7AReviewAns

# 7AReviewAns - Physics 7A Final Review Christopher Smallwood...

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Physics 7A Final Review Christopher Smallwood December 14, 2008 1. (In John Long’s book...) If we examine the bottom carabiner in both situations (a) and (b), we note that a free body diagram in either case must be identical. Consequently, the tension in the strands of webbing running up to the anchor carabiners must also be the same in cases (a) and (b). However, in the American Triangle setup, there is another force acting on the upper carabiners caused by the tension in the strand of webbing running across the top. In the worst-case scenario we could assume no friction between the upper carabiners and the webbing, in which case the tension in the strand across the top would be equal to the tension in the strands running down to the lower carabiner. The resultant stress on the upper carabiner could then be as much as twice the necessary stress. 2. (A block of mass m 1 is set into motion...) (a) Energy is conserved at all points in time and space except the collision between the block and bullet, where heat is generated by the sticking force that keeps the two objects together. Momentum in the x-direction is conserved throughout the entire problem because there are no external net forces in this direction. Momentum in the y-direction is conserved until the block begins to slide up the ramp. At the point it begins its ascent, the system’s center of mass begins to move upward, which it cannot do unless there is a net external force in the y-direction. This is caused by the normal force of the ground. Note that while there are normal forces between the block and ramp, they have no effect on the entire system’s momentum because they are internal forces. (b) The system’s initial momentum p 0 is given by only the bullet’s momentum. Thus p 0 = m 2 v 0 . (1) After the bullet collides with the bullet, the two items have a combined momen- tum p 1 = ( m 1 + m 2 ) v 1 . (2) 1

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Momentum is conserved, and thus p 0 = p 1 and v 1 = m 2 m 1 + m 2 v 0 . (3) (c) From this point forward consider the block and bullet to be one system with mass m and initial velocity v 1 . To solve we rely on two physical concepts: (1), Both energy and momentum are conserved as the block slides up the ramp, and (2), the block and ramp have the same velocity as the block reaches its highest point. According to momentum conservation mv 1 = ( M + m ) v 2 (4) and according to energy conservation 1 2 mv 2 1 = 1 2 ( M + m ) v 2 2 + mgh. (5) Solving these gives the result h = M 2 g ( M + m ) v 2 1 . (6) Note that the limiting cases of this correspond to exactly what you would expect. In the limit M 0 the block just pushes the ramp forward without ever going up the ramp. In the limit M → ∞ the block rises to the same height it would if the ramp were glued to the ground. (d) One way to solve this is to analyze the equations you would need to solve for the motion of the block and ramp exactly. The free body diagrams for the ramp and block would be as shown. From these we obtains the following equations, ma x = - N sin θ (7) ma y = N cos θ - mg (8) Ma x = N sin θ (9) N = Mg + N cos θ, (10) 2
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