conserved Pf = Pi. o Two groups of canoeists meet in the middle of a lake when a person in canoe 1 pushes on canoe 2 with 46 N to separate the canoes. If the mass of canoe 1 is 130 kg and the mass of canoe 2 is 250 kg, what is the momentum of each canoe after 1.2 s of pushing? First, find a using a2x = F/m = 46/250 = 0.18 m/s2 and a1x = F/m = -46/130 = -0.35 m/s2. Now, find v after 1.2 s using v = at. This tells us that v1x = -0.42 m/s and v2x = 0.22 m/s. Using this, P1x = m1v1x = (130)(-0.42) = -55 kg m/s. Also, P2x = m2v2x = (250)(0.22) = 55 kg m/s. Recoil is the effect of moving in the opposite direction of the force you exerted. Collisions are the acts of two objects hitting each other when the external forces are 0 or negligible. With inelastic collisions, the momentum is conserved (P f = Pi). In completely inelastic collisions, the objects stick together after the collision. Inelastic Collision Equations o After a collision when objects move together with a common velocity, we can find this velocity: o o At the Superbowl, Tom Brady (95-kg) runs at 3.75 m/s and gets sacked by Tedy Bruschi (111-kg) running at 4.10 m/s. They stick together, so what is their velocity after the collision? What are their initial and final kinetic energies? First, set Pf = Pi, so m1v1 + m2v2 = (m1 + m2)vf. Now, find vf: Now to find their kinetic energies, Ki = ½ m1v12 + ½ m2v22 and Kf = ½(m1 + m2)vf2. o A Mercedes of 950-kg going 16 m/s collides with a beat-down Chevrolet of 1300-kg going at 21 m/s in the same intersection. They collide and stick together. Find the speed and direction of the wrecked vehicles after the collision. First, if you think of it as a triangle, the x component is m1v1, the y component is m2v2 and the hypotenuse is (m1 + m2)vf. Therefore, to find the x component of the momentum, m1v1 = (m1 + m2)vf cos θ and to solve for the y component, m2v2 = (m1 + m2)vf sin θ. We can solve for this algebraically, but now, divide the y equation by the x equation and get Now, o can find the speed using the x-direction equation for momentum: In an elastic collision where m1 is moving with vo initially and m2 is at rest initially, their velocities are: o o we An apple of 0.130-kg and an orange of 0.160-kg collide with speeds of 1.11 m/s and 1.16 m/s, respectively. The orange makes an angle of 42o after collision with respect to its original direction of motion. What is the final speed and direction of the apple assuming an elastic collision? First, find KE i. then KEf in terms of v1f, then set the KE’s equal: The center of mass is one point of any system that has special significance. o o o The same goes for the YCM coordinate. Find the center of mass when an arm is held perpendicular to the table when the upper arm has a mass of 2.5-kg and the CM is 0.18m above the elbow and the lower arm has a mass of 1.6 kg and has a CM 0.40m to the right of the elbow and the hand has a mass of 0.64-kg and a center of mass 0.40m to the right of the elbow. To find velocity of the center of mass, replace the x’s with v’s. To find acceleration of the center of mass, replace the x’s with a’s. o An air cart of mass m and speed vo moves towards a second, identical air cart at rest. When the carts collide they stick together and move as one. Find the velocity of the center of mass of this system before and after the collision. First, use the equation to find the velocity of the mass before collision and then use momentum conservation in the x-direction to find the speed after the collision. Then calculate the velocity of the center of mass of the two carts after the collision.