Solutions To Jaan Kalda’s Problems In Kinematics
Solutions to Jaan Kalda’s Problems inKinematicsWith detailed diagrams and walkthroughsEdition 2.0.0Ashmit Dutta, Daniel Yang, Kushal Thaman, QiLin Xue, Tarun AgarwalhbFacUpdated April 5, 2020
Kalda Kinematics1PrefaceJaan Kalda’s handouts are beloved by physics students both in for a quick challenge, to students preparing forinternational Olympiads. As of writing, the current kinematics handout (ver 2.0) has 66 unique problems and45 main ‘ideas’.This solutions manual came as a pilot project from the online community at artofproblemsolving.com.Although there were detailed hints provided, full solutions have never been written. The majority of thesolutions seen here were written on a private forum given to those who wanted to participate in makingsolutions. In an amazing show of an online collaboration, students from around the world came together todiscuss ideas and methods and created what we see today.This project would not have been possible without the countless contributions from members of the community. Online usernames were used for those who did not wish to be named:Rakshit, Evan Kim, Ameya Deshmukh, Alan Abraham, dbs27, Heramb Podar, Anant Lunia, Sumgato, VirajJayamStructure of The Solutions ManualEach chapter in this solutions manual will be directed towards a section given in Kalda’s kinematics handout. There are six major chapters: velocities, accelerations/displacements, optimal trajectories, rigid bodies/hinges/ropes, miscellaneous topics, and revision problems. If you are stuck on a problem, cannot makeprogress even with the hint, and come here for reference, look at only the start of the solution, then try again.Looking at the entire solution wastes the problem for you and ruins an opportunity for yourself to improve.Contact UsDespite editing, there is almost zero probability that there are no mistakes inside this book. If there are anymistakes, you want to add a remark, have a unique solution, or know the source of a specific problem, thenplease contact us at hello@physoly.tech. The most current and updated version can be found on our websitephysoly.techPlease feel free to contact us at the same email if you are confused on a solution. Chances are that manyothers will have the same question as you.1
Kalda Kinematics12Solutions to Velocities ProblemsThis section will consist of the solutions to problems from problem 1-9 of the handout. In this section we willbe analyzing the usage of reference frames. Reference frames are defined as point or perspective of an actionor motion from different objects or people. For example, two different reference frames of a car moving on around could be one of an observer standing right by the road who sees the car to be moving on the road or theperson inside the car who does not see the car to be moving at all. Reference frames are extremely importantin physics as they allow us to solve seemingly complex problems which become greatly simplified with the useof reference frames.pr 1. In a time 2t, the barge moved a distance of 6 km in the ground frame, so this implies that thespeed of the water is:6 km d 4 km/hvwater 2t1.5 hoursIn the water’s reference frame, the barge is stationary and the boat travels at a constant speedvboat rel water relative to the water, where:vboat rel water d30 6 km 16 km/h2t1.5 hourspr 2. Moving into the frame of the red plane, we see the blue plane with a diagonally directed velocity.80060010005dThe closest approach would be when the faster plane’s path makes a perpendicular line with the slowerplane. This turns out into a geometry problem where we have two similar right triangles. We can breakup the velocity of the blue plane into components (since the displacement is in the same direction asvelocity, this is also the components of its displacement). The top triangle is a 3 4 5 right triangleso the bottom right triangle must also be a 3 4 5 right triangle.Now all we need to know now is to determine how far away the blue plane is when it is directly overheadthe red plane. The time it takes to reach this point is:t 20 km 0.025 h800 km/hand the vertical distance it travels during this time is: y (600 km/h)(0.025 h) 15 km/h2
Kalda Kinematics3meaning the vertical separation is 5 km/h. Therefore:d 4 kmSolution 2: Let us work in the lab frame this time, but break the velocities of the two planes into adirection perpendicular and towards the other plane. We only need to worry about this radial component.Originally, the two planes will be nearing each other but will eventually get farther apart. The point atwhich this happens is when the radial component of their velocities are directed in the same directionand have the same magnitude. If we measured their radial acceleration at this point, it would be zero.xα800yα600Thus, we must have:600 sin α 800 cos α tan α 43Due to similar triangles, we must also have:y4 3xLet t 0 be when the fast plane is directly above the slow plane. The vertical separation at this pointis 5 km. Therefore, we have x 800t and y 5 600t We want:5 600t4 800t3Solving for t and plugging it into x and y can give you the separation, which turns out to be 4 km .Solution 3: We can use generalized coordinates. The distance between the two planes is:d (20 800t)x̂ ( 20 600t)ŷ s v twhere s 20x̂ 20ŷ and v 800x̂ 600ŷ, which represents the relative velocity. As with before, we One way of doing it is maximizingwant the relative velocity to be perpendicular to the displacement d.the dot product: d v s v v v t s v v . Therefore, all we need is to evaluate:At the maximum, this cross product has to be equal to d d s v 4 km v 3
Kalda Kinematics4Solution 4: Here’s a standard calculus method. The distance between the two planes after a time t is:d2 (20 800t)2 (20 600t)2d is maximized when d2 is maximized or when: d(20 800t)2 (20 600t)2 2(20 800t)( 800) 2(20 600t)( 600)dt0 4(20 800t) 3(20 600t)0 80 3200t 60 1800tt 7/250Plugging in t 7/250 into the distance formula gives:d 4 kmpr 3. Consider the following diagram:vMoving into the reference frame moving leftward at velocity v/2 gives us the following diagram (whereu is the velocity of intersection):v/2ura/2Using SAS similarity we find thatru p2v/2r (a/2)2vvu p p222r r (a/2)2 1 (a/2r)24
Kalda Kinematics5Solution 2: We start similarly to Solution 1 and move into a reference frame moving left at v/2.pLet y r2 a2 /4 be the height of of the intersection above the centers of the hoop. We then see thatdy da1vdy · p·2dtda dt(2r/a) 1 2Therefore,squ u2x u2y u v2 v21 ·244 (2r /a) 1vp2 1 (a/2r)2pr 4. We notice that the graph is quadratic so we can fit it to the equation π602α (t 7) 6018049π (t 7)2 π/3147where α is in radians and t is in minutes.Since we know that the upward ascending velocity is constant, it is 14π0vy Lα (0) 1000147 299 m/min 4.99 m/sThe height is simplyh vy t 2000 mAt t 7 min, the change in elevation angle is momentarily 0, which means that the velocity vector alsopoints at 60 degrees.Thus we can getvx vy tan(30 ) 2.8 m/sYou don’t need the equation of the curve to perform calculations, but even without it, the answer can appear a bit off.e.g. the initial slope you get could be:4 /0.2 min 0.0698 rad/12 sec 0.00582 sec 15
Kalda Kinematics6pr 5.uvFfIn the board’s frame of reference, there is only a horizontal force (the friction force), which has a constantdirection that is anti-parallel to the velocity. Thus, the chalk moves in a straight line .pr 6. Let us assume that the block is originally pushed leftwards in the frame of the ground and theconveyor is travelling upwards. In the frame of the conveyor belt, the block is moving with a speed of 5 m/s. This is represented bythe red vector. Due to friction opposing the motion, the direction of motion relative to the belt will beconstant. The magnitude will steadily decrease to zero.To move back to the frame of the ground, we can add back the velocity of the conveyor belt, as shownbelow.θ 5 m/s1 m/s2 m/sThe blue vector shows the velocity of the block relative to the ground. Initially, it is 2 m/s but as frictionreduces the magnitude of the red vector (which represents the velocity relative to conveyor belt), theblue vector will decrease to a minimum. This minimum occurs when it forms a right angled triangle(represented by the dotted lines).Therefore, the minimum velocity of the block relative to the ground is2v 1 sin θ 56
Kalda Kinematics7pr 7.vuLv cos αutαv cos αsDenote the dashed line by the wall which is a distance L away from the point source.We express the lateral displacement of the ball as the sum of two components: lateral displacement inthe air’s frame of reference, and the lateral displacement of the moving frame.In the air’s frame the displacement is given byutair uLv cos αand the lateral displacement in the moving frame is given by s. This gives usut sLuL s t v cos αu v cos αpr 8. Draw a right trapezoid as follows:We decompose v into parallel and perpendicular components, v v x v y ; let us mark points A, B andC so that AB v x and BC v y (then, AC v ).Next we mark points D, E and F so that CD v y0 v y , DE v x , and EF 2u x ; then, CF v y0 v x 2u x v 0 and AF 2v y 2u x 2 u.Due to the problem conditions, ACF 90 . v xD2u xEv y0 v y2v y 2u x 2 uCαβv yαBv xA7F
Kalda Kinematics8We now can see that ACF is an isoceles triangle containing the lengths provided in the figure below.Let us also mark point G as the centre of AF ; then, F C is both the median of the right trapezoid ABDF(and hence, parallel to AB and the x axis), and the median of the triangle ACF .F uCαβ u vαABy splitting ACF into it’s median, we find,u cos α vv u .22 cos αFor this to also happen, we see that β 180 2α because ACF is an isoceles triangle.pr 9. We move into the reference frame that is rotating clockwise at ω about the center of the mirror(i.e. the mirror is stationary).S0ωaωSIn this frame of reference, the image S 0 has angular velocity ω clockwise about the center of the mirror.Moving back into the reference frame where S is stationary, we see that S 0 is moving with angular velocity2ω about the center of the mirror, so the image has speedv 2ωa8
Kalda Kinematics29Solutions to Accelerations/Displacements ProblemsThis chapter will contain problems 10-16 of the handout. In this section, we will be analyzing the usage ofnon-constant velocity and graphs. When an object does not have a constant velocity, there is an accelerationpresent meaning that the object is either speeding up or slowing down. The implementation of acceleration andgraphs allows us to figure out many problems. In this section, we will also be looking at the implementation ofcutting into tiny pieces and integrating to derive a result.pr 10. Consider the following graph:v (m/s)642501015t (s) 2Since the particle starts from the origin, the distance graph is simply the area under the velocity graph:d (m)18.75161412108642t (s)051015We need to find the maximum displacement, so our answer is 18.75 m920
Kalda Kinematics10pr 11. Let us divide the displacement into tiny pieces, s P s where s v t.If the function v(t) were known, the last formula would have been completed our task, becauseis the sum of rectangles making up the area under the v t-graph.Pv(t) tHowever, the acceleration is given to us as a function of v, hence we need to substitute t with v.While trying to do that, we can introduce the acceleration (which is given to as a function of v): t v ·In other wordss v v t . v v/ taXvaZ v vdv.a(v)This tells us that the answer is equal to the area under a graph which depictsvas a function of v.a(v)Applying a quartic least-squares fit to some of the discernible data points, we can see that the curve a(v) iswell approximated by the function 0.00617211v 4 0.0301639v 3 0.0581573v 2 0.0546369v 0.000715828.Taking the integral, we can see thatZ 4Z 4vvdv dv4320 a(v)0 0.00617211v 0.0301639v 0.0581573v 0.0546369v 0.000715828 39 mDon’t worry if your answer isn’t exactly the same as ours, as this result may be difficult to determine by hand with thegraph provided. A rough approximation (within reasonable limits) would suffice.pr 12.ABxααIn the reference frame of ball A, ball B accelerates to the left withaB 2g sin α cos αWe can find that the initial length AB is g t21 t22 sin α210
Kalda Kinematics11Therefore, g t21 t22 sin α cos αx 2Since there is no relative acceleration in the y-direction, we needaB t2 x2 2 t2 sin α cos αgt12gt2 sin α cos α 2r22t1 t2t 2Solution 2: Each ball will accelerate with the same acceleration down their platform, meaning thatthey will travel the same distance in the same timeframe.Let x be the distance traveled by the individual balls and k be the distance between the two balls. Letthe height of the ball at point A be h1 and the height of the ball at point B be h2 .If you draw a diagram you will find that there is a triangle formed by the position of the two balls andh1 h2, k.the intersection of the planks. The lengths of the triangle are x, x sin αAh1 h2 xsin αkBxOααBy the Law of Cosines, we have22k x Let β h1 h2 xsin α 2 2x h1 h2 x cos(2α)sin αh1 h2for simplicity.sin αSimplifying the expression, we get thatpk(x) 2x2 (1 cos (2α)) 2xβ(1 cos (2α)) β 2After taking the derivative of the quadratic and setting it equal to zero, we get thatxm Bβ 2A211
Kalda Kinematics12Using acceleration along the ramp we can also find thath2h1gt2 sin αgt2 sin α 1 2sin α2sin α2gt2m sin αh1 h2xm 22 sin αPlugging in everything we find thatstm h1 h2g sin2 αvu 2 2u gt1 sin α gt22 sin2 αu t22 g sin2 αrt21 t22 2pr 13.g sin αdg cos αLet t be the time it takes the ball to hit the ramp. Therefore, we find thats2d1 21d at d g cos αt2 t .22g cos αNow, we note that the total time T 3t because the ball travels a distance d to collide with the ramp,bounces up a distance d to the vertex of its parabolic trajectory, and then falls back down for the finaldistance d.This means that the distance between both bouncing points s is found by1s g sin α(T 2 t2 )2 12d2s g sin α(8t ) 4g sin α2g cos αs 8d tan α12
Kalda Kinematics13Solution 2: Consider the following diagram:d90 αLαWe rotate the plane by α counterclockwise such that gravity now has acceleration g cos α in the y-directionand g sin α in the x-direction.r2gdWhen the ball hits the plane, it strikes with velocity v0 at an angle 90 α to thesin (90 α)inclined plane.Then, we use modified projectile motion to get thatt 2v0 sin (90 α)2v0 g cos (α)gWe also have that1L v0 cos (90 α) t g sin (α) t2222v cos (90 α) 4v02 g sin (α) 0 g2g 2222v sin (α) 2v0 sin (α) 0 gg 8gd tan α13
Kalda Kinematics14pr 14.βαv0When on the plane, the puck experiences no change in its x-velocity, which isv0 cos β 5 m/sHowever, it experiences an acceleration parallel to the plane with magnitudea g sin αWe note from the trajectory given that the puck drops 2.5 m below the apex of its trajectory whileundergoing a horizontal displacement of x 5 m.The time it takes to complete this motion ist x 1sv0 cos βTherefore, we have thatgt2 sin α 2.525sin α 2gtα 30 pr 15. Due to symmetry the turtles meet at the centroid of the triangle formed, and form an equilateraltriangle at any instant. The velocity of the first turtle with respect to the other is obviously0.1 cos (60 ) m/sThus, the relative velocity of separation isv(1 cos (60 )) 3v2Since this is constant, the time taken for the turtles to meet ist d3v2 2d 6.7 s3v14
Kalda Kinematics15Solution 2: The path length of any turtle in the motion is simplys rdθ 2ds dr 1 drUsing polar coordinates, one can deduce thatdr 10 sin (60 )dt10dθ 5vdt2Hence we haveSince t dθ2dr1 ds drr 33ds, we find the total time by integrating this expression:10Z 0Z T2drdt 3v 130T 2d 6.7 s3vpr 16. Note thatdk dsvdt .L(t)L utIntegrating both sides gives us1ZZdk 00TvdtL utUsing a s substitution s L ut dx udt and rearranging the integral gives usv1 uZ0Tdxu ln(ut L) T0sv uT Lu lnvLSubstituting the values given in the problem tells us uT LuT100 ln e100 1LLT e100 115
Kalda Kinematics316Solutions to Optimal Trajectories ProblemsThis section will contain problems 17-22 of the handout. In this section, we will be analyzing trajectories. Youshould have a good understanding of conics (parabolas in general) and projectile motion before getting intothis section. There are a lot of proofs and sometimes more they have more math than what you would callphysics. However, don’t be demotivated, there are a lot of great problems in this section and some of theseproblems have ties to other areas of physics.pr 17.MPhAlαOOne extreme case that we must consider first is directly travelling along the path AO, which gives l2 h2t vWe’ll deal with this later, but we first use fact 5, as shown in the following diagram:MPφαxφx0αO16A
Kalda Kinematics17We use φ as defined above to make calculations easier and we get thatsin φv sin αu φ arcsinv sin αu Sincex0 l h tan φx (l h tan φ) cos αWe also have thatx tan αh v cos φuh(l h tan φ) sin α v cos φuh sin φ sin α l sin αh v cos φu cos φu2hh sin φ l sin α v cos φv cos φuh cos φ l sin α vu lHowever, we must also note that, when φ arctan, the boy never actually reaches side OP .hTherefore, our answer ist lif φ arctanhh cos φ l sin αx (l h tan φ) cos α, t vu 22l hx 0, t votherwisepr 18. Solution 1 (Fermat’s Principle):Aaαn2 v wun1 1Let us move into the reference frame of the river such that everywhere in the water, the boy is travellingat a constant speed u. This might seem troublesome at first because his destination would be moving,but that doesn’t trouble us at all. As it will soon be made clear, the angle α the boy makes with theshore-line will be independent of how far away the target is.Consider a light beam that starts off from A and ends up travelling with a speed of v w parallel to the17
Kalda Kinematics18shoreline. Since it will take the fastest path, the boy will need to mimic this behavior. Snell’s Law gives:un1 sin(90 ) n2 sin α sin α α arcsinv w uv w The total time the boy spends swimming is:t da uu cos αRelative to the water, the boy swims a horizontal distance a tan α. The water during this time flows awadistance wt u cosα in the opposite direction. Therefore, the horizontal distance x is:x a w tan αu cos αSolution 2 (Huygens Principle):AutθθOB(w v)tConsider the same setup as before by moving into the frame of the river. This time however, the pathis reversed. The boy starts running from point O along the shore and eventually starts swimming tolocation A. Imagine the boy emitting Chernenko radiation as he moves as shown in the diagram. Thewave speed is u while the speed of the boy is w v u. Physically, the outlines of all the circles representthe superposition of all the points in which the boy can be at after a time t.Due to Huygen’s Principle, we can see that this forms a wavefront that is moving towards A at a speed ofu. We can let this wavefront evolve until a part of it eventually reaches the point A. The path that thispart of the wave takes will represent the optimal path of the boy, that is, perpendicular to the wavefront.We can determine the angle θ by considering two extreme paths the boy can take.First, the boy can start swimming immediately and reach a distance ut after a time t. During this period,the boy can also run a distance (w v)t. The angle θ is thus given by:cos θ uw vand thus the angle α 90 θ normal to the shore is α arcsinuw v This is the same angle found in the first solution and as a result we can copy the exact ste
Apr 05, 2020 · Kalda Kinematics 1 Preface Jaan Kalda’shandoutsare beloved by physics students both in for a quick challenge, to students preparing for international Olympiads. As of writing, the currentkinematicshand
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