SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM I

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1 University of California, Berkeley Physics 7A Sprin 009 (Yury Kolomensky) SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM I Maximum score: 100 points 1. (15 points) Race Stratey Two swimmers need to et from point A on one bank of the river to point B directly across the river on the opposite bank. One of them decides to swim alon the line AB. The other swims perpendicular to the current to the opposite bank, and then walks with velocity u upstream to B from the point where he ets out of the water. Find the value of u such that both swimmers reach B at the same time. Assume that the speed of each swimmer relative to the water is v =.5 km/h, and the speed of the water is v 0 =.0 km/h. 1. Solution v 0 v 1 B v 1 A v 0 v v This is a problem about relative velocities and addition of vectors. The first swimmer moves alon line AB. His velocity, relative to the bank, is a vector sum C v 1 = v 1 + v 0 and its manitude, accordin to the picture above is v 1 = v v0 The amount of time it takes to cross the river is t 1 = AB = AB. v 1 v v0 The second swimmer crosses the river with velocity v = v + v 0 relative to the bank. Its component in the direction of AB is always v, so it takes him t swim = AB v to cross the river. After that, he needs to walk from C to B, which takes t walk = BC u The total time is, therefore, t = AB v + BC u From eometry, it is clear that (1) BC = AB v 0 v. Insertin that into Eq. (1) and requirin t 1 = t we et an equation for u: AB ( 1 v + v 0 uv ) = AB v v0 and after some alebra v v0 u = v 0 = 3.0 m/s. v v v0. (0 points) John Kruer and his parachute There is a scene in the movie Eraser, where the hero John Kruer (Arnold Schwarzeneer) is fallin head first from an airplane, tryin to catch up with a folded parachute. Let s see if Hollywood ot some basic physics riht. In the scene, Kruer drops the parachute first. We will assume, that the parachute is fallin with the constant vertical velocity v p = 0 m/s (that s rouhly the terminal velocity for a folded parachute). t = 7 seconds later, after dodin a few bullets from the bad uys, Kruer lets o of the plane, and falls head first after the parachute. We will assume that the dra force actin on John is neliible.

2 (a) (10 points) At what altitude, relative to the plane, does John catch up with the parachute? (b) (5 points) What is his velocity at that point? (c) (5 points) If he continues to fall for another 0 seconds, while tryin to put on the parachute, how much farther does he fall? Aain, assume that dra is neliible. (d) (Bonus 5 points) Do you see anythin wron with this scenario?. Solution (a) Let s start the clock at t 0 = 0 when John starts fallin and use y 0 = 0 to be the altitude of the plane. Then, the parachute starts fallin at t p0 = t. Let s also follow the tradition and point the Y axis up (so that neative altitude means, as usual, that the object is below the plane). You may have chosen to point your Y axis downward, in which case all positions and velocities below would switch sin. The equations of motion for John and the parachute are y J (t) = t () y p (t) = v p (t + t) (3) (notice sins). We want to find such time t c so that y J (t c ) = y p (t c ). The equation for t c follows from Eqs. (,3): t c v p t c v p t = 0 which has one allowed (positive) solution t c = v p + vp + v p t = 7.7 s. At the instance t = t c, John would be at the elevation (b) y J (t c ) = t c = 90 m. v J (t c ) = t c = 77 m/s. This velocity is hih, but even at 80 m/s, the dra force actin on John is not sinificant, if he is fallin headfirst. His cross sectional area is about A 0.04 m, and the dra force is approximately F d 0.5 (k/m 3 )Av J = 60 N or about 5% of Schwarzeneer s weiht. So inorin dra for Kruer is justified. (c) Between t = t c and t = t c + 0 sec, John and his parachute have fallen by y = v J (t c )(t t c ) (t t c ) = 3.5 km So John s final altitude at that time t would be y = y J (t c ) + y = t = 3.8 km. Now, how realistic is that? First off, in the movie John spent about 0 sec (instead of 8) to catch the parachute. I uess that adds to the dramatics, but it just can t be riht, unless someone added a lead brick to John s parachute ba (which would make v p sinificantly larer). Also, in 0 sec, John would have fallen by km and would likely have created a sinificant crater on the round. The total fall distance of almost 4 km is also problematic. In the movie, the pilot can be heard sayin Descendin to 3000 feet 0 sec before John jumps. Indeed, when John opens the door, there is no sinificant pressure difference between the plane and the outside, the oxyen masks don t appear, etc. This means the plane was probably flyin below 8000 feet, which is only.4 km. Yet, when John opens the parachute, he appears to still be fairly hih in the air at least 3000 feet or so. The final oof that amused me is the motion alon plane s path (which we inore for this problem, but have to account for in real life). Indeed, in 7 seconds, the jet, which should have had a cruisin speed of at least 00 km/h, would have traveled by about 400 meters. The parachute, on the other hand, would quickly reach the horizontal velocity of v p (it can be seen flyin backwards relative to the plane). When John jumps, he still has a horizontal velocity alon the plane s direction so he is flyin away from the chute. How can he ever catch up? Schwarzeneer may be the Eraser, but he s no Superman (5 points) Ski Jumper Ski jumpin is a nordic sport to be featured in the upcomin Winter Olympics next year. Imaine an athlete slidin down a ski jump of heiht H, startin at rest. Consider the jump to be a ramp with the constant incline until it becomes horizontal near the release point

3 (see picture). Inorin friction, what elevation of the jump point h would provide the maximum fliht distance s max, and what is that distance? H 3. Solution First, let s express distance s in terms of parameters H and h, and then find when s is maximum. Suppose velocity of the skier at the release point is v. The velocity is horizontal. After the release, the skier will move under the influence of ravity, and his x and y coordinates will depend on time as h x(t) = vt (4) y(t) = h t. (5) Pluin in x(t) = s and y(t) = 0 (condition for landin), we et h s = v. (6) Now we need to find the velocity v at the release point. Let s consider the motion of the skier alon the ramp. First, we draw the free-body diaram: H h m α s N Y X If the lenth of the ramp is S, the velocity of the skier at the bottom of the ramp is related to acceleration throuh v = a xs (which is one of the kinematic formulae derived in class), which means v = a x S = S sinα = (H h) (7) where we used H h = S sinα from the properties of the riht trianle. Pluin the expression from Eq. (7) into Eq. (6), we find s = h(h h). In order to find the maximum distance, we need to require that the first derivative of s w.r.t. to h (the only variable in the problem) is zero: The solution is ds dh = H h h(h h) = 0 h = H/;s max = H. Notice that Eq. (7) follows very straihtforwardly from the work-enery theorem we discussed in class this week. Even thouh you are not expected to use the work-enery theorem on the exam, it is an allowed tool, if you are comfortable with it. 4. (40 points) Three Blocks What should the horizontal acceleration a of block A be such that small blocks 1 and are not movin relative to it? Masses of blocks 1 and are equal, and the coefficient of friction between them and block A is equal to µ. Inore pulley and strin masses and friction in the pulley. In the Y direction, the normal force is balanced by the Y projection of the ravity: N = m cos α In the X direction, the acceleration is created by the unbalanced X projection of the ravity force: ma x = m sinα a x = sin α A F 1 N 1 T T m 1 F N a m

4 4. Solution This problem would be much too involved for the actual exam, but it is, nonetheless, an excellent practice example. The trick in this problem is to realize where the normal force that creates friction between block and block A comes from and to set up force equations correctly. We also should not be concerned with forces actin on block A too much it is movin with acceleration a but it is irrelevant what force is causin this acceleration. It is also important to draw the force diarams correctly. It actually turns out to be quite relevant: the answer chanes dependin on the direction of the friction forces, as we will show below. Let s first look at block 1. Since it is not movin relative to A, it is movin relative to the LAB frame with acceleration a. The forces actin on it are strin tension T, friction F 1, ravity m where m is the mass of the block, and normal force N 1. It is clear which way T, m, and N 1 should point (riht, down, and up, respectively) but what about friction F 1? Intuitively, if A is at rest and there is no friction, the block would start to fall and pull block 1 to the riht. Therefore, it is natural to assume that for small accelerations friction would point to the left opposite to the direction of motion. We will examine this assumption a bit later. Now back to the force equations: (X projection) : ma = T F 1 (8) (Y projection) : 0 = N m. (9) From Eq. (9), N = m (no surprise here). Therefore, F 1 = µn = µm. Now let s look at block. It is in contact with block A and is movin with the same acceleration a. Therefore, there will be a force that block A exerts on block. This is the normal force N, since it is normal to the surface of, and it is eneratin a non-zero friction force F. Other forces actin on 1 are ravity m and tension T. The block is not movin in the vertical direction. So the nd Newtons law for block is (X projection) : ma = N (10) (Y projection) : 0 = T m + F. (11) The friction force F is directed upwards in the direction opposite to where the block would have moved in its absence. F = µn = µma from Eq. (10). We can now subtract Eq. (11) from Eq. (8) to et rid of tension T and find a: ma = T µm T + m µma a = 1 µ 1 + µ Now let s come back to our assumption about the direction of the friction forces. What happens if we increase acceleration a beyond a = 1 µ? At some point block 1 would start slidin to the left relative to A. At that point, the direction of the friction force F 1 should clearly be to the riht. How about F? If block 1 wants to slide left, block will be pulled upwards and F should be directed downwards. Let s rewrite our force equations for these directions and consider acceleration a such that the blocks 1 and are still at rest relative to A, but just barely. (Block 1, X projection) : ma = T + F 1 (1) (Block 1, Y projection) : 0 = N m (13) (Block, X projection) : ma = N (14) (Block, Y projection) : 0 = T m F.(15) You may notice that Eq. (1) and Eq. (15) are similar to Eq. (8) and Eq. (11) if we substitute µ µ. Therefore, the solution in this case is a = 1 + µ 1 µ If you d like, you can show this explicitly usin Eqs. (1-15). What will happen if acceleration a is between the two extremes, a = 1 µ and a = 1 µ? The blocks will still be at rest relative to A, but the friction forces would not be maximum (in other words, F 1 µn 1 and F µn ). In particular, for a =, both friction forces are zero. F 1 points left and F points up for 1 µ 1 + µ a < and F 1 points riht and F points down for < a 1 + µ 1 µ. Notice that both forces chane their direction at the same value of a =. In other words, you will never find

5 a situation where F 1 points left and F points down (or the other way around, F 1 points riht and F points up). If that were possible, it would mean that in the absence of friction, blocks would be movin either towards or away from each other, makin tension force either zero or infinite. This is not physical.