Exam2 practice solution

Exam2 practice solution - Mar06 LeClair PH 105-3 Exam 2...

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Mar06 LeClair PH 105-3 Exam 2 Solutions Instructions: There are 8 problems below. Choose any 6 problems to solve. All questions have equal point values. ± Problem 1: Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (see below). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. Find the original elastic potential energy in the spring if M = 0.250 kg. What we want in the end is to relate the initial energy (±gure part a) stored in the spring to the ±nal energy which is purely kinetic. Thus at some point conservation of total energy is needed. What we ±rst have to realize is that once the cord has been burned, the spring pushes away both masses, and the masses are moving apart in the ±nal state. This means that each block has kinetic energy after the cord is cut, not just the one with mass 3M. Therefore, we need to ±nd the velocity of the block of mass M before we proceed. Finding the velocities after the cord has been cut can be most easily done by conservation of momentum: 1
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- * p before = - * p after - * p before = 0 - * p after = 3M v 3M + M v M v M = - 3 v 3M = - 6m / s After we have both velocities, we can use conservation of energy: K i + U i = K f + U f 0 + U spring = K f + 0 K f = 1 2 M v 2 M + 1 2 (3M) v 2 3M K f = (0 . 5)(0 . 25)( - 6) 2 J + (0 . 5)(3)(0 . 25)(2) 2 J K f = 4 . 5J + 1 . 5 J = 6J U spring = 6J ± Problem 2: A bead slides without friction around a loop-the-loop (below). The bead is released from a height h = 2 . 95 R . What is its speed at point A? Answer in terms of R and g , the acceleration of gravity. Since there is no friction present, we are only dealing with conservative forces, i.e., forces indepen- dent of the path taken. Basically, this means that we can use conservation of energy and make this a very simple calculation. Since the object starts out at rest at a height h oF of the ground, its initial energy is purely potential: E i = K i + U i = mgh When it reaches the top of the loop-the-loop, it has some velocity v and a height oF the ±oor of 2 R . The total energy is now: 2
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E f = K f + U f = 1 2 mv 2 + mg (2 R ) The most common mistake here was to forget that at point A on the loop, the height oF of the ground is not zero and miss the 2 R . Equating E i and E f , mgh = 1 2 mv 2 + 2 mgR mg ( h - 2 R ) = 1 2 mv 2 g ( h - 2 R ) = g (2 . 95 R - 2 R ) = 0 . 95 gR = 1 2 v 2 v 2 = 2(0 . 95) gR v = p 1 . 9 gR Note that if we had measured potential energy relative to the top of the loop instead of the ±oor, we could just write down E i = mg ( h - 2 R ) and E f = 1 2 mv 2 and the result is the same. It has to be, gravitational potential energy only depends on height diferences .
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This note was uploaded on 04/28/2008 for the course PHY 1102 taught by Professor Sawyer during the Spring '08 term at FIT.

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Exam2 practice solution - Mar06 LeClair PH 105-3 Exam 2...

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