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Long-Answer HW 18 - Motion in a Plane

# Long-Answer HW 18 - Motion in a Plane - Solution for...

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Unformatted text preview: Solution for Long-answer Homework 18 Motion in a Plane problems Solution to Long-answer Homework Problem 18.1() Problem: A box leaves the edge of a table ( 1 . 00m above the ground) with an initial velocity of 1 . 00 m s in the horizontal direction. (a)How long does it take to hit ground? (b)How far does the box travel horizontally before it lands? (c)What is the velocity of the box when it lands? (d)Suppose the box is put on an incline (with angle of inclination θ ) on the table top so that it can gain the velocity it needs to leave the table. As it leaves the table, there is a frictionless curvey bit that changes the direction of the velocity to horizontal, without changing its size. What is the vertical height up the incline at which the box must be released in order to have the desired initial horizontal velocity of 1 m s when it leaves the table? You may assume that the incline is frictionless. Solution to Part (a) The velocity of the box in the vertical direction is zero. To find the time the box takes to hit the ground, consider only the vertical direction, taking positive upward. Δ y =- 1 2 gt 2 t = radicalBigg 2(Δ y )- g = radicalBigg 2(- 1m)- 9 . 81 m s 2 = 0 . 451s Grading Key: Part (a) 5 Points Solution to Part (b) Now, only consider motion in the horizontal direction. The horizontal velocity doesn’t change over the time the box falls. Δ x = v x Δ t = parenleftBig 1 . m s parenrightBig (0 . 451s) = 0 . 451m Grading Key: Part (b) 3 Points Solution to Part (c) Now, we must consider the magnitude and direction of the velocity using the x- and y-components. v y =- g Δ t = parenleftBig- 9 . 81 m s 2 parenrightBig (0 . 451s) =- 4 . 43 m s vectorv = v x ˆ x + v y ˆ y = 1 m s ˆ x- 4 . 43 m s ˆ y Grading Key: Part (c) 4 Points Solution to Part (d) 1 There are a number of ways to solve this problem; here’s one. Acceleration down an incline is given by a = g sin θ (you can easily derive this by drawing a picture and decomposing the force of gravitation into components along the incline and perpendicular to the incline) This is the acceleration that the box experiences down the incline . Since we only know what the final velocity needs to be (to derive this you can eliminate t between the constant-acceleration equations for position and velocity) or, we can just use conservation of energy: The vertical distance needs to be such that- Δ U g = 1 2 mv 2 : Δ y = parenleftbigg 1 9 . 8 × 2 v 2 f parenrightbigg = 5 . 1cm Something to think about: originally when I gave this problem as part of a test, I had put friction on this incline, right up to the curvey bit, so you had to figure out a loss of energy due to the normal force times the friction coeffficient times the distance down the incline related to the distance fallen. I gave both the angle and the coefficient of friction. Somehow this got lost in typing up the solutions. My guess is Jennifer got tired and didn’t want to have to draw the free body diagram which was a good part of the points.part of the points....
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Long-Answer HW 18 - Motion in a Plane - Solution for...

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