Physics 18 Spring 2011 Homework 2 - Solutions Wednesday January 26, 2011

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1 Physics 18 Sprin 011 Homework - s Wednesday January 6, 011 Make sure your name is on your homework, and please box your final answer. Because we will be ivin partial credit, be sure to attempt all the problems, even if you don t finish them. The homework is due at the beinnin of class on Wednesday, February nd. Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions durin the discussion session or durin office hours. 1. True or false (inore any effects due to air resistance): (a) When a projectile is fired horizontally, it takes the same amount of time to reach the round as an identical projectile dropped from rest from the same heiht. (b) When a projectile is fired from a certain heiht at an upward anle, it takes loner to reach the round than does an identical projectile dropped from rest from the same heiht. (c) When a projectile is fired horizontally from a certain heiht, it has a hiher speed upon reachin the round than does an identical projectile dropped from rest from the same heiht. (a) True. The rate at which the objects fall is independent of their horizontal velocity. (b) True. The projectile fired up starts headin up first, before it starts headin toward the round. So, it will take loner to hit the round than an object just dropped from rest. (c) True. The object just dropped picks up speed from ravitational acceleration. The fired projectile also picks up this speed, but it also had the initial velocity in the horizontal direction. The final speed includes both these contributions, and so the fired projectile hits with a hiher speed. 1

2 . To withstand -forces of up to 10 s, caused by suddenly pullin out of a steep dive, fihter jet pilots train on a human centrifue. 10 is an acceleration of 98 m/s. If the lenth of the centrifue arm is 1 m, at what speed is the rider movin when she experiences 10 s? The centripetal acceleration is just a cent = v /r, where v is the velocity, and r is the distance from the center of the orbit. So, solvin for the velocity ives v = ar. Pluin in the numbers ives which is about 77 miles per hour. v = ar = 98 1 = 34.3 m/s,

3 3. A wall clock has a minute hand with a lenth of 0.50 m and an hour hand with a lenth of 0.5 m. Take the center of the clock as the oriin, and use a Cartesian coordinate system with the positive x axis pointin to 3 o clock and the positive y axis pointin to 1 o clock. Usin unit vectors î and ĵ, express the position vectors of the tip of the hour hand ( A) and the tip of the minute hand ( B) when the clock reads (a) 1:00, (b) 3:00, (c) 6:00, (d) 9:00. (a) Both hands point up (alon the y axis) at 1:00, and so A = 0.5ĵ, and B = 0.5ĵ. (b) The hour hand points horizontally (alon the x axis), while the minute hand points up, so A = 0.5î, and B = 0.5ĵ. (c) The hour hand points down, while the minute hand points up, so A = 0.5ĵ, and B = 0.5ĵ. (d) The hour hand points to the left, while the minute hand points up, so A = 0.5î, and B = 0.5ĵ. 3

4 4. A particle has a position vector iven by r = (30t) î+(40 5t ) ĵ, where r is in meters, and t is in seconds. Find the instantaneous velocity and instantaneous acceleration vectors as functions of time. The instantaneous velocity is iven by v = d dt r r, where the dot over the quantity denotes the time derivative. So, takin the derivative with respect to time ives r = 30î 10tĵ. The instantaneous acceleration is the second derivative of the position, a = r, or the first derivative of the velocity, a = v. So, takin another derivative ives v = r = 10ĵ, which is a constant acceleration alon the vertical direction. So, this position vector could describe a particle fallin in a (slihtly stroner) ravitational field. 4

5 5. A cat is chasin a mouse. The mouse runs in a straiht line at a speed of 1.5 m/s. If the cat leaps off the floor at a 30 anle and a speed of 4.0 m/s, at what distance behind the mouse should the cat leap in order to land on the poor mouse? Model: Use the particle model for the cat and apply the constant-acce Visualize: We can picture the situation as seen in the fiure to the riht. The cat has a rane of R (θ) = v 0 sin θ. For the iven numbers, the cat jumps a distance R (θ) = 4 sin 60 = 1.41 m. 9.8Solve: The relative velocity of the cat from the mouse s reference frame is It takes him a time T = v 0 sin θ = 8 sin = 0.41 (4.0cos30 seconds to travel 1.5) this m/sdistance. = m/s = v 0 Durin this time, the mouse moves a distance d = v mouse T = = 0.61 meters. So, in order for Thus, the vcat 0x = tov pounce 0 = onm/s theand mouse, v 0 he y = vshould 0 sin30 = start (4.0 back m/s)sin30 a distance =.0 m/s. Th D = R d = = 0.80 meters, or 80 centimeters. floor is found as follows: y = y + v ( t t ) + a ( t t ) 0 m= 0 m + (.0 m/s) t y 1 0 y The horizontal distance covered in time t 1 is x x v t t a t t t 1 = 0 s (trivial solution) and s 1 1 = 0 + 0x( 1 0) + ( x 1 0) = 0 m + (1.964 m/s)(0.408 That is, the cat should leap when he is 80 cm behind the mouse. 5

6 6. Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from the pulsars at reular intervals equal to the period that they rotate. Some of these pulsars rotate with periods as short as 1 ms! The Crab Pulsar, located inside the Crab Nebula in the constellation Orion, has a period currently of lenth ms. It is estimated to have an equatorial radius of 15 km, an averae radius for a neutron star. (a) What is the value of the centripetal acceleration of an object on the surface at the equator of the pulsar? (b) Many pulsars are observed to have periods that lenthen slihtly with time, a phenomenon called spin-down. The rate of slowin of the Crab Pulsar is s per second, which implies that if this rate remains constant, the Crab Pulsar will stop spinnin in s (about 3000 years from today). What is the tanential acceleration of an object on the equator of this neutron star? (a) The centripetal acceleration is iven by a cent = v /r, where v is the tanential velocity of a point at the radius r. The speed of the point is the total distance the point oes around (the circumference), divided by the time it takes to o around (in other words, the period). So, v = πr/t. Thus, a = v r = (πr) rt with the numbers. This is a hue acceleration! = 4π r T = 4π ( ) = m/s, (b) The tanential velocity was iven by v = πr/t. The tanential acceleration is a tan = dv. The radius of the star isn t chanin, but the period is chanin. So, dt takin the time derivative ives a tan = dv ( ) πr dt dt = T dt, where dt/dt is the spin-down rate. So, pluin in the values ives, usin the chain rule, ( ) πr dt a tan = T dt = π ( ) ( ) = m/s, where the spin-down rate is neative because it s slowin down. The tanential acceleration is very much smaller than the centripetal acceleration! 6

7 7. Human blood contains plasma, platelets, and blood cells. To separate the plasma from other components, centrifuation is used. Effective centrifuation requires subjectin blood to an acceleration of 000 or more. In this situation, assume that blood is contained in test tubes that are 15 cm lon and are full of blood. These tubes ride in the centrifue tilted at an anle of 45.0 above the horizontal. (a) What is the distance of a sample of blood from the rotation axis of a centrifue rotatin at 3500 rpm, if it has an acceleration of 000? (b) If the blood at the center of the tubes revolves around the rotation axis at the radius calculated in Part (a), calculate the accelerations experienced by the blood at each end of the test tube. Express all accelerations as multiples of. (a) The acceleration that the blood experiences is the centripetal acceleration, a cent = v. Now, the velocity is v = πr/t, where T is the period. The period can be r rewritten in terms of the frequency, f = T 1, and so v = πfr. So, a = 4π f r. Since the centrifue rotates at 3500 rpm, it makes 3500/60 = 58.3 rotations per second, which is the frequency, f. Thus, solvin for the distance, and pluin in the numbers ives r = a = = 0.15 m = 15 cm. 4π f 4π 58.3 This value seems lare, but the centrifue is spinnin really fast! (b) Because the test tube is tilted, the top and bottom of the tube are at different radii from the axis. This leads to different accelerations. Since the center of the test tube is at 15 centimeters, and is also 15 centimeters lon, the top of the test tube is at r = sin 45 = 9.70 cm. The bottom of the test tube is at r + = sin 45 = 0.3 cm. The acceleration is, from before, a ± = 4π f r ±. Pluin in each value, and dividin by = 9.8 to express the acceleration in terms of, we find a = 4π f r = 4π = 138 a + = 4π f r + = 4π = 780 So, the accelerations rane from about 1330, to about

8 8. Wile E. Coyote (Carnivorus hunribilus) is chasin the Roadrunner (Speedibus cantcatchmi) yet aain. While runnin down the road, they come to a deep ore, 15.0 m straiht across and 100 m deep. The Roadrunner launches himself across the ore at a launch anle of 15 above the horizontal, and lands with 1.5 m to spare. (a) What was the Roadrunner s launch speed? (b) Wile E. Coyote launches himself across the ore with the same initial speed, but at a different launch anle. To his horror, he is short of the other lip by 0.50 m. What was his launch anle? (Assume that it was less than 15.) For a iven launch speed and anle, the rane of a projectile is iven by R = v 0 sin(θ). (a) The roadrunner jumps at an anle of θ = 15, and reaches a rane R = 16.5 meters. So, solvin for the velocity ives R v 0 = sin(θ) = = 18 m/s. sin 30 (b) The poor coyote only jumps 14.5 meters. Solvin for the anle ives θ = 1 ( ) R sin 1 = 1 ( ) v0 sin 1 =

9 9. You are asked to consult for the city s research hospital, where a roup of doctors is investiatin the bombardment of cancer tumors with hih-enery ions. The ions are fired directly toward the center of the tumor at speeds of m/s. To cover the entire tumor area, the ions are deflected sideways by passin them between two chared metal plates that accelerate the ions perpendicular to the direction of their initial motion. The acceleration reion is 5.0 cm lon, and the ends of the acceleration plates are 1.5 m from the patient. What acceleration is required to deflect an ion.0 cm to one side? The plates add in a vertical acceleration, a y. The horizontal acceleration is always zero, a x = 0. The ion experiences a constant acceleration between the plates, and is deflected a heiht H 1 = 1a yt 1 where t 1 is the amount of time the ion spends in between the plates. Since the ion is travelin at a constant speed, v x, then if the lenth of the plates is d, then t 1 = d/v x. So, the net displacement from the plates is H 1 = ayd. vx After emerin from the plates, the ion travels alon at a constant speed, in a straiht line. The ion then travels up a distance H = v y t where v y is it s velocity, while t is the time the ion spends oin between the plates to the taret. If the distance away is L, then t = L/v x. Furthermore, the velocity is just what it s picked up from bein between the plates, v y = a y t 1 = a y d/v x. So, H = aydl. vx These two displacements need to add up to the h = centimeter deflection. So, h = H 1 + H = ayd + aydl. Solvin for the acceleration ives vx vx Pluin in our numbers ives a y = hv x d + dl. a y = hv x d + dl = (0.0) (5 106 ) (0.05)(1.5) = m/s, which is a hue acceleration, rouhly 100 billion times that due to ravity! 9

10 10. A toy cannon is placed on a ramp that has a slope of anle φ. (a) If the cannonball is projected up the hill at an anle of θ 0 above the horizontal and has a muzzle speed of v 0, show that the rane R of the cannonball (as measured alon the ramp) is iven by R = v 0 cos θ 0 (tan θ 0 tan φ). cos φ Inore any effects due to air resistance. The path of the cannonball can be expressed in terms of the distance x as ( ) y(x) = (tan θ 0 )x x, v0 cos θ 0 where we have set the initial y 0 = 0, since we set the cannon at the oriin. Now, if the distance alon the ramp is R, then the cannonball hits the ramp at (x, y) = (R cos φ, R sin φ). Pluin in these values to the rane expression ives ( ) R sin φ = (tan θ 0 )R cos φ R cos φ, v0 cos θ 0 or, dividin by R cos φ ives ( ) tan φ = tan θ 0 R cos φ. v0 cos θ 0 Now, rearranin and solvin for R ives or R cos φ = v 0 cos θ 0 which is the answer we were lookin for. (tan θ 0 tan φ), R = v 0 cos θ 0 cos φ (tan θ 0 tan φ), 10