ic_sol_w08d1_1 - Problem 24: Two acrobats. An acrobat of...

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Problem 24: Two acrobats. An acrobat of mass m A jumps upwards off a trampoline with an initial velocity v 0 . At a height h 0 , the acrobat grabs a clown of mass m B . Assume that the time the acrobat takes to grab the clown is negligibly small. a) What is the velocity of the acrobat immediately before grabbing the clown? b) What is the velocity of the acrobat immediately after grabbing the clown? c) How high do the acrobat and clown rise? How high would the acrobat go in the limit that the mass of the acrobat is much heavier than the mass of the clown? d) If the acrobat missed the clown, how high would the acrobat go? How does this compare to your limit in part c)? Solution: Modeling the problem: The first important observation to make is that there is a collision between the acrobat and the clown. This collision is completely inelastic in that two bodies collide and “stick’ together after the collision. Since we are not given any specific details of the collision, we cannot describe the microdynamical description of how the acrobat slows down and how the clown speeds up. However if we choose as our system the acrobat, clown, and the earth, then the details of the collision are determined by internal forces. We can identify four states in this problem. We need to specific the position and velocities of the acrobat and the clown in each of these states. Since this is essentially one dimensional motion, let’s choose an origin at the trampoline and the positive y-axis upwards.
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r State 1: Acrobat A just leaves trampoline with initial velocity v 1, A = v 0 ˆ j , and initial position y 1, A = 0 . Denote this time by t 1 = 0 . Clown B is at rest at the position y B = h 0 . State 2: Acrobat A just arrives at platform located at y 2, A = y 2, B = h 0 with velocity r v A = v A ˆ j , immediately before grabbing Clown B. Denote this time by t 2 . State 3: The collision lasts a time t col . During this time interval acrobat A grabs Clown r B and the two acrobats and at the end of the interval rise together with velocity v 3 = v 3 ˆ j . Denote the time at the end of this interval by t 3 = t 2 +∆ t col .The key assumption is that the collision time is instantaneous t col 0 . State 4: The two acrobats rise to a final height y = h f and hence are at rest, v r 4 = 0 r . f Model: Since there are no external forces doing work between states 1 and 2, and states 3 and 4 the mechanical energy is constant between these states. There are no external forces acting during the time interval between states 2 and 3. Therefore during the collision, the total momentum of the system is constant. The change in the earth’s momentum is negligible but there is some slowing down of the acrobat A during the collision. If the collision lasts a significant length of time, then we need to calculate this effect. However by assuming the collision is instantaneous, we can ignore this slowing down, and then the change in momentum of the two acrobats before and after the collision is zero.
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This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.

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ic_sol_w08d1_1 - Problem 24: Two acrobats. An acrobat of...

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