Hello can you please just do part 3

Here all the answers for part 2

Part 2: Conservation of Ener Pinball machines work by pulling a pin that compresses a spring. When the pin is released, the spring extends and pushes the ball. See diagram below: Knowns o The spring constant on pinball machines can vary from 60.0 300.0 me. Choose a k value for your machine . 0 Estimate how far back someone would pull the pin. lam estimating cm. 0 Research to find a reasonable frictional co-efficient between the pinball and the playing surface. [1 = _ Note: Source the site you found your frictional co-efficient o The mass of a pinball is 80.0 g. Solve Using the dimensions you estimated in part 1, and the values you have above, determine the speed of the pinball when it reaches the top of the table after you shoot it. Note: To make fife simpler, assume when the spring is compressed, that is horizontal and at against the front of the machine. Aiso, use ENERGY methods. Do not use only knowiedgefrom our dynamics unit. Part 3: Momentum The ball you just shot had an elastic collision with the back wall and has rolled halfway back down the table, calculate its new velocity. Include the effects of friction. The initial velocity at the top of the table is the final velocity you calculated in part 2. Note: Use ENERGY methods. Do not use only knowiedge from our dynamics unit. Another ball comes at [insert birth month + 2 Ex. October - 10 + Z = 5] m;'s at an angle of 6 [insert birth day Ex. 31] degrees below the horizontal. It leaves with a velocity of [cut the original velocity in half ex. 2.5] mfs, and an angle of (1 [your birthday * 2 ex. 62] degrees below the horizontal. 1) What is the velocity of the first ball after they collide? Note: We are using our birthdays to make up numbers for velocity and angles, these numbers may not be realistic. As a result, the numbers we calculate may not seem super realistic either. 2) Was this collision elastic or inelastic? Give both mathematical and conceptual explanations. 3) Does this abide by the conservation of energy? Explain. 4) Would a pinball be able to hit just the right features on the pinball table top such that it continued moving forever? Explain. Part 4: Reflection How do pinball machines use the concepts of energy and momentum? How would you suggest they improve? Note: Ideas for improvement should be related to energy and momentum.mKnowns: - -va|ue for the machine: use 100 Nlrn how far back would you compress the ball: x=3 cm coefficient of friction: assume wood table (Source: https://bri|Iiant.org/wiki/identifying-action-reaction-forces-on-free-body/ Static friction: 0.3, Kinetic friction: 0.2 mass of the pinball: 80.0 g _ l 2 - Potential Energy: The potential energy of the system is the elastic potential energy by the spring. It has the formula: U T 2 I\" where k is the spring constant and x is how far you compressed the ball. 2 1 Em\" where m is the mass of the ball and v is the speed of the ball after we . Kinetic Energy: The kinetic energy of the system is the usual. K : shoot it. W: . Work due to friction: The work due to friction is equal to _Ff\"c($)where Ff'icis the frictional force given by ch : MN : J"'km':(1"where iu'kis the kinetic friction, rn is the mass of the pinball and g is the acceleration due to gravity. N is called the normal force on the ball which is equal to the magnitude of the weight mg. The value of work is negative because the direction of the frictional force is opposite the displacement of the ball. For the system at hand: Potential Energy U(0.03m) : %(100N/m)[0.03m}2 : 0.045J Energy loss (work done by friction) W = p,kmgx = (0.2)(0.08kg) (9.8m/32)(0.03m) = 0.0047J Speed calculation: Using the law of conservation of energy: T:U+K+W:consmnt where T is the total energy, U is the potential energy, K is the Kinetic Energy and W is the energy loss: Using the law of conservation of energy: T = U+K +W = constant where T is the total energy, U is the potential energy, K is the Kinetic Energy and W is the energy loss: E = U +W E = 0.045J - 0.0047J E = 0.0403J Using E = mv 2E 2(0.0403J) U = 0.08kg = 1.003m / s m