Physics I Chapter C Solutions

Chapter C - Problems Problem C.1 Two points have cartesian coordinates H3 m, -8 mL and H-5 m, 5 mL. (a) Convert these points to polar coordinates. (b) What is the distance between these points? Solution to C.1 (a) To convert from cartesian coordinates to polar use: x2 + y2 r= q = : y tan-1 when x > 0 x y 180 ° + tan-1 K O when x < 0 x -8 O = -69.4 ° 3 Hx1 , y1 L = H3, -8L ï r1 = 32 + 82 = 8.54 m and q1 = tan-1 K Hx2 , y2 L = H-5, 5L ï r2 = 52 + 52 = 7.07 m and q2 = 180 ° + tan-1 K 3 O = 135 ° -3 (b) The distance between two points is: HDxL2 + HDyL2 = Hx2 - x1 L2 + Hy2 - y1 L2 = 233 = 15.3 m. Problem C.2 Two points have polar coordinates Hr, qL = H5 m, 25 °L and Hr, qL = H8 m, 120 °L. (a) Convert these points to cartesian (rectangular) coordinates. (b) What is the distance between these points? Solution to C.2 (a) To convert from polar to cartesian coordinates use: x = r cos q and y = r sin q Hr1 , q1 L = H5 m, 25 °L ï Hx1 , y1 L = H4.53 m, 2.11 mL Hr2 , q2 L = H8 m, 120 °L ï Hx2 , y2 L = H-4 m, 6.93 mL (b) The distance between points is Dx2 + Dy2 = Hx2 - x1 L2 + Hy2 - y1 L2 = 9.77 2 Chapter C - Problems Dx2 + Dy2 = Hx2 - x1 L2 + Hy2 - y1 L2 = 9.77 Problem C.3 A car drives around a traffic circle with a radius of 20 m. (a) When the car drives half way around the circle, what is the magnitude of the displacement vector and the total distance driven? (b) When the car drives once around a full circle, what is the magnitude of the displacement vector and the total distance driven? Solution to C.3 (a) The displacement vector is the vector from the starting position to the stopping point. It's magnitude is then the distance between these two point. In this case the distance is 2 r = 2 µ 20 = 40 m. The total distance driven is half the circumference. 1 2 ˛ r = ˛ 20 m = 62.8 m 2 (b) For a full circle the magnitude of the displacement is 0, so its magnitude is 0. The total distance is the circumference. 2 ˛ r = 2 ˛ 20 m = 126 m Problem C.4 (a) Obtain a component expression for a position vector when its magnitude and direction are given by r = 40 m and q = 155 °. (b) A car moves at 70 mi ê hr in the direction 12° south of west. Obtain a component expression for the velocity of the car. (c) A rope pulls with a force of 40 N in the direction given by q = -75 °. Obtain a component expression for the force vector. Solution to C.4 The magnitude and direction of a position vector in two dimensions are just its polar coordinates and the component expression for it is ` ` ” r = x x + y y = Xx, y\ = Xr cos q, r sin q\. For any other 2D vector A we can write it in terms of its magnitude A and direction q ` ` A = Ax x + Ay y = YAx , Ay ] = XA cos q, A sin q\. ” (a) r = 40 m, q = 155 ° ï r = Xr cos q, r sin q\ = X-36.3, 16.9\ m (b) v = 70 mi ê hr. If the direction is 12° south of west then the angle, measured counterclockwise from the positive x axis (taken as East) is q = 192 °. This gives ” v = Yvx , vy ] = Xv cos q, v sin q\ = X-68.5, -14.6\ mi ê hr (c) F = 40 N, q = -75 ° ï F = YFx , Fy ] = XF cos q, F sin q\ = X10.4, -38.6\ N Chapter C - Problems 3 Problem C.5 A person walks in successive displacements of 20 m west and then 30 m in the direction 25 ° East of North. What is the person's net displacement? Solution to C.5 ” s1 = X-20, 0\ s2 = 30 m and q2 = 65 ° gives ” s2 = Ys2x , s2y ] = Xs2 co s q2 , s2 sin q2 \ = X12.679, 27.189\. The net displacement is the sum of the two: ” ” ” s = s1 + s2 = X-7.32, 27.2\ m Problem C.6 Define the vectors A and B by ` ` ` A = X5, -6, 2\ = 5 x - 6 y + 2 z and ` ` ` B = X-3, 5, -7\ = -3 x + 5 y - 7 z ` (a) What is the vector C = A + B? Also give the magnitude of C and specify its direction by finding C, a unit vector in the direction of C. ` (b) What is the vector D = A - 2 B? Also give the magnitude of D and specify its direction by finding D, a unit vector in the direction of D. Solution to C.6 (a) C = A + B = X5, -6, 2\ + X-3, 5, -7\ = X2, -1, -5\ = YCx , Cy , Cz ] C = °C¥ = C2x + C2y + Cz2 = 5.47 ` C C= = X0.365, -0.183, -0.913\ °C¥ (b) D = A - 2 B = X5, -6, 2\ - 2 X-3, 5, -7\ = X11, -16, 16\ = YDx , Dy , Dz ] D = °D¥ = D2x + D2y + D2z = 25.2 ` D D= = X0.437, -0.636, 0.636\ °D¥ 4 Chapter C - Problems Problem C.7 A soccer ball is kicked in four consecutive displacements as shown. What is the resultant displacement of the ball? Also give the magnitude and direction of the resultant. Solution to C.7 y R 35° 30 ft x 40 ft 60 ft 80 ft 30° Call the four displacements A, B, C and D. The resultant R is the sum R=A+B+C+D A is in the positive x-direction and B is in the negative y-direction. A = X30, 0\ and B = X0, -60\ Properly defined angles are measured counterclockwise from the positive x axis. The angles for the vectors C and D are defined differently. Relative to the positive x-axis we get qC = 150 ° and qD = 55 °. The vectors, in component form, can be found using the usual formulas. C = 80 and qC = 150 ° ï C = YC cos qC , C sin qC ] = X-69.282, 40\ D = 40 and qD = 55 ° ï D = XD cos qD , D sin qD \ = X22.943, 32.766\ To find R we then perform the sum: R = A + B + C + D = X-16.339, 12.766\ = YRx , Ry ]. The resultant displacement is R = X-16.3, 12.8\ ft. To find the magnitude and direction of R use the usual formulas. R= R2x + R2y = 20.7 ft and q = 180 ° + tan-1 Ry Rx = 142.0 ° Problem C.8 A golfer putts a ball toward a hole 5 yards to the north. If his first putt is 6 yards in the direction 15 ° east of north, then what displacement vector is needed for the second putt to hit the hole. Also give the magnitude and direction of the second putt's displacement. Chapter C - Problems 5 Solution to C.8 y ”r 2 5 yds ”r ”r 1 x ” ” ” r = r1 + r2 ” ” ” Here we are given r and r1 and we are looking for r2 . ” ” r = X0, 5\ and r1 = X6 cos 75 °, 6 sin 75 °\ = X1.5528, 5.7956\ ” ” ” r2 = r - r1 = Xx2 , y2 \ = X-1.5528, -.7956\ = X-1.55, -.796\ yards For the magnitude and direction: r2 = y x22 + y22 = 1.74 yards and q = 180 ° + tan-1 2 = 207.1 ° x2 Problem C.9 (a) Show that if the angle between vectors A and B is q, then the magnitude of their sum is: ±A + Bµ = A2 + B2 +2 A B cos q (Hint: Since this is independent of the choice of coordinates, one can, without loss of generality, choose coordinates so that A is in the x direction.) (b) The law of cosines says that if a triangle has sides of lengths A, B and C, and the angle opposite the side of length C is g then: C2 = A2 + B2 -2 A B cos g Use the result of part (a) to prove this. 6 Chapter C - Problems Solution to C.9 y C A B g B sin q q B cos q x (a) Define C = A + B and refer to the diagram. C2 = @A +HB cos qLD2 + HB sin qL2 = A2 +2 A B cos q +B2 Icos2 q + sin2 qM= A2 + B2 + 2 A B cos q (b) Writing this in terms of the angle g gives the law of cosines. g = 180 ° - q ï cos q = -cos g This proves the result. Problem C.10 A car drives West at 20 m ê s for 5 min, then drives 30 m ê s to the North for 2 min and, finally, drive in the direction 30 ° East of South at 40 m ê s for 3 min. (a) What is the net displacement of the car? (b) What is the average velocity of the car? (c) What is the average speed of the car? Solution to C.10 (a) The distance for each segment is the speed multiplied by the time in seconds. ” D r = X-20 µ 300, 0\ + X0, 30 µ 120\ + X40 µ 180 cosH-60 °L, 40 µ 180 sinH-60 °L\ ” ï D r = X-2400, -2635.4\ = X-2400. - 2640\ m. (b) The average velocity is v = ” Dr . Dt v= ” X-2400,-2635.4\ Dr m = = X-4.00, -4.39\ Dt 300+120+180 s (c) The average speed is defined by: Average Speed = Total Distance . Total Time This becomes: Average Speed = Problem C.11 20µ300+30µ120+40µ180 m = 28 . 300+120+180 s Chapter C - Problems 7 Problem C.11 The displacement of a body as a function of time in SI units is given by: ” rHtL = Yt3 + 9 t - 5, 2 t4 - 5 t2 + 18]. (a) What is the average velocity of the body between 1s and 3s? (b) What is the speed of the body at 2s? (c) In what direction is the body moving at 2s? (d) What is the acceleration at 2s? Solution to C.11 ” ” (a) To find the average velocity, we need the position vecotors at 1 s and 3 s: rH1L = X5, 15\ and rH3L = X49, 135\ v= ” ” ” rH3L-rH1L Dr m = = X22, 60\ Dt 3-1 s (b) The speed is the magnitude of the velocity. Differentiate the position vector to get the velocity, plug in the time of 2 s and then find the magnitude. ” „r” v= = Y3 t2 + 9, 8 t3 - 10 t] „t ” vH2L = X21, 44\ ï v = v2x + v2y = 212 + 442 = 48.8 m s (c) The direction of motion is the direction of the velocity vector. q = tan-1 vy vx = tan-1 J 44 N = 64.5 ° 21 (d) The derivative of the velocity vector is the acceleration. a= ” „v m = Y6 t, 24 t2 - 10] ï aH2L = X12, 86\ „t s2 Problem C.12 At time zero a particle starts at the origin with a velocity of X30, -20\ m ê s. It moves with a constant acceleration so that after 5 s the velocity is X-20, 10\ m ê s . ” ” (a) Obtain expressions for the velocity as a function of time vHtL and the position vector as a function of time rHtL. (b) What is the speed after 12 s? Solution to C.12 (a) The velocity as a function of time for constant acceleration has the general form. ” ” vHtL = v0 + a t 8 Chapter C - Problems ” ” vHtL = v0 + a t ” ” m vH5L = v0 + a t ï X-20, 10\ = X30, -20\ + a 5 ï a = X-10, 6\ s2 The velocity as a function of time becomes: ” ” vHtL = v0 + a t = X30, -20\ + X-10, 6\ t = X30 - 10 t, -20 + 6 t\. (in SI units) The position as a function of time is: ” ” ” 1 1 rHtL = r0 + v0 t + a t2 = X0, 0\ + X30, -20\ t + X-10, 6\ t2 = Y30 t - 5 t2 , -20 t - 3 t2 ] (in SI units) 2 2 (b) The speed at 12 s is the magnitude of the velocity at that time. ” ” v = vH12L = X30 - 10 µ 12, -20 + 6 µ 12\ = X-90, 52\ v2x + v2y = ï v= 902 + 522 = 104 m s Problem C.13 A bartender slides a beer mug to Joe at the end of a bar. Joe, not being at his best, misses the mug which then tragically slides off the bar onto the floor below. Suppose the bar is height h and the mug lands a distance d from the base of the bar. (a) With what speed did the mug leave the bar? (b) What was the speed of the mug when it hit the floor? (c) In what direction was the mug moving when it hit the floor? (d) What are the answers to parts (a), (b) and (c) when h = 80 cm and d = 1.2 m? Solution to C.13 If a projectile is shot with a horizontal initial velocity, v0y = 0, then the height, initial velocity and horizontal range, h, v0x and D x are related by: h= 1 2 gt 2 (since D y = -h) D x = v0 x t This represents a common class of problems. Of the three variables h, v0x and D x two will be given and the third will be asked for. This problem has an horizontal initial velocity and is in this class. (a) Since velocity is purely horizontal when it leaves, the initial speed is just v0x . h= 1 2 gt ï t= 2 D x = v0x t ï v0x = 2h g Dx =d t g 2h (b) The speed is the magnitude of the velocity, se we first need the velocity when the mug hit the floor? vx = v0x = d g 2h and vy = v0y - g t = 0 - g 2h =g 2gh Chapter C - Problems vx = v0x = d g 2h and vy = v0y - g t = 0 - g v2x + v2y = v= d2 g 2h =g 9 2gh +2gh 2h (c) The direction of motion is just the direction of the velocity vector. vy q = tan-1 vx = tan-1 - 2gh d g 2h = -tan-1 2h d (d) Turning to the numerical part of this problem, we have: d = 1.2 m, h = 0.80 m and g = 9.8 g v0x = d d2 v= 2h g q = -tan-1 s2 m s = 2.97 + 2 g h = 4.95 2h m m s 2h = 53.1 ° d Problem C.14 An astronaut walking on some new planet finds that with an ititial speed of 3 m he can jump a maximum distance of 6 m. What is the s acceleration due to gravity on this planet? Solution to C.14 This is a simple application of the Range formula for projectiles: R= v20 g sinH2 qL The maximum range is at 45 °. Rmax = v20 g ï g= v20 Rmax = 32 m = 1.5 6 s2 Problem C.15 A projectile is shot so that its horizontal range is five times its maximum height. What initial angle is needed achieve this? You have enough information to give a numerical answer. Solution to C.15 We must find at which angle R = 5 ymax . The maximum height is ymax = Dy when vy = 0 . 10 Chapter C - Problems The maximum height is ymax = Dy when vy = 0 . v2y - v20 y = -2 g D y ï 0 - v20 y = -2 g ymax ï ymax = v20 y 2g We could use the range formula but this would just complicate the trig. We will solve for the range in terms of the components: 2 v0 y 1 2 gt ï t= g 2 2 v0 x v0y ï R = D x = v0 x t = g 0 = D y = v0 y t - R = 5 ymax ï 2 v0x v0y g =5 v20y 2g ï q = tan-1 ï v0y v0 sin q 4 = = = tan q 5 v0x v0 cos q 4 = 36.7 ° 5 Problem C.16 A football is kicked at a speed of 25 m ê s at an angle of 50 ° above horizontal. (a) What is the maximum height reached by the football? (b) At t = 2.5 s, what is the displacement, velocity, speed and acceleration of the football? (c) What is the total horizontal distance the football travels in the air? (d) Suppose this kick is toward a crossbar that is 3 m high. For the kick to be a field goal if must pass over the crossbar. What is the longest field goal (the horizontal distance to the crossbar) that can be made by this kick? Solution to C.16 v0 = 25 m and q = 50 ° s ” v0 = Yv0x , v0y ] = Yv0 cos q, v0 sin q] = X16.070, 19.151\ (a) At the highest position vy = 0. ymax = D y at that position. v2y - v20y = -2 g D y ï 0 - v20y = -2 g ymax ï ymax = v20y 2g = 18.7 m (b) For the entire time the football is in the air it is a projectile and a = X0, -g\. At t = 2.5 s the displacement, velocity and speed are: ” 1 D r = XD x, D y\ = Zv0x t, v0y t - g t2 ^ = X40.2, 17.3\ m 2 ” m v = Yvx , vy ] = Yv0x , v0y - g t] = X16.070, -5.3489\ = X16.1, -5.35\ s v= v2x + v2y = 16.9 m s (c) Since we want D x when D y = 0, this is a simple application of the range formula. R= v20 g sin 2 q = 252 sin 100 ° = 62.8 m 9.8 Chapter C - Problems R= v20 g sin 2 q = 11 252 sin 100 ° = 62.8 m 9.8 (d) Here we must first solve for the time the football is at the height of the crossbar and then solve for the horizontal distance travelled in that time. D y = v0y t ï t= 19.151± 1 2 g t ï 3 = 19.151 t - 4.9 t2 2 19.1512 -4µ4.9µ3 2µ4.9 ï t = 0.16349 and t = 3.4745 The later time is the one we want. Now we need the horizontal distance travelled in this time. D x = v0x t = 16.070 µ 3.4745 = 60.2 m Problem C.17 In making a movie a car drives at a speed of 40 m ê s off a 30 m high cliff. How long after leaving the cliff should the sound of the crash be added? Include the time it takes for the sound to return to the top of the cliff. Sound travels at 343 m ê s. Solution to C.17 This is a two step problem involving a horizontal initial velocity. The height of the cliff is h = 30 m. This gives the time of fall of the rock, t1 . h= 1 2 g t ï t1 = 2 1 2h = g 2µ50 = 2.4744 s 9.80 We now need to solve for the horizontal distance the car travels. D x = v0x t = 40 µ t1 = 98.974 The distance the sound travels is d = h2 + HDxL2 = 110.89. It follows that the return time t2 and the total time t are: t2 = d 110.89 = = 0.24652 s vsound 343 The total time is t1 + t2 . t1 + t2 = 2.4744 + 0.24652 = 2.71 s Problem C.18 A golfer hits a ball at 40m/s at an angle of 55 ° above horizontal up an inclined fairway with a 10 ° slope. How far up the incline does the ball travel before hitting? (Hint: Solve of the trajectory (y as a function of x) of the ball and equate that to the equation describing the incline.) Solution to C.18 To simplify the equation for the trajectory (y vs. x) take the initial position to be the origin. This gives: D x = x and D y = y. x = Iv0 cos qM t ï t = x v0 cos q 12 Chapter C - Problems x = Iv0 cos qM t ï t = y = Iv0 sin qM t - x v0 cos q g 1 2 g t ï y = Htan qL x x2 2 2 v20 cos2 q The equation for an incline of angle f is: y = Htan fL x. Eliminating y from the above expressions gives a quadratic expression for x. Htan q - tan fL x = g 2 v20 cos2 q x2 Dividing by x gives the unique nonzero solution: x= 2 v20 cos2 q g Htan q - tan fL This is the horizontal distance. the distance up the incline is d, where d= x = cos f d 2 v20 cos2 q x = Htan q - tan fL cos f g cos f Plugging in the numbers gives: d= 2ÿ402 ÿcos2 55 ° Htan 55 ° - tan 10 °L = 137 m. 9.80ÿcos µ 10 ° Problem C.19 A car travelling at 100 km ê hr is 300 m behind a slower car moving at 80 km ê hr. If both cars continue at the same rate how long does it take for the faster car to catch the other? Solution to C.19 Let us study the motion of the faster car with respect to the slower car and the road. Take v = 100 km ê hr to be the velocity of the faster car with respect to the road and v0 = 80 km ê hr to be the velocity of the slower car with respect to the road. The relative velocity of the faster with respect to the slower is v£ = v - v0 = 20 km 1000 m 1 hour 20 m µ µ = hr 1 km 3600 s 3.6 s The relative distance between the bumpers of the car is d £ = 300 m. d£ 300 d £ = v£ t ï t = £ = = 54 s v 20ê3.6 Problem C.20 A 160 m wide river has a steady current with a uniform speed of 1.5 m/s. A boat crosses this river by maintaining a bearing, the direction in which the boat points, that is perpendicular to the river's flow and with a speed of 2 m/s relative to the water. (a) What is the speed of the boat relative to the shore? (b) How far downstream from the starting point does the boat land? Solution to C.20 ” ” ” ” ” ” v = v£ + v0 , where v0 is the velocity of the moving frame with respect to the stationary frame. v and v£ represent some velocity with respect to the fixed and moving frames, respectively. Chapter C - Problems 13 ” ” ” ” ” ” v = v£ + v0 , where v0 is the velocity of the moving frame with respect to the stationary frame. v and v£ represent some velocity with respect to the fixed and moving frames, respectively. D 1.5 mês v”0 160 m 2 mês ”£ v q Path of boat v” q Here, we are considering the motion of the boat with respect to the shore (the fixed frame) and the water (the moving frame.) The ” ” ” river flows in the direction of v0 . The direction of the boats bearing is v£ , the boat's velocity relative to the water. v, the boat's velocity relative to the shore, which is the direction of the bait's trajectory. (a) Using the Pythagorean theorem we can solve for v, the speed of the boat relative to the shore. v= v£2 + u2 = 22 + 1.52 = 2.5 m s (b) Using similar triangles we can find D, the downstream distance of the boat. D 1.5 = ï D = 120 m 160 2 Problem C.21 The pilot of a plane reads his compass and sees his heading is due south. He also sees that the plane's speed with respect to the air is 200 mi ê hr. Suppose there is a 50 mi ê hr wind blowing to the west. (a) What is the velocity of the plane with respect to the ground? (b) What are the speed and direction of the plane relative to the ground? Solution to C.21 (a) u = -50 mi ` x is the velocity of the wind. hr ” mi ` v£ = -200 x is the velocity of the plane relative to the wind. hr ` ` km ” ” km v = v£ + u = I-200 y -50 xM = X-50, -200\ is the velocity of the plane relative to the ground. hr hr (b) Use the usual rules to get the magnitude and direction of the plane relative to the ground. v= 502 + 2002 = 206 km -200 and q = 180 ° + tan-1 K O = 256.0 ° hr -50 14 Chapter C - Problems Problem C.22 A car moving at a speed v hits a puddle in the road and sprays water. Relative to the car, the spray is shot backward at speed v and at an angle q above horizontal. What is the velocity of the spray relative to the road? Take the direction of the car's motion to be the x direction and take y to be vertical. Solution to C.22 Car to Road v” 0 v”£ ` y v” Spray to Car Spray to Road q ` x ` ” v0 = v x is the velocity of the car, the moving frame. ` ` ” v£ = -v cos q x +v sin q y is the velocity of the spray relative to the car. So, the velocity of the spray relative to the road is: ` ` ” ” ” v = v£ + v0 = Hv - v cos qL x +v sin q y ” Comment: This is counterintuitive but true; the spray is actually thrown forward relative to the road. The magnitude of v£ is the same as the car's speed because that is the speed of the rim of the tire relative to the road.