course PHY 121 the practice major quiz is as follows, and I attempted a practice test 2 with poor results, but an error message came up and I couldn't finish it. [~^~Zn堚Oڞassignment #001
......!!!!!!!!...................................
21:46:01 Problem Number 1 A coasting cart has velocities 4.5 m/s, 10 m/s and 2.25 m/s at clock times t = 0 s, 9 s and 13 s. Approximately how far does it move and at what average rate does it accelerate over each of the two time intervals (i.e., 0- 9 seconds and 9 - 13 seconds)? Show how you could use a graph of velocity vs. time to obtain your results. Directly reason out your results using the concept of rate. If the automobile is coasting, and if air resistance is not a significant factor, then over which time interval do you think the average slope of the road is greatest?
......!!!!!!!!...................................
RESPONSE --> A graph with clock time on the x axis and velocity on the y axis gives us a change in velocity vs. clock time. dt = 9s dv = 5.5m/s a = 5.5m/s / 9s = .61m/s^2 ds = vAve * dt ds = 5.5m/s /2 * 9s = 24.75m dt = 4s dv = -7.75m/s a = -7.75m/s / 4s = -1.94m/s^2 ds = -3.9m/s * 4s = -15.6m The greatest slope would be on the second time interval with the negative acceleration and distance.
.................................................
......!!!!!!!!...................................
21:47:58 STUDENT ANSWER: The average acceleration for the cart from time 0s to 9s, the average velocity is (4.5+10)m/s/2=7.25 m/s so displacement is 7.25 m/s * 9s =65.25m. The displacement for the cart between clock time 9s and 13s is (2.25+10)m/s / 2 * 4s=24.5m The acceleration of the cart between clock time 0 - 9s is a = `dv / `dt ? (10-4.5)m/s / (9 s) =.6m/s^2. The acceleration of the cart between clock time 9-13s is (2.25-10( m/s / (4s) = -1.93m/^2. You could use a graph of velocity vs. time to obtain the results. By placing the information on a graph of v vs. t you could see the rise and the run between the points. For example looking at 0,4.5 and 9,10 you could see that it had a run of 9 s and a rise of 5.5 m/s and then to find the acceleration you would have ave accel = slope = 5.5 m/s / (9 s) = .61m/s^2. If the automobile is coasting then the slope of the road is the greatest when the magnitude of the acceleration is greatest. This occurs on the second slope, where the magnitude of the acceleration is 1.93 m/s^2. INSTRUCTOR NOTES FOR ALL STUDENTS: Be sure you use units correctly in every calculation, and be sure you are using the definitions of velocity (rate of change of position, so vAve = `ds / `dt) and acceleration (rate of change of velocity, so a = `dv / `dt).
......!!!!!!!!...................................
RESPONSE --> I did not take the average of the two velocities, but the average of the change in velocity.
.................................................
......!!!!!!!!...................................
21:59:48 Problem Number 2 We set up a straight ramp and measure the time required to coast down the ramp, from rest, at various slopes. The differences in elevation between one end and the other, for the different slopes, are 1.8, 4.2 and 6.7 cm. The time required for a cart to coast 78 cm down the ramp, starting from rest, is 2.722297 seconds on the first incline, 2.518101 seconds on the next, and 2.6606 seconds on the last. How well do these results support the hypothesis that acceleration is linearly dependent on slope?
......!!!!!!!!...................................
RESPONSE --> 1.8cm 4.2cm 6.7cm ds 78cm dt = 2.722297, 2.528101, 2.6606 Vo = 0 a = dv/dt ds = vAve * dt 78cm = vAve * 2.722297 vAve = 28.65cm/s 28.65 * 2 = 57.3cm/s = vf dv = 57.3cm/s a = 57.3cm/s / 2.722297 = 21cm/s^2 78cm/2.528101 = 30.2cm/s^2 30.2cm/s * 2 = 60.4cm/s 60.4cm/s / 2.528101 = 23.9cm/s^2 78cm/2.6606s = 29.3cm/s 29.3cm/s * 2 = 58.6cm/s 58.6cm/s / 2.6606s = 22cm/s^2
.................................................
......!!!!!!!!...................................
22:00:48 STUDENT RESPONSE: 1.8cm 2.722297s 4.2cm 2.518101s 6.7cm 2.6606s 'ds=78. These results show that the smallest slope the time to coast is the slowest The middle ramp has the fastest time down the ramp These results show that acceleration is fastest down the middle ramp (4.2cm) INSTRUCTOR COMMENT ON STUDENT RESPONSE: ** To support the idea that acceleration is linearly dependent on slope you first have to calculate the accelerations, as you did in the Video Experiments. You then need to plot a graph of acceleration vs. ramp slope and see if it forms a straight line. **
......!!!!!!!!...................................
RESPONSE --> My last calculation was smaller than the second one, and it wouldn't be a straight line.
.................................................
......!!!!!!!!...................................
22:07:04 Problem Number 3 A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?
......!!!!!!!!...................................
RESPONSE --> ds = 50cm dt = 3.8s vf = 6.31579cm/s Vave = 50cm/3.8s = 13.2cm/s 13.2cm/s * 2 = 26.4cm/s 26.4cm/s - 6.31579cm/s = 20.1cm/s = vo 6.31579 - 20.1cm/s = -13.8cm/s = dv a = -13.78cm/s / 3.8s = -3.6cm/s^2
.................................................
......!!!!!!!!...................................
22:08:11 STUDENT RESPONSE: .'ds=50cm vf=6.31579cm/s 'dt = 3.8s The average velocity on the ramp is 50cm/3.8s =13.16m/s The initial velocity will be v0 = 2 'ds /dt, v0 = 2 *50 /3.8, v0 = 26.31579 -vf which is 6.31579, so the v0 is 20.00 ** You got it but this step isn?t correctly stated. v0 = 2 `ds/dt is not a correct statement. v0 = 2 vAve ? vf, or v0 = 2 `ds/`dt ? v0 (which is the same thing since vAve = `ds/`dt) is the correct equation. The reasoning is that ave velocity is the average of initial and final velocities, since acceleration is uniform. So you have vAve = (v0 + vf) / 2.. You know vAve from your previous calculation and vf is given. So you solve this equation for v0 and then substitute these values and simplify. You get v0 = 2 vAve ? vf, then substitute. This is actually what you did; just be careful you state it this way. If you don?t state it right there?s a chance you might not do it right. ** The accleration is 6.31579-20/3.8 = -3.601m/s/s If your v0 was correct this would be right **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:14:00 Problem Number 4 A projectile leaves the edge of a table at 18 cm/s, then falls freely a distance of 95 cm to the floor. How far does it travel in the horizontal direction during the fall?
......!!!!!!!!...................................
RESPONSE --> vo = 0 vf = 18cm/s ds = 95cm ds = 0 + .5a dt^2 95cm = .5(980cm/s^2) dt^2 .19 = dt^2 dt = .4s ds = 9cm/s *.4 s = 3.6cm
.................................................
......!!!!!!!!...................................
22:14:44 STUDENT RESPONSE: .'ds=95cm a=980cm/s/s v0=0 First, we take the equation 'ds = v0'dt + .5(980) 'dt^2 95cm = .5 (980) 'dt^2 'dt = sqrt 95cm/.5(980)=.194 So, 'dt = sqrt .194 'dt=.44s Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s INSTRUCTOR COMMENT TO ALL STUDENTS: Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.
......!!!!!!!!...................................
RESPONSE --> I took the 18cm/s as the final velocity and divided to get the average velocity.
.................................................
......!!!!!!!!...................................
22:18:44 Problem Number 5 What are the average acceleration and final velocity of an object which accelerates uniformly from rest, ending up with velocity 13.9 cm/sec after traveling a distance of 40 cm from start to finish?
......!!!!!!!!...................................
RESPONSE --> vO = 0 vF = ds = 40cm a = ? v = 13.9cm/s vf = 13.9cm/s 6.95cm/s = 40cm/dt 40cm/ 6.95cm/s = 5.8s a = 13.9cm/s / 5.8s = 2.4cm/s^2
.................................................
......!!!!!!!!...................................
22:19:29 STUDENT RESPONSE: 'ds=40cm v0=0 vf=13.9cm/s If vf is 13.9cm/s then the using the equation a 'ds = vf^2 - v0^2)/2 will be , a * 40 cm = ( (13.9 cm/s)^2 ? 0^2) /2 40 * a = 193.21 cm^2/s^2 / 2 a = 96.61 cm^2/s^2 / (40 cm) a = 2.415cm/s^2 ** If vf^2 = v0^2 + 2 a `ds then a `ds = (vf^2 ? v0^2) / 2, as you say, so a = (vf^2 ? v0^2) / (2 `ds). This is what you did (good job) but be careful to state it this way. ** Then use the equation 'ds = v0 'dt + .5 a 'dt^2 ** I wouldn?t use this equation to solve for `dt, since the equation is quadratic in `dt. At this stage you know all variables except `dt. The equation vf = v0 + a `dt would be much simpler. Or you could just figure average velocity and divide into displacement. Either way would be fine. ** 40= +.5 2.415cm/s/s dt^2 dt = sqrt 40/ ** You could have used the fact that vAve = (13.9 cm/s + 0) / 2 = 6.95 cm/s to obtain `dt: Since vAve = `ds / `dt, we have `dt = `ds / vAve = 40 cm / (6.95 cm/s) = 5.8 sec, approx. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
wKGUϒł assignment #003 X{wǭp Liberal Arts Mathematics I 07-28-2006
......!!!!!!!!...................................
22:23:13 Practice Test 2 Problem Number 1. Explain how we used a rubber band and a rail to demonstrate the conservation of the F `ds quantity.
......!!!!!!!!...................................
RESPONSE --> The force of the rubber band times the distance moved equals the ?
.................................................
......!!!!!!!!...................................
22:23:43 GOOD STUDENT ANSWER: We calculate the potential energy of the system when the rubber band is fully stretched, and compare with the F `ds total as the rail slides across the floor to see if all the potential energy was dissipated against friction.
......!!!!!!!!...................................
RESPONSE --> PE against friction
.................................................
......!!!!!!!!...................................
22:28:29 Problem Number 2 An Atwood machine consists of 34 paper clips, each of mass .4 grams, suspended from each side of a light pulley. If 4 clips are transferred from one side to the other, what will be the total gravitational force on the system?
......!!!!!!!!...................................
RESPONSE --> 34 * .4g = 13.6g 13 * .4 = 5.2g 21 * .4 = 8.4g F = m * g F = 8.4g * 980cm = 8232N
.................................................
......!!!!!!!!...................................
22:29:41 GOOD STUDENT SOLUTION: Well Im assuming that there are 34 paper clips on each side. So; 34 * .0004 kg= .0136 kg on one side when even 4 * .0004 kg = .0016 kg less on the side that has 4 removed so that gives you .0152 kg on one side and .012 kg on the other now to get the gravitational force on the system you multiply both by 9.8 m/sec^2 and that will give you the force acting on each one, which are: .14896 N and .1176 N respectively. Now to get the net gravitational force just subtract and that will give you a net gravitational force of GravFnet = .03136 N
......!!!!!!!!...................................
RESPONSE --> I thought there were 17 on each side and then 4 were taken from one side and moved to the other?
.................................................
......!!!!!!!!...................................
22:33:01 If the frictional force exerted by the pulley is .05 times the total weight of the system, then what is the net accelerating force?
......!!!!!!!!...................................
RESPONSE --> fnet = m * a fnet = 13.6g * .05 = .68g cm/s^2
.................................................
......!!!!!!!!...................................
22:35:42 well take the weights given above and combine them, then multiply by the .05, that will give us: [(.0136 kg + .0016 kg) * 9.8 m/sec^2] * .05= .007448 N for the frictional force now subtract GravFnet Ff = Fnet, or 0.03136 N - .007448 N = .023912 N net force.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:38:18 What therefore is the acceleration of the system?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:39:00 Divide net force by the total mass of the system to get the acceleration of the system .023912 N / .0152 kg = 1.573 m/sec^2
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:44:03 Problem Number 3 What will be the tension in the string holding a ball which is being swung in a circle of radius 1 meters, if the ball is making a complete revolution every .3 seconds? Assume that the system is in free fall (e.g., in a freely falling elevator, in orbit, etc.)?
......!!!!!!!!...................................
RESPONSE --> 9.8m/s * .3s = 2.94N
.................................................
......!!!!!!!!...................................
22:48:23 STUDENT ATTEMPT: Alright Im not very good at this YET, so I could use some help!!!!!!!! I know what is given in the problem. From this I can figure: angular velocity = 2pi*(f) where f = 3.33333 rev/sec so angular velocity = 20. 94359102 sec^-1 and because 1m* omega = linear velocity is 20. 944 m/sec AND THE REST I GET LOST ON, IM HAVING PROBLEMS WITH ALL THESE RELATIONSHIPS!!!!!!!!!!!!!!!! INSTRUCTOR NOTE: ** Centripetal acceleration is v^2 / r. That's the key thing you're forgetting here. This gives you aCent = (20.94 m/s)^2 / (1 m)^2 = 440 m/s^2. The centripetal force would therefore be mass_ball * 440 m/s^2. **
......!!!!!!!!...................................
RESPONSE --> v^2 / 4 = centripetal acc. angular v = 2 pi 3.3333(rev/s) = 20.9s^-1 1m * omega = 20.9m/s (20.9m/s)^2 / (1m)^2 = 440m/s^2 Centripetal force = mass of ball * 440m/s^2
.................................................
......!!!!!!!!...................................
22:52:47 What would be the tension in the string if the system was on and stationary with respect to the surface of the Earth, with the ball being swung in a vertical circle, when the ball is at the top of its arc?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:53:57 The net force on the ball is always equal to the centripetal force Fcent = mass_ball * 440 m/s^2. Gravity exerts a force equal to mass_ball * 9.8 m/s^2. If the ball is at the top of its arc the net force consists of the tension plus the gravitational force, both of which act in the same direction, which is downward. So using downward as the positive direction we have tension + mass_ball * 9.8 m/s^2 = mass_ball * 440 m/s^2. We find that tension = mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = mass_ball * 430.2 m/s^2. You could substitute a reasonable mass (e.g., .4 kg) for the ball and obtain all quantities in Newtons. Note that these calculations are not in fact accurate to 4 significant figures. The numbers shown here are intended to demonstrate how the tension differs from the centripetal force.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:54:50 What if the ball is at the bottom of its arc?
......!!!!!!!!...................................
RESPONSE --> Then the gravitational force would be negative?
.................................................
......!!!!!!!!...................................
22:55:16 If the ball is at the bottom of its arc the net force consists of the tension plus the gravitational force, with tension up and gravity down. Again using downward as the positive direction and noting that the upward centripetal acceleration is with this assumption negative we have tension + mass_ball * 9.8 m/s^2 = -mass_ball * 440 m/s^2. We find that tension = -mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = -mass_ball * 449.8 m/s^2.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:55:24 What if the ball is at its halfway height?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:56:15 At the halfway height the centripetal force and gravitational acceleration are perpendicular and hence independent. The only force acting toward the center is the tension in the string, which must therefore supply the entire centripetal force. The tension is mass_ball * 440 m/s^2 toward the center.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:58:57 Problem Number 4 If we set the expression G M / r^2 for the gravitational acceleration at distance r from the center of a planet of mass M equal to the centripetal acceleration v^2 / r and solve for v in terms of r, what is the result?
......!!!!!!!!...................................
RESPONSE --> GMR = V^2R^2
.................................................
......!!!!!!!!...................................
22:59:20 Setting the two expressions equal we have the equation G M / r^2 = v^2 / r. To solve for v we first multiply both sides by r to get G M r = v^2, Taking the square root of both sides and reversing sides of the equation gives us v = sqrt(G M r).
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:03:34 Problem Number 5 A disk of negligible mass and radius 30 cm is constrained to rotate on a frictionless axis about its center. On the disk are mounted masses of 7 gram at a distance of 23.1 cm from the center, 24 grams data distance of 16.8 cm from the center and 47 grams at a distance o 9 cm from the center. A uniform force of .06983 Newtons is applied at the rim of the disk in a direction tangent to the disk.
......!!!!!!!!...................................
RESPONSE --> 7g * 23.1cm = 161.7g cm 24g * 16.8cm = 403.2g cm 47 * 9cm = 423g cm ???
.................................................
......!!!!!!!!...................................
23:05:56 What will be the angular acceleration of the disk?
......!!!!!!!!...................................
RESPONSE --> alpha = tau/I I = mr^2 uniform disk is 1/2 MR^2
.................................................
......!!!!!!!!...................................
23:07:09 ** The moment of inertia of the system is the sum of all the m r^2 contributions of the individual particles. The net torque is the product of the net force and the moment arm. Newton's 2d law F = m a, expressed in in angular form, is `tau = I `alpha, where `tau is net torque (analogous to force), I is moment of intertia (sum of all mr^2, analogous to mass) and `alpha is angular acceleration in rad/s^2. **
......!!!!!!!!...................................
RESPONSE --> how do I find net torque and the moment of arm?
.................................................
......!!!!!!!!...................................
23:08:24 If the force is applied for 4 seconds with the disk initially at rest, what angular velocity with the disk attain?
......!!!!!!!!...................................
RESPONSE --> acceleration * 4s
.................................................
......!!!!!!!!...................................
23:08:30 ** Once you know angular acceleration it's easy to find change in angular velocity **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:10:13 What then will be the speed of each of the masses?
......!!!!!!!!...................................
RESPONSE --> velocity times radius
.................................................
......!!!!!!!!...................................
23:10:32 ** You know angular velocity and distance of each mass from the axis of rotation. Angular velocity is the velocity of the mass along the arc divided by the radius. So what is the velocity of each of the masses along its arc? **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:11:41 What will be their total kinetic energy?
......!!!!!!!!...................................
RESPONSE --> .5mv^2 = KE
.................................................
......!!!!!!!!...................................
23:11:45 *&*& Add up the individual kinetic energies.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:11:56 Compare the total kinetic energy to the change in the quantity .5 I `omega^2.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:12:00 *&*& The two quantities should be the same.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:12:34 Problem Number 6 A gun fires a bullet of mass 33 grams out of a barrel 38 cm long, from which it exits at 267 m/s. Assuming that the bullet accelerates uniformly from rest along the length of the barrel
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:18:24 How long does it take the bullet to exit the barrel after the powder ignites?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:19:10 Well I know that v0= 0 and vf= 267 m/sec and `ds= .38 m and from there I can use vf^2 = v0^2 + 2a`ds so I figure a= 93801 m/ sec^2 From there I can find the change in time to be 2.846441948 * 10^- 3 ** OK but it's easier to average init and final vel then divide into displacement (e.g., vAve = (0 + 267) / 2 m/s = 133.5 m/s so to move .38 meters requires .38 m / (133.5m/s) = .028 sec approx. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:19:27 Using the Impulse-Momentum Theorem determine the average force exerted on the bullet as it accelerates along the length of the barrel.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:20:09 By the impulse-momentum theorem Fave `dt = m `dv we get Fave = (m`dv)/ `dt so from there I can substitute in the values already determined and find the force to be 3095.4434 N ** Good work. Impulse-momentum is appropriate here. It would also have been possible to find acceleration and multiply by mass, but impulse-momentum is more direct. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:20:25 Using Newton's Second Law determine the average force exerted on the bullet as it accelerates along the length of the barrel.
......!!!!!!!!...................................
RESPONSE --> f = m * a
.................................................
......!!!!!!!!...................................
23:20:33 i dont know what to do here. ** Sure you do. You found the acceleration above when you didn't need it. You need it now. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"