You may wish to closely review the following figures, examples, and the text sections that discuss them:

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1 Physics 1061 Fall 007, Temple University C. J. Martoff, Instructor Midterm Review Sheet The midterm has 7 or 8 questions on it. Each is a "problem" as opposed to definitions, etc. Each problem has several parts to allow easier award of partial credit. The exam is not multiple choice. Any and all material in the 80 pages of reading which make up the first six chapters of the textbook is fair game for this exam and for the final exam. You may wish to closely review the following figures, examples, and the text sections that discuss them: Fig..8, Ex..6 Fig. 3.3, 3.4, 3.5, Ex. 3.5 Ex. 4.4 Ex. 5.1, Sec. 5.5 Fig Fig. 6.5 Good luck! ========================================== Physics 1061 Fall 007, Temple University C. J. Martoff, Instructor First Midterm Exam 1. A locomotive on a straight track accelerates from rest at 0.5 m/s and reaches a speed of 15 m/s. It then travels at constant speed for 1 minute. Finally it slows to a stop by undergoing a uniform (de-)acceleration during 15 seconds. Use the graph above to show the position, velocity and acceleration vs. time for the locomotive during these movements. You must label the scale on the time axis, and the features of your graphs (changes of slope, passages through zero, etc.) must be accurately aligned with the time axis. Label the scales on the v and a axes, but you need not do this for the position axis.

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3 Review Item: Fig..8 and accompanying text. A particle with constant acceleration has a straight-line velocity vs. time graph and a parabolic position vs. time graph. If the acceleration is positive(negative): the velocity vs. time graph slopes upward(downward), since acceleration is the derivative of velocity the position vs. time graph is concave upward(downward), since acceleration is the second derivative of position To label the time axis one must also be able to figure out long it takes to reach 15 m/s with acceleration 0.5 m/s. Since v=v 0 +at one finds t=(15m/s)/(0.5m/s ) = 30 sec. Note that the constant-speed part of the motion lasts for 60 seconds i.e. starts at 30 seconds and ends at 90 seconds. Note also that there can be no "kinks" (discontinuities of slope) in the position vs. time graph, since the velocity would be undefined at such a "kink".. Consider the vectors: B = i + 3 j C = -i - j D = 4i a) Find B - C Review Item: Fig. 3.5 and accompanying text p. 34. B- C = 3i + 4 j b) Find the angle between C and D Review Item: Fig. 6.5 and accompanying text Eqn C D = (-)((4) + (-1)(0) = -8. This is equal to C D cosθ CD = cos CD =4 5cos CD. Solving for cosθ CD gives cosθ CD = -.894, so θ CD = A block of mass m is placed and then released from rest, halfway up a frictionless ramp that slants upward at θ=15. The mass is initially a distance h = 0.15 m above the flat floor that the ramp sits on. a) What is the magnitude and direction of the normal force exerted by the ramp on the mass? Review Item: Ex Magnitude of the normal force (n y in Ex. 5.1) is mgcosθ and its direction is upward perpendicular to the ramp surface. Any number that evaluated to mgcosθ was given credit i.e..966mg, 9.46m, etc. b) What is the magnitude and direction of the acceleration of the mass? As shown in Ex a = gsinθ =.54 m/s and its direction is downward parallel to the ramp surface. Any number that evaluated to gsinθ was given some credit. c) What is the final speed of the mass when it reaches the bottom of the ramp?

4 The easiest way to do this is to use the Work Energy Theorem as described in Review Item Fig Since the work done is by gravity on the way down is mgh, this is equal to the final kinetic energy 1/ mv. Solving for v gives v= gh = 1.71 m/s. d) What is the work done on the mass by gravity during its slide? Review Item Fig shows that the work done is by gravity on the way down is mgh, e) What is the work done on the mass by the normal force during its slide? The normal force is perpendicular to the displacement, so it does no work on the way down because W = F Δx, and the dot product of two perpendicular vectors is zero. 4. Two blocks of mass m 1 =1 kg and m =3 kg are in contact with one another, being pushed across a floor with kinetic friction coefficient μ = 0.. The pushing force is horizontal, it is large enough that it causes the blocks to accelerate at a =.5 m/s, and it is applied to m 1. Review Item: Ex. 4.4 plus basic ideas about friction from Ch a) What is the magnitude of the net force on the blocks? F net =ma = (m 1 + m )a for the two blocks together. Solving for F net gives (.5 m/s)*(4 kg) = 10N. b) What is the magnitude of the pushing force? F net = F push + F friction. In 1-D with the push and the friction oppositely directed this can be written F net =F push - F friction = 10 N. The magnitude of the friction force is μmg = μ(m 1 + m )g = 7.84 N, so solving for F push gives 17.8 N. c) What is the magnitude of the friction force on m 1? The magnitude of the friction force on m 1 is μm 1 g =.0 N. d) What is the magnitude of the net force on m 1? F net =m 1 a = (1 kg)(.5 m/s ) =.5 N. e) Why is the answer to d) not just the difference between b) and c)? Evidently m 1 is pushing on m because m is accelerating. Newton's third law then requires that m is pushing back on m. This force must be included along with b) and c)

5 when calculating the net force on m 1. f) What is the magnitude of the force that m 1 exerts on m? From d), F net =m 1 a = (1 kg)(.5 m/s ) =.5 N. The net force is the sum of all the forces acting on m 1 so F net = F push - F friction -F m on m1. Solving for F m on m1. gives 17.8 N -.0 N -.5 N = 13.3 N. 5. A. cal bullet of mass.0035 kg is shot vertically upward from ground level on the Earth at 340 m/s. Study Item: Ex..6. Also need to know the Work Energy Theorem, equation a) Neglecting air resistance, what maximum height will it reach? Per Ex..6 h = y 0 + v 0 /g = 0 + (340m/s) /[(9.8m/s )] = 5.9 x 10 3 m. b) Neglecting air resistance, what speed will it have when it returns to the ground? Exactly the same calculation as Ex..6 but initial and final state are reversed. Final speed on the way down will be the same as initial launch speed, v 0 = 340 m/s. c) Suppose that it is found that due to air resistance, the bullet only rises to 0.8 of the height calculated in part a). Use the Work Energy Theorem to calculate how much work was done by the air resistance on the way up. W net = ΔK = Δ(1/ mv ). The net work is the work done by gravity = -0.8 mgh, plus the work done by air resistance. This must equal Δ(1/ mv ), which is just equal to 1/ m v 0 since the bullet now stops at 0.8 h. So we have -0.8 mgh + W air = 1/ m v 0. This could be solved numerically for W air by plugging in information already available from previous parts. But it is easier if we realize that applying the Work Energy Theorem to part a) shows that 1/ m v 0 must equal minus the (negative) work done by gravity while the bullet rose to the full height h in part a). So the Work Energy equation becomes -0.8 mgh + W air = mgh. Solving for W air we find 0.mgh = 40.5 J. d) What will be the ratio of the kinetic energy when the bullet returns to the ground, to its initial kinetic energy, when this air resistance is included? (Hint; answer can only be approximate) This was the toughie. The work done by air resistance on the way up is given as 0. mgh = 0. times the initial kinetic energy. One might guess that on the way down it would be the same 0. mgh, in which case the kinetic energy on return to the ground be 0.6 of its original value. However the work done by air resistance on the way down will be less than. mgh, because the force of air drag increases with the speed. The bullet has lost some energy to air drag on the way up, so it will not get going as fast as v 0 on the way down. Therefore the work done by air resistance will be somewhat less than 0.mgh on the way down. The most that can be said without a detailed numerical calculation is 0.8 > K final/k initial > 0.6.