PHSX 211 10th Edition Lecture 16 Outline of Last Lecture Practice ProblemsI. If you have a car on a banked curve traveling at a fixed velocity, describe the possible forces of friction.II. Write an equation for finding the velocity for the general case above.III. What are the angular velocity and the tangential acceleration for the objects below?Outline of Current Lecture I. Impulse: The amount of force exerted over a period of timea. Impulse (J) = F*tb. J = ΔpII. Momentum: The force described by an object’s mass and velocitya. Momentum (p) = m*vIII. What is the impulse final velocity of an object of 2 kg with a force of -2 N over a time period of .5 seconds with an initial velocity of 1 m/s?IV. What is the initial momentum and final momentum and impulse of an object of 2 kg starting with a velocity of -1 m/s and a final velocity of 2 m/s?V. A bouncy ball of 0.1 kg is traveling at 10 m/s right before it hits a ground and bounces back at 8 m/s. It has an impulse period of .05 seconds. What is the force the ball exerts on the ground?Current LectureVI. Impulse: The amount of force exerted over a period of timea. Impulse (J) = F*tb. J = ΔpVII. Momentum: The force described by an object’s mass and velocityc. Momentum (p) = m*vThese 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.VIII. What is the impulse final velocity of an object of 2 kg with a force of -2 N over a time period of .5 seconds with an initial velocity of 1 m/s?d. J = F*te. J = (-2)(.5) = -1 N*sf. Δp = Jg. pf – pi = Jh. mvf- mvi = Ji. (2)vf – (2)(1) = -1j. Vf = .5 m/s RightIX. What is the initial momentum and final momentum and impulse of an object of 2 kg starting with a velocity of -1 m/s and a final velocity of 2 m/s?k. J = Δpl. mvf- mvi = Jm. (2)(2)-(2)(-1) = Jn. J = 6 N*s, pi = -2 N*s, pf = 4 N*sX. A bouncy ball of 0.1 kg is traveling at 10 m/s right before it hits a ground and bounces back at 8 m/s. It has an impulse period of .05 seconds. What is the force the ball exerts on the ground?o. J = Δpp. mvf- mvi = Jq. (.1)(8) – (.1)(-10) = Jr. .8 + 1 = J = 1.8 N*ss. F = J/tt. F = 1.8/.05 u. F = 36
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