PHSX 211 10th Edition Lecture 19 Outline of Last Lecture Practice ProblemsI. A 50 kg archer standing on frictionless ice shoots a 100 g arrow at a speed of 100 m/s. What is the recoil speed of the archer?II. A particle initially at rest explodes into three directions. One momentum vector is expressed by p1= 3i and the second by p2 = -2i – 2j. What is the third momentum vector?III. Try to create your own 2-dimensional collision problem that is solvable to help you understand what information is needed in able to solve these types of problems.Outline of Current Lecture Practice ProblemsI. What is the initial and final gravitational potential energy of a 1 kg ball that starts at 1 m above the ground and ends at 2 m above the ground?II. A roller coaster car rolls down a frictionless track, reaching speed vf at the bottom. If youwant the car to go twice as fast as the bottom, by what factor must you increase the height of the track? Does your answer depend on whether the track is straight or not?III. A car is going 15 m/s when it runs out of gas at the bottom of a 10 m high hill. Can it coast to the top of the hill? Assume friction is negligible.IV. Three balls of equal mass are fired with equal speeds from the same height above the ground. Rank the speeds as they hit the ground.V. A block of mass 100 kg slides from rest down a frictionless incline. Find the speed at the bottom using:a. Energy methodsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.b. Newton’s laws force methodsCurrent LecturePractice ProblemsI. What is the initial and final gravitational potential energy of a 1 kg ball that starts at 1 m above the ground and ends at 2 m above the ground?a. Gravitational Potential Energy (U) = mgyb. U(initial) = (1kg)(9.8)(1m) = 9.8 Jc. U(final) = (1kg)(9.8)(2m) = 19.6 JII. A roller coaster car rolls down a frictionless track, reaching speed vf at the bottom. If youwant the car to go twice as fast as the bottom, by what factor must you increase the height of the track? Does your answer depend on whether the track is straight or not?a. Assume the initial velocity is 0 m/s and the final height is 0 m.b. Ki + Ui = Kf + Ufc. (1/2)mv2 + mgy = (1/2)mv2 + mgyd. Mgy = (1/2)mv2 e. Look at just y and v relationship: y = v2f. So if v is doubled then y will need to be 4 times as much or as high.III. A car is going 15 m/s when it runs out of gas at the bottom of a 10 m high hill. Can it coast to the top of the hill? Assume friction is negligible.a. Ki + Ui = Kf + Ufb. (1/2)mv2 + mgy = (1/2)mv2 + mgyc. (1/2)(15)2 + (9.8)(0) = (1/2)(0)2 + (9.8)yd. [(1/2)(15)2 ]/9.8 = ye. Y = 11.5f. Yes (the maximum distance the car can travel in the vertical direction, so it can make it to the top of a 10 m hill)IV. Three balls of equal mass are fired with equal speeds from the same height above the ground. Rank the speeds as they hit the ground.a. All three balls hit the ground with the same final velocity, just in different ways. This is because the distribution of energy has to be the same if they are starting at the same height, have the same mass, and are thrown with the same initial velocity. V. A block of mass 100 kg slides from rest down a frictionless incline. Find the speed at the bottom usinga. Energy methodsi. Ki + Ui = Kf + Ufii. (1/2)mv2 + mgy = (1/2)mv2 + mgyiii. (1/2)(0)2 + gy = (1/2)v2 + mg(0)iv. Vfinal = 31.3 m/sb. Newton’s laws force methodsi. Fy: may = n – mgcos20 = 0ii. Fx: max = mgsin20iii. ax = gsin20 = 3.35 m/s2iv. Using law of sines find length x = 146.2 m100 kgx m20 degrees50 mv. Using kinematics equations: vf = 31.3