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## Chapter 19 Answers and Explanations to Practice SAT Physics Subject Test 1

1 A Momentum is mass times velocity. Since the mass of the object is just a positive constant, the graph of momentum should have the same shape as the graph of the velocity.

2 D If the velocity vs. time graph has a corner, then the acceleration vs. time graph will be discontinuous (have a jump). As shown in the diagram, the first part of the velocity vs. time graph is a straight line with a positive slope. Thus, the corresponding acceleration graph should be a horizontal line above the axis. Next on the velocity vs. time graph there is a curvy part, decreasing but concave up (like the left half of a cup). The slope of that is negative to begin with and then becomes less negative as we approach the center of the cup. Thus a straight line on the acceleration vs. time graph–negative but getting less negative.

3 B Kinetic energy is proportional to v2. Since the first part of the v versus t graph is a straight line, it must have the form v = at for some constant, a. Squaring this gives us something proportional to t2, the graph of which is parabolic. This eliminates A and D. Next, since v drops to 0 in the original graph, the kinetic energy must also drop to 0, so now E is eliminated. Finally, we can eliminate the graph in C, because if it were correct, it would mean that the object had a constant kinetic energy for the latter part of its motion (since the graph is flat); but the original graph shows us that v is never constant.

4 E Since the given graph of v versus t is always above the t axis, that means v is never negative. From this we can conclude that the object never changes direction (because the velocity would change from positive to negative if this were true). If the object is always traveling in the same direction, its distance from the starting point must always increase. This behavior is only illustrated by the graph in E.

5 E Kinetic energy is a scalar, equal to one-half an object’s mass times the square of its speed. Like potential energy and work, kinetic energy does not have a direction associated with it.

6 C Newton’s second law is Fnet = ma, so if we know Fnet and m, we can calculate the acceleration, a.

7 D Impulse is equal to force multiplied by time, so its units are N-s. Because 1 N = 1 kg-m/s2, we see that 1 N-s = 1 kg-m/s, which we immediately notice is the same as the units of mass (kg) times those of velocity (m/s), and mv is linear momentum, D. You may also have remembered the impulse–momentum theorem, which says that the impulse delivered to an object is equal to the resulting change in its linear momentum. Since impulse gives the change in linear momentum, it must be true that impulse can be expressed in the same units as linear momentum.

8 C If an object’s speed is changing, it must be moving, so its displacement is changing. Since velocity, v, is speed plus direction, a change in speed automatically means a change in velocity. Now, since the velocity is changing, the linear momentum, p, must be changing also, since p =mv. Finally, because kinetic energy is equal to times the square of the speed, a changing speed implies a changing kinetic energy. The answer must be C. An object undergoing a change in speed may certainly be subject to a constant acceleration (gravitational acceleration, for example).

9 B The red shift of light refers to the increase in the wavelength (or, equivalently, the decrease in the frequency) of light from a distant source when it’s measured here. This change in wavelength (and frequency) is the Doppler effect and implies that the source and detector are moving away from each other. This provides evidence for the expansion of the Universe.

10 E Stars are huge nuclear-fusion reactors. When nuclei fuse, the mass of the product nucleus is less than the combined masses of the original nuclei. The “missing” mass has become energy, which is radiated away. Einstein’s famous mass-energy equivalence equation, E = mc2, can be used to calculate the amount of energy resulting from a fusion reaction.

11 B Refraction, the change in the direction of a wave when it enters a new medium, is caused by the change in the speed of the wave as it travels within the new medium.

12 C Only a transverse wave, defined to be a wave in which the oscillation is perpendicular to the direction of wave propagation, can be polarized. Since sound waves are longitudinal, they cannot be polarized.

13 B The gravitational force on an object of mass m can be expressed either by mg or by . Setting mg equal to , we get g = , which is the object’s (free-fall) acceleration. Notice that the mass of the object cancels out, so whether we’re asked for the acceleration of the feather or the hammer, the answer would be the same. (By the way, the experiment described in this question was actually performed by Apollo 15 astronaut David Scott on July 30, 1971. Both the feather and the hammer hit the lunar surface at the same time, verifying the fact—first stated by Galileo—that under conditions of no air resistance, all objects fall with the same acceleration, regardless of their mass.)

14 A Let m be the mass of a satellite orbiting the earth in a circular orbit of radius r at a constant speed of v. Since the centripetal force is provided by the gravitational force due to the earth (mass M), we can write . Solving for v gives us v = . This result tells us that the mass of the satellite is irrelevant; only the mass of the earth, M, remains in the formula. Since v is inversely proportional to the square root of r, the satellite that’s closer will have the greater speed. In this case, since Satellite #1 has the smaller orbit radius, it has the greater speed, and, since the radius of its orbit is the radius of Satellite #2’s orbit, its orbit speed is greater by a factor of .

15 E Using the equation p = mv, we can figure out that before the collision, the momentum of the left-hand block was (4 kg)(8 m/s) = 32 kg-m/s, and that of the right-hand block was zero (since it was at rest), so the total momentum before the collision was 32 kg-m/s. Since total momentum is conserved in the collision, the total momentum after the collision must also be 32 kg-m/s.

16 D Using the equation K = mv2/2, we can figure out that before the collision, the kinetic energy of the left-hand block was (4 kg)(8 m/s)2/2 = 128 J and that of the right-hand block was zero (since it was at rest), giving a total kinetic energy before the collision of 128 J. Since kinetic energy is conserved in an elastic collision, the total kinetic energy after the collision must also be 128 J.

17 A If the blocks stick together after the collision—a perfectly inelastic collision—then conservation of momentum gives us 32 kg-m/s = (4 + m)v, where v denotes the common speed of the blocks after the collision. If m = 12 kg, then v = 32/(4 + 12) = 32/16 = 2 m/s.

18 A If the mass of the +q charge is m, then its acceleration is The graph in A best depicts an inverse-square relationship between a and r.

19 E Since Coulomb’s law is an inverse-square law (that is, F is inversely proportional to r2), if r decreases by a factor of 2, then F increases by a factor of 22 = 4.

20 A The period is the reciprocal of the frequency: T = 1/f = 1/(2.5 Hz) = 0.4 sec.

21 A We use the equation that relates wavelength, frequency, and wave speed 22 C When a wave enters a new medium, its frequency does not change, but its wave speed does. Since λf = v, the change in wave speed implies a change in wavelength also.

23 C While the cannonballs are in flight, the only force they feel is the gravitational force, so the acceleration of each cannonball is equal to g.

24 D For an ideal projectile, the horizontal velocity while in flight is constant and equal to the initial horizontal velocity. In this case, the initial horizontal speed of the cannonball shot from ground level is ( v0)cos θ0. Now, multiplying this rate by the time of flight, T, gives the total horizontal distance covered.

25 B As the cannonball falls, it accelerates downward and its speed increases; this eliminates A, C, and D. The mass of the cannonball does not change, eliminating E. The answer is B. As the cannonball falls, its height decreases, so its gravitational potential energy decreases.

26 A B is false, because a solid must absorb thermal energy in order to melt. C is false since it generally requires much more energy to break the intermolecular bonds of a liquid to change its state to vapor than to loosen the intermolecular bonds of a solid to change its state to liquid. And D and E are false: While a substance undergoes a phase change, its temperature remains constant. The answer must be A.

27 E We know that like charges repel and opposite charges attract. So, we can put Charges 1, 2, 3, and 4 into two “camps.” Because Charge 1 attracts Charge 2, these charges must be in opposite camps

1    2

Next, since Charge 2 repels charge 3, Charge 3 is in the same camp as Charge 2

1    2
3

And, finally, since Charge 3 attracts Charge 4, these charges are in opposite camps, giving us

1    2
4    3

We now see that only (E) can be correct.

28 B The current entering the parallel combination containing Resistors b, c, and d will split evenly among the resistors since all their resistances are the same. Because there are 3 resistors in the parallel combination, each resistor in this combination will get 1/3 of the current. Another way of saying that the current through Resistor b is 1/3 the current through Resistor a is to say that the current through Resistor a is 3 times the current through Resistor b.

29 C Using Ohm’s law in the form I = V/R, we find that the current through resistor a is I = V/(10R/3) = (3/10)(V/R).

30 D The power dissipated by a resistor carrying current is given by P = IV or by P = I2R. Since resistors e and f carry the same current I (because they’re in series) and have the same resistance R, they dissipate the same power.

31 E The maximum net force on the object occurs when all four forces act in the same direction, giving Fnet = 4F = 4(10 N) = 40 N, and a resulting acceleration of a = Fnet/m = (40 N)/(5 kg) = 8 m/s2. These four forces could not give the object an acceleration greater than this.

32 E The impulse-momentum theorem says that the change in momentum is equal to the impulse, which is the area under the force versus time graph. The region under the graph from t = 0 to t = 0.4 sec is composed of a right triangle (from t = 0 to t = 0.2 sec) plus a rectangle (from t = 0.2 sec to t = 0.4 sec), so the total area under the graph from t = 0 to t = 0.4 sec is (1/2)(0.2)(100) + (0.4 – 0.2)(25) = 10 + 5 = 15 kg-m/sec.

33 B Apart from sign, the magnification factor, m, is equal to i/o, where i is the image distance from the lens and o is the object distance. In this case then, we have m = i/o = (8 cm)/(20 cm) = 2/5.

34 C The nth harmonic frequency is equal to n times the fundamental frequency, f1. Therefore 35 C If both the source and detector travel in the same direction and at the same speed, there will be no relative motion and hence no Doppler shift.

36 C The frequency does not change, so the wavelength must change (because the wave speed changes). 37 E Gamma rays and X-rays are very high-energy, short-wavelength radiations. Ultraviolet light has a higher energy and shorter wavelength than visible light. Within the visible spectrum, the colors are—in order of increasing frequency—ROYGBV, so orange light (“O”) has a lower frequency—and thus a longer wavelength—than blue light (“B”).

38 D All linear dimensions within the plate—including the radius and circumference of the hole—will increase by the same amount during thermal expansion. (To see that the hole does indeed get bigger, imagine that it was filled with a flat circular plug of metal. This plug would get bigger as the entire plate expanded, so if the plug were removed, it would leave behind a bigger hole.)

39 B The net force acting on the block is 80 N – 60 N = 20 N. Dividing the net force by the object’s mass gives the acceleration (Newton’s second law), so we find that a = Fnet/m = (20 N)/(40 kg) = 0.5 m/s2.

40 C Since momentum, p, is equal to mv, we just multiply the two entries (m and v) in each row of the table and see which one is the greatest. This occurs in Trial 3, where p = (2 kg)(2 m/s) = 4 kg-m/s. So, the answer must be either C or D. To decide which, we only need to find the kinetic energy of the object in Trial 2 and Trial 3 and choose the one that’s greater. Since mv2 = (1 kg)(3 m/s)2 = J in Trial 2, but only (2 kg)(2 m/s)2 = 4 J in Trial 3, we see that the object’s kinetic energy is greater in Trial 2, so the answer is C.

41 C A and E were known before Rutherford conducted this series of experiments, and B and D concerning the electrons were proposed later by Bohr. What Rutherford discovered with these experiments was that an atom’s positive charge was not uniformly distributed throughout the entire atom but was instead concentrated into a very small volume at the atom’s center (the nucleus).

42 B Use the ideal gas law. 43 B We use the equation for the power dissipated by a resistor, P = IV. 44 D To balance the superscripts, we write 2 + 2 = 3 + A, and get A = 1. Now, to balance the subscripts, we write 1 + 1 = 2 + Z, so Z = 0. Therefore, the particle X has a mass number of 1 and a charge of 0; it’s a neutron.

45 B We use the equation q = mcΔT to find ΔT. 46 D From the kinetic theory of gases, we know that the average kinetic energy of the molecules of an ideal gas is directly proportional to the absolute temperature. This eliminates B and C. Furthermore, the fact that KEavgT implies that the root-mean-square speed of the gas molecules,vrms, is proportional to the square root of the absolute temperature. This eliminates A and E.

47 A Ice melts at 0°C and boils at 100°C. During these phase transitions, the temperature remains constant. Therefore, the graph of the sample’s temperature must be momentarily flat at both 0°C and at 100°C.

48 C Relative to the central maximum, the locations of the bright fringes on the screen are given by the expression mL(λ/d), where λ is the wavelength of the light used, L is the distance to the screen, d is the separation of the slits, and m is an integer. The width of a fringe is, therefore (m + 1)L(λ/d) – mL(λ/d) = λL/d. One way to increase λL/d is to decrease d.

49 E If the photons of the incident light have insufficient energy to liberate electrons from the metal’s surface, then simply increasing the number of these weak photons (that is, increasing the intensity of the light) will do nothing. To produce photoelectrons, each photon of the incident light must have an energy at least as great as the work function of the metal.

50 B The energies emitted during electron transitions are equal to the differences between the allowed energy levels. A, 17 eV, is equal to the energy emitted by the photon when an electron drops from the –21 eV level to the –38 eV level. C, 64 eV, is equal to the energy emitted by the photon when an electron drops from the –21 eV level to the –85 eV level. D, 255 eV, is equal to the energy emitted by the photon when an electron drops from the –85 eV level to the –340 eV level. And E, 302 eV, is equal to the energy emitted by the photon when an electron drops from the –38 eV level to the –340 eV level. However, no electron transition in this atom could give rise to a 42 eV photon.

51 A Call the top wire (the one carrying a current I to the right) Wire 1, and call the bottom wire (carrying a current 2I to the left) Wire 2. In the region between the wires, the individual magnetic field vectors due to the wires are both directed into the plane of the page (use the right-hand rule with your right hand wrapped around the wire and your right thumb pointing in the direction of the current), so they could not cancel in this region. Therefore, the total magnetic field could not be zero at either Point 2 or Point 3. This eliminates B, C, D, and E, so the answer must be A. (Because the magnetic field created by a current-carrying wire is proportional to the current and inversely proportional to the distance from the wire, the fact that Point 1 is in a region where the individual magnetic field vectors created by the wires point in opposite directions and that Point 1 is twice as far from Wire 2 as from Wire 1 imply that the total magnetic field there will be zero.)

52 C The proximity of the charged sphere will induce negative charge to move to the side of the uncharged sphere closer to the charged sphere. Since the induced negative charge is closer than the induced positive charge to the charged sphere, there will be a net electrostatic attraction between the spheres. 53 E The capacitance of a parallel-plate capacitor is C = Kε0A/d, where K is the dielectric constant, A is the area of each plate, and d is their separation distance. Decreasing d will cause C to increase.

54 B The resistance of a wire made of a material with resistivity ρ and with length L and cross-sectional area A is given by the equation R = ρL/A. Since Wire B has the greatest length and smallest cross-sectional area, it has the greatest resistance.

55 A The amplitude of a wave is the maximum displacement from equilibrium (the position of zero displacement). Since the distance between the maximum positive displacement and the maximum negative displacement is 0.16 m, the amplitude is half this: 0.08 m.

56 D Each “hump” along the wave represents half a cycle, or half a wavelength. The drawing given with the question shows three such consecutive humps having a total length of 0.6 m, so each one has a length of 0.2 m. Since this is half a wavelength, the full wavelength must be 0.4 m.

57 D The period, T, is the reciprocal of the frequency, and the frequency, f, always satisfies the equation λf = v. So, we find that 58 A Because each half-life is 1.5 hours, a time interval of 6 hours is equal to 4 half-lives. After each half-life elapses, the mass of the sample is cut in half, so after 4 half-lives, the mass of the sample decreases from 2 grams to 1 gram to 0.5 grams to 0.25 grams and, finally, to 0.125 grams.

59 E The electrical potential energy of a pair of charges is given by kq1q2/r, where r is the distance between the charges. Therefore, the electrical potential energy of the pair of charges +Q and +Q will be greatest when the distance between them is the smallest. Of the five diagrams given, the charges are closest together in diagram E.

60 A By the principle of superposition, the total electric force on the charge in the lower right-hand corner is simply the sum of the individual electric forces produced by each of the charges in the other three corners. The first diagram below shows the directions of the individual electric forces that each of the other three charges exerts separately on the lower right-hand charge. The second and third diagrams then show how these three vectors add together to give the net electric force. Note that F2 and F3 are each larger than F1 since FE = . Therefore, F2 + F3 is also larger than F1.

61 C The torque is equal to rF, where F = mg. In this case, then, we find that torque = rmg = (0.5 m)(0.2 kg)(10 N/kg) = 1 N-m.

62 B Because the rocks are dropped from rest simultaneously and air resistance is negligible, both rocks will accelerate at the same rate and have the same speed, v, at impact. So, the ratio of Rock 1’s momentum to Rock 2’s momentum will be (M1v)/(M2v) = M1/M2.

63 B Because the forces N and w are vertical while the displacement of the block is horizontal, the work done by each of these forces is zero, so they contribute nothing to the total work performed. This eliminates C, D, and E. Since the force f is opposite to the direction of the displacement, the work it does is negative; in fact, it’s –fL. The work done by the force F is FL, so the total work performed on the block is –fL + FL = (Ff)L.

64 A B, C, D, and E all change the magnetic flux through the loop containing the light bulb, thus inducing an emf and a current. However, just swinging the handle over as described in choice A changes neither the area presented to the magnetic field lines from the bottom coil nor the density of the field lines at the position of the loop. No change in magnetic flux means no induced emf and no induced current.

65 E If the efficiency of the engine is 25%, then the energy that is output for useful work is 25% of the input energy, which is (25%)(800 J) = 200 J. The remaining 800 J – 200 J = 600 J is expelled as waste heat.

66 C Label the charges 1, 2, and 3, in order from left to right. We’re asked to find the force on Charge 1 due to the other two charges. The electrostatic force on Charge 1 due to Charge 2 is a repulsive force, of magnitude , and the electrostatic force on Charge 1 due to Charge 3 is a weaker, attractive force, of magnitude . Since these individual forces act in opposite directions, the magnitude of the net force is found by subtracting the magnitudes of the individual forces. So, the total electric force on Charge 1 has magnitude 67 A In order for total internal reflection to occur, the beam must be incident in the medium with the higher index of refraction and strike the interface at an angle of incidence greater than the critical angle. Since v1 < v2, the refractive index of Medium 1, n1 = c/v1, must be greater than the index of Medium 2 (n2 = c/v2), and the critical angle is θcrit = sin−1 (n2/n1) = sin−1 (v1/v2).

68 E The frequency of the oscillations, f, can be found from the equation f = . Since k is known, all we need is f in order to calculate m. If we know the quantity given in E, we can double it to get the period, then take its reciprocal to obtain f.

69 B When the block is at Point Y, all its energy is due to the potential energy of the stretched spring, which is kd2. At the moment the block passes through the equilibrium position, O, this energy has been converted entirely to kinetic. By setting mv2 equal to kd2, we find that 70 D Because the particle travels in a circular path with constant kinetic energy (which implies constant speed), the net force on the particle is the centripetal force. 71 A The change in the electrical potential energy as a charge q moves from position X to position Z is equal to q times the difference in potential between the points: ΔPE = qV = qΔφ = q(φZφX). Because positions X and Z are equidistant from the source charge (+Q), the potentials at these locations are the same. Since Δφ = 0, there is no change in potential energy, so the work done by the electric field is also zero.

72 E The electric field strength is given by E = . Therefore 73 C The distance from the mirror to the image is equal to the distance from the mirror to the object. Therefore, the distance from the object to the image is 100 cm + 100 cm = 200 cm.

74 D The impulse delivered to the ball is equal to the ball’s change in momentum, and the presence of the mitt does not change this, eliminating A and C. The extra area may decrease the pressure on the catcher’s hand, but not the force (eliminating B). The padding of the mitt causes the time during which the ball comes to a stop to increase (just as an air bag increases the time it takes for an automobile passenger involved in an accident to come to a stop), which decreases the magnitude of the ball’s acceleration (since a = Δvt). A decrease in the magnitude of the acceleration means a decrease in the magnitude of the force (since a = F/m).

75 C The second postulate of Einstein’s theory of special relativity states that the speed of light is a universal constant (c = 3 × 108 m/s), regardless of the motion of the source or the detector. So despite the fact that the spaceship is approaching the planet at a speed of c/2, occupants of the ship will still measure the speed of light to be c. The time it takes the light pulse to travel a distance of 4.5 × 108 m is The Princeton Review Practice SAT Physics Subject Tests Scoring Grid

Recall from the Introduction that your raw score is equal to the number of questions you answered correctly minus of the number of questions you answered incorrectly,

(number of correct answers) – (number of wrong answers)

then rounded to the nearest whole number. Questions that you leave blank do not count toward your raw score. ﻿