pre lab 3

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1401

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Physics

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Dec 6, 2023

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PRE-LAB Instructions: Print out these pages. Feel free to refer to the lab instructions and other materials, your physics textbook, other students, etc. to help you to ponder, understand, and work out answers to the following question(s). Show your work & answers in the space(s) provided. PRE-LAB Questions: A car is driven along a straight road. Its distance relative to a cross street is graphed at the right: 1)draw CIRCLES around groups of points that indicate zero velocity for the car. 2) draw PLUS symbols near group points that indicate a positive velocity.
3) draw MINUS symbols near group points that indicate a negative velocity. 4) draw BOXES around groups of points that indicate either positive or negative acceleration for the car
5) In the space below, sketch a graph of the velocity vs. time corresponding to the above chart.
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6) Consider a ball that is dropped from rest from a height of 1.0 meter above the floor. It starts at rest, then speeds up, suddenly reverses direction, and then slows as it rises. Suppose it bounces off the floor and rises back up to a height of 1.0 m (it is a super superball). Sketch x, y, vx, and vy vs. time graphs of this motion (on a separate sheet of paper),as illustrated in the theory section of this report. X vs time graph: the ball travels x-direction Y VS time graph: Vx Vs time graph: Vy vs time graph:
7) In the diagram below, a cart is released from rest at a distance x1 from the first of two photogates. The cart then rolls down the ramp through both photogates. The distance between the photogates is x2. For this problem, you may assume that the cart is frictionless and the acceleration of the cart a = g sin θ a) If x1 = 20 cm, x2 = 50 cm, and θ = 20°, what is the average velocity of the cart between the photogates (over the distance x2)? a= gsin ∅ a=9.8sin20=3.352 vx1= √ 0 + 2 ( 3.352 ) ( 20 ) =0.1158 m/s Vx2=(0.1158)^2+2(3.352)(50) =0.2166m/s ´ v = ¿ (0.1158+0.2166)/2 =0.166m/s b) If x1 is reduced to 19 cm and everything remains the same, what is the average velocity of the cart between the photogates (over the distance x2)? T1= √ 2 ( 0.19 m )/ ( 9.8 m s 2 ) sin 20 =0.337s T2= √ 2 ( 0.70 m )/ ( 9.8 m s 2 ) sin 20 =0.646s V1=(0.51m)/(0.646s-0.337) =1.65 m/s
8) A cart is pushed up a ramp with an initial speed vo = 2.5 m/s. For this problem, you may assume that the cart is frictionless and the acceleration of the cart a = g sin θ. a) If the angle of the ramp θ = 20°, what is the maximum distance d that the cart will travel up the ramp? S=v^2-v0^2/(2gsintheta) =(0)^2-(2.5)^2/2(-(9.8)sin20) =0.93 m c) If θ is increased to 45°, what is the maximum distance d that the cart will travel up the ramp? S=v^2-v0^2/(2gsintheta) =(0)^2-(2.5)^2/2(-(9.8)sin45) =0.45 m
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