PHYS_2014_Homework_2_soln

PHYS_2014_Homework_2_soln - PHYS2014 Benton Fall 2010 OSU...

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PHYS2014 Fall 2010 Benton OSU Physics Dept. 1 Physics 2014: General Physics I Solutions to Homework Assignment 2 Due: Friday, 10 September 2010 at the beginning of your Recitation session. Problems: 1. A pair of ice skaters are practicing a figure skating move on the ice. The two skaters are separated by a distance of 15 m. If the first skater skates at a constant velocity of 5.0 m/s along a line initially making a 40 ° angle with respect to the line connecting the two skaters, with what velocity will the second skater need to skate at so that she meets the first skater at a right angle? Solution: We are asked to find the velocity, v 2 , for skater 2 such that she covers a distance s 2 while skater 1 covers a distance s 1 skating at a velocity of v 1 . Skater 1 will take time t to skate a distance s 1 at velocity v 1 . Skater 2’s velocity, v 2 , must be such that she skates a distance s 2 in that same time t . Let s 1 be the distance that skater 1 must travel and s 2 be the distance skater 2 must travel as shown in the diagram below. From this diagram, it is clear that: 1 2 15 m cos40 11.49 m 15 m sin40 9.64 m s s = = = = r r Time t is then: 1 m 5.0 s v = a θ = 40 ο 1 5 m 2 ? v = a skater 1 skater 1 1 m 5.0 s v = a θ = 40 ο 2 ? v = a skater 1 skater 1 1 s a 2 s a
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PHYS2014 Fall 2010 Benton OSU Physics Dept. 2 1 1 11.49 m 3.3 s m 5.0 s t v s = = = And the velocity v 2 is: 2 2 9.64 m m 4.2 s 3.3 s s v t = = = 2. I set up my inclined plane apparatus at an angle θ . Then I roll my billiard ball up the track with an initial velocity of 3 m/s as shown in the diagram above. It takes the ball a total of 1.2 s to roll up the track, come to a stop, and then roll back down into my hand. Neglecting frictional effects, find the angle of the track? Solution: The problem is symmetrical in that it will take exactly the same amount of time for the ball to roll up the ramp and come to a stop as it will for the ball to roll down the ramp starting from rest and reaching a final speed of 3 m/s. Thus we only need to consider half the problem, for which the time is 0.6 s. Let just consider that part of the problem where the ball starts at velocity v 0 = 3 m/s, rolls up the ramp while it is being slowed down by the acceleration due to gravity, then comes to a stop after 0.6s. We can rotate the axis by θ° so that the velocity vector is parallel to the x-axis. In that case, the ball will be accelerated by that component of gravity parallel to the x-axis. sin x a g = − We can use our first kinematic equation f v 0 0 1 1 0 2 sin m 3 s sin sin 30.7 m 9.80 0.6 s s v at v g t v gt = + = − = = = r 3. You are on the target range preparing to shoot a new rifle when the range safety officer comes up to you and asks you what the muzzle velocity is of your rifle. (Muzzle velocity is the velocity
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PHYS_2014_Homework_2_soln - PHYS2014 Benton Fall 2010 OSU...

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