course PHY 201 ??????j{??????z??assignment #017017. collisions
......!!!!!!!!...................................
11:37:44 `q001. Note that this assignment contains 5 questions. . A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.
......!!!!!!!!...................................
RESPONSE --> The initial momentum (p) of the object will be p=mv ip = 10 kg ( 5 m/s) ip = 50 kg m/s The final momentum (fp) of the object will be fp = mv fp = 10 kg * 3 m/s fp = 30 kg m/s So 'dp = 30 kg m/s-50 kg m/s = -20 kg m/s 'dp = Fnet ('dt) Fnet = 'dp/'dt Fnet = -20 kg m/s /.03 s Fnet = -666.667 Newtons of Force lost by the moving object so the stationary object must have exerted equal but opposite amount of force against the moving object so it would have exerted 666.67 Newtons of Force not counting any other Forces. confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:42:10 By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.
......!!!!!!!!...................................
RESPONSE --> I didn't use the right theorum but I did come up with the answer of -667 Newtons, but I thought that this would be the force lost by the first object and not the force exerted by the second. self critique assessment: 2 I understand the mechanics of how to answer the question but was unsure about what I was finding.
.................................................
......!!!!!!!!...................................
12:24:26 `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.
......!!!!!!!!...................................
RESPONSE --> The force exerted on the second object is going to be equal but opposite the force that it exerted on the first objecst so it will be 667 Newtons in the direction of original fravel. We can use this to find the final velocity. m v = Fnet('dt) v = Fnet('dt)/ m v = 667 kg m/s^2(.03 s)/ 2kg v = 10 m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:25:23 Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second. This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object. A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s. Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.
......!!!!!!!!...................................
RESPONSE --> Same thing that I got. self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:36:50 `q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?
......!!!!!!!!...................................
RESPONSE --> The KE before the collision is .5 m v^2= .5(10 kg)(5 m/s)^2 = 125 Joules ( kg m^2/s^2) The KE after will be the combination of the KE of the 2 masses .5(10 kg)(3 m/s)^2 = 45 Joules and .5(2 kg)(10 m/s)^2 = 100 Joules The total would there fore be 145 Joules. This answer is saying there is more Kinetic energy within the system but I don't see how this would be possible unless there is more forece being applied somwhere. Gravity or air pushing aginst one or both as they move down the incline. confidence assessment: 2 I don't think that I should have gained KE so I must have did something wrong.
.................................................
......!!!!!!!!...................................
12:39:33 The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved.
......!!!!!!!!...................................
RESPONSE --> I was right in my answer and even saw that this shouldn't be possible under normal circumstances. self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:44:57 `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?
......!!!!!!!!...................................
RESPONSE --> The momentum before the collision will be that of the first object because the second is not moving so it has a velocity of 0. p = 10 kg ( 5 m/s) =50 kg m/s The momentum after will be the total of the 2 momentums. p1 = 10 kg (3 m/s) = 30 kg m/s plus p2 = 2 kg ( 10 m/s) = 20 kg m/s The momentum is the same after the collision as it is before the collision. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:45:28 The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision. The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second. The total momentum after collision is therefore equal to the total momentum before collision.
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:53:01 `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?
......!!!!!!!!...................................
RESPONSE --> The impluse momentum theorem is based on net fore over a period of time. Fnet('dt) = 'dp We know that forece is not lost. When force is applied by one objecst it loses force but this force is gained by the object it is applied against. If the force within an enclosed system is there fore unchanging then the momentum must also be unchanging. confidence assessment: 2 Not sure if this is enough or if I need to use formulas to show my ideas.
.................................................
......!!!!!!!!...................................
12:54:27 Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.
......!!!!!!!!...................................
RESPONSE --> This is the same thing that I was saying but I was generalizing beyond just this instance. self critique assessment: 3
.................................................
course PHY 201 €??????~???assignment #017
......!!!!!!!!...................................
02:14:20 ANSWERS/COMMENTARY FOR QUERY 17
......!!!!!!!!...................................
RESPONSE --> Ok, I don't know what you want for this or if it is just a title line.
.................................................
......!!!!!!!!...................................
02:26:59 prin phy and gen phy 6.33: jane at 5.3 m/s; how high can she swing up that vine?
......!!!!!!!!...................................
RESPONSE --> We know that 'dKE + 'dPE = 0 or that 'dKe = - 'dPE also that 'dKE = .5mv^2 and 'dpe = mg'dy so .5mv^2 = -(gmh) where h is the change in height for a constant mass then .5 v^2 = -gh so h = .5v^2/-g We know that her velocity is 5.3 m/s and gravity if you are working against it is going to be -9.8 m/s^2 so we have h = .5(5.3^2)/-(-9.8 m/s^2) = 1.433 meters in the upward direction. confidence assessment: 3
.................................................
......!!!!!!!!...................................
02:28:32 Jane is going to convert her KE to gravitational PE. We assume that nonconservative forces are negligible, so that `dKE + `dPE = 0 and `dPE = -`dKE. Jane's KE is .5 M v^2, where M is her mass. Assuming she swings on the vine until she comes to rest at her maximum height, the change in her KE is therefore `dKE = KEf - KE0 = 0 - .5 M v0^2 = - .5 M v0^2, where v0 is her initial velocity. Her change in gravitational PE is M g `dy, where `dy is the change in her vertical position. So we have `dKE = - `dPE, or -5 M v0^2 = - ( M g `dy), which we solve for `dy (multiply both sides by -1, divide both sides by M g) to obtain `dy = v0^2 / (2 g) = (5.3 m/s)^2 / (2 * 9.8 m/s^2) = 1.43 m.
......!!!!!!!!...................................
RESPONSE --> I explained it a little differently but worked through it the same way and got the same answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
02:58:24 prin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball
......!!!!!!!!...................................
RESPONSE --> First we will use the formula for calculating the PE of a compresed spring. PE = .5k('dx)^2 where k is the spring constant and 'dx is how far it is compressed. PE = .5(950 N/m)(0.150m)^2 = 10.6875 N m (Joules) We also know that PE = mgh 10.6875 kg m^2/s^2 = (0.30 kg)(9.8 m/s^2) h divide both sides by 0.30 kg and 9.8 m/s^2 to get h = 3.94 m To calculate the final velocity we will use PE = - KE 10.6875 kg m^2/s^2 = .5mv^2 v = sqrt ( 10.6875 kg m^2/s^2)/(0.5*.30 kg) v = sqrt ( 71.25 m^2/s^2) v = +- 8.44 m/s This will be determined by which direstion you want to use as positive speed. confidence assessment: 3
.................................................
......!!!!!!!!...................................
03:12:04 We will assume here that the gravitational PE of the system is zero at the point where the spring is compressed. In this situation we must consider changes in both elastic and gravitational PE, and in KE. The PE stored in the spring will be .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 107 J. When released, conservation of energy (with only elastic and gravitational forces acting there are no nonconservative forces at work here) we have `dPE + `dKE = 0, so that `dKE = -`dPE. Since the ball is moving in the vertical direction, between the release of the spring and the return of the spring to its equilibrium position, the ball t has a change in gravitational PE as well as elastic PE. The change in elastic PE is -107 J, and the change in gravitational PE is m g `dy = .30 kg * 9.8 m/s^2 * .150 m = +4.4 J. The net change in PE is therefore -107 J + 4.4 J = -103 J. Thus between release and the equilibrium position of the spring, `dPE = -103 J The KE change of the ball must therefore be `dKE = - `dPE = - (-103 J) = +103 J. The ball gains in the form of KE the 103 J of PE lost by the system. The initial KE of the ball is 0, so its final KE is 103 J. We therefore have .5 m vv^2 = KEf so that vf=sqrt(2 KEf / m) = sqrt(2 * 103 J / .30 kg) = 26 m/s. To find the max altitude to which the ball rises, we return to the state of the compressed spring, with its 107 J of elastic PE. Between release from rest and max altitude, which also occurs when the ball is at rest, there is no change in velocity and so no change in KE. No nonconservative forces act, so we have `dPE + `dKE = 0, with `dKE = 0. This means that `dPE = 0. There is no change in PE. The initial PE is 107 J and the final PE must also therefore be 107 J. There is, however, a change in the form of the PE. It converts from elastic PE to gravitational PE. Therefore at maximum altitude the gravitational PE must be 107 J. Since PEgrav = m g y, and since the compressed position of the spring was taken to be the 0 point of gravitational PE, we therefore have y = PEgrav / (m g) = 107 J / (.30 kg * 9.8 m/s^2) = 36.3 meters. The ball will rise to an altitude of 36.3 meters above the compressed position of the spring.
......!!!!!!!!...................................
RESPONSE --> The first part of this explination is wrong. .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 107 J This should be 10.7 not 107. I didn't think that we would need to take into account the work that the spring will have to do to overcome gravity. I thought this would be taken into account in the spring constant that is given. I thought all the force of the spring would be transfered to the ball. Is this wrong. self critique assessment: 2 I was confussed by this explination.
.................................................
......!!!!!!!!...................................
03:40:10 gen phy problem A high jumper needs to be moving fast enough at the jump to lift her center of mass 2.1 m and cross the bar at a speed of .7 m/s. What minimum velocity does she require?
......!!!!!!!!...................................
RESPONSE --> I am not sure about this problem but I will give it a shot. I am going to use the motion formulas. looking for vo vf will be .7 m/s a = -9.8 m/s^2 'ds = 2.1 m vf^2 = vo^2 + 2a'ds v0^2 = vf^2 - 2a'ds v0 = sqrt ((0.7 m/s)^2 -2( -9.8 m/s^2)(2.1m)) v0 = sqrt (0.49 m^2/s^2 +41.16 m^2/s^2) v0 = sqrt(41.65) =+-6.45 m/s confidence assessment: 1 iI DON'T THINK THAT I DID THIS RIGHT
.................................................
......!!!!!!!!...................................
03:42:55 FORMAL SOLUTION: Formally we have `dPE + `dKE = 0. `dPE = M * g * `dy = M * 20.6 m^2 / sec^2, where M is the mass of the jumper and `dy is the 2.1 m change in altitude. `dKE = .5 M vf^2 - .5 M v0^2, where vf is the .7 m/s final velocity and v0 is the unknown initial velocity. So we have M g `dy + .5 M vf^2 - .5 M v0^2 = 0. Dividing through by M we have g `dy + .5 vf^2 - .5 v0^2 = 0. Solving for v0 we obtain v0 = sqrt( 2 g `dy + vf^2) = sqrt( 2* 9.8 m/s^2 * 2.1 m + (.7 m/s)^2 ) = sqrt( 41.2 m^2/s^2 + .49 m^2 / s^2) = sqrt( 41.7 m^2 / s^2) = 6.5 m/s, approx.. LESS FORMAL, MORE INTUITIVE, EQUIVALENT SOLUTION: The high jumper must have enough KE at the beginning to increase his PE through the 2.1 m height and to still have the KE of his .7 m/s speed. The PE change is M * g * 2.1 m = M * 20.6 m^2 / sec^2, where M is the mass of the jumper The KE at the top is .5 M v^2 = .5 M (.7 m/s)^2 = M * .245 m^2 / s^2, where M is the mass of the jumper. Since the 20.6 M m^2 / s^2 increase in PE must come at the expense of the initial KE, and since after the PE increase there is still M * .245 m^2 / s^2 in KE, the initial KE must have been 20.6 M m^2 / s^2 + .245 M m^s / s^2 =20.8 M m^s / s^2, approx. If initial KE is 20.8 M m^s / s^2, then .5 M v0^2 = 20.8 M m^s / s^2. We divide both sices of this equation by the jumper's mass M to get .5 v0^2 = 20.8 m^2 / s^2, so that v0^2 = 41.6 m^2 / s^2 and v0 = `sqrt(41.6 m^2 / s^2) = 6.5 m/s, appprox.
......!!!!!!!!...................................
RESPONSE --> This is the same answer that I got. self critique assessment: 3
.................................................
......!!!!!!!!...................................
03:43:09 query Univ. 7.42 (7.38 in 10th edition). 2 kg block, 400 N/m spring, .220 m compression. Along surface then up 37 deg incline all frictionless. How fast on level, how far up incline?
......!!!!!!!!...................................
RESPONSE --> Not in this class. confidence assessment: 3
.................................................
......!!!!!!!!...................................
03:43:19 ** The spring exerts a force of 400 N / m * .220 m = 84 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 84 N) / 2 = 42 N, so the work done in the compression is `dW = Fave * `ds = 42 N * .220 m = 5.0 Joules, approx. If all this energy is transferred to the block, starting from rest, the block's KE will therefore be 5.0 Joules. Solving KE = .5 m v^2 for v we obtain v = sqrt(2 KE / m) = sqrt(2 * 5.0 Joules / (2 kg) ) = 2.2 m/s, approx.. No energy is lost to friction so the block will maintain this speed along the level surface. As it begins to climb the incline it will gain gravitational PE at the expense of KE until the PE is 5.0 J and the KE is zero, at which point it will begin to slide back down the incline. After traveling through displacement `ds along the incline the height of the mass will be `ds sin(37 deg) = .6 `ds, approx., and its gravitational PE will be PE = m g h = m g * .6 `ds = .6 m g `ds. Setting this expression equal to KE we obtain the equation .6 m g `ds = KE, which we solve for `ds to obtain `ds = KE / (.6 m g) = 5.0 Joules / (.6 * 2 kg * 9.8 m/s^2) = .43 meters, approx. **
......!!!!!!!!...................................
RESPONSE --> Not in this class. self critique assessment: 3
.................................................
......!!!!!!!!...................................
03:43:32 query univ phy 7.50 62 kg skier, from rest, 65 m high. Frict does -10.5 kJ. What is the skier's speed at the bottom of the slope? After moving horizontally over 82 m patch, air res 160 N, coeff frict .2, how fast is she going? Penetrating 2.5 m into the snowdrift, to a stop, what is the ave force exerted on her by the snowdrift?
......!!!!!!!!...................................
RESPONSE --> Not in this class. confidence assessment: 3
.................................................
......!!!!!!!!...................................
03:43:44 ** The gravitational PE of the skier decreases by 60 kg * 9.8 m/s^2 * 65 m = 38 kJ, approx. (this means 38 kiloJoules, or 38,000 Joules). The PE loss partially dissipated against friction, with the rest converted to KE, resulting in KE = 38 kJ / 10.5 kJ = 27.5 kJ. Formally we have `dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the skier against friction. Since friction does -10.5 kJ of work on the skier, the skier does 10.5 kJ of work against friction and we have `dKE = -`dPE - `dWnoncons = - (-38 kJ) - 10.5 kJ = 27.5 kJ. The speed of the skier at this point will be v = sqrt( 2 KE / m) = sqrt( 2 * 27,500 J / (65 kg) ) = 30 m/s, approx. Over the 82 m patch the force exerted against friction will be .2 * 60 kg * 9.8 m/s^2 = 118 N, approx., so the force exerted against nonconservative forces will be 118 N + 160 N = 280 N approx.. The work done will therefore be `dWnoncons = 280 N * 82 m = 23 kJ, approx., and the skier's KE will be KE = 27.5 kJ - 23 kJ = 4.5 kJ, approx. This implies a speed of v = sqrt( 2 KE / m) = 12 m/s, approx. To stop from this speed in 2.5 m requires that the remaining 4.5 kJ of KE be dissipated in the 2.5 m distance. Thus we have `dW = Fave * `ds, so that Fave = `dW / `ds = 4500 J / (2.5 m) = 1800 N. This is a significant force, about 3 times the weight of the skier, but distributed over a large area of her body will cause a good jolt, but will not be likely to cause injury.**
......!!!!!!!!...................................
RESPONSE --> Not in this class. self critique assessment: 3
.................................................
????g?????I? assignment #018 018. `query 18 Physics I 10-21-2007
......!!!!!!!!...................................
14:42:29 Query intro problem sets Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
......!!!!!!!!...................................
RESPONSE --> If we know the vertical displacement and the inital vertical velocities we can calculate the time it will take for the object to complete the vertical displacement and then with this time and the horizontal velocite we can determine how far the object will travel in the horizontal direction. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:43:31 ** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **
......!!!!!!!!...................................
RESPONSE --> I didn't give the formulas but this is the same thing that i said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:46:03 Query class notes #17 Why do we expect that in a collision of two objects the momentum change of each object must be equal and opposite to that of the other?
......!!!!!!!!...................................
RESPONSE --> because momentum is unchanging. If it changes in one it must change by an equal and opposite amount in the other object. The total momentum in a system will be the same before an impact as it will be after given that there is no extra force added to the system from chemical or other reaction within the system. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:47:52 **COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change. Momentum has nothing directly to do with energy. Two colliding object exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. **
......!!!!!!!!...................................
RESPONSE --> I understood what was ment and gave close to the same answer but I didn't use the term impulse. I did know that momentum would be conserved. self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:51:16 What are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related?
......!!!!!!!!...................................
RESPONSE --> We need the mass of both objects. The velocity of the objects including their direction before the colistion. We could aslo use the force and the time this force is exerted. We would also use or find the velocities of the objects after the collision. confidence assessment: 2 Not sure if i got what you were looking for or not.
.................................................
......!!!!!!!!...................................
14:53:55 ** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. **
......!!!!!!!!...................................
RESPONSE --> I did name all that you were looking for but I also added in force and duration of time that force is exetted to find the change in momentum or impulse. self critique assessment: 3 I will know next time I am ask.
.................................................
......!!!!!!!!...................................
15:10:43 `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision?
......!!!!!!!!...................................
RESPONSE --> We first convert 95 km/h to m/s = 26.38889 m/s Then we find the KE of each car .5mv^2 .5(7650 kg)(26.689 m/s)^2 = 2663628.472 Joules if each has this amoung of energy when they colide and they become have a velocity of 0 afterwards. Then the energy has to go somewhere. Some will be lost to friction, sound, and wind resistance but most will be converted to therman energy. So when they colide 2.66 * 10^6 J *2 will be converted to thermal energy = 5.32 *10^6 Joulses self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:11:14 There is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy. The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 265,000 Joules, so that their total KE is 2 * 265,000 J = 530,000 J. This KE is practially all converted to thermal energy.
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:01:05 `1Query* gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the track)?
......!!!!!!!!...................................
RESPONSE --> v0 = 1.7 m/s y1 = 35 m given in the book y2 = 0 g = 9.8 m/s^2 vf = ? The work of negative nonconductive surfaces (frictional forces) is equal to the change in the KE plus the change in the PE -Wnc = ((.5mvf^2) - .5m v0^2) + (mg'dy2 -mg'dy1) wnc is said to be 1/5 wt = 1/5* (mg) times the 'dy over which it acts to get work.
.................................................
......!!!!!!!!...................................
16:09:50 **GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION: Until just now I did not think I could work that one, because I did not know the mass, but I retried it. Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0. PE is all gravitational so that `dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) It looks like the M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.
......!!!!!!!!...................................
RESPONSE --> This is close to what I had. I don't know where they come up with the 28 m they used for 'dy. I used the 'dy in the picture with y1 = 35 m and y2 = 0 m so 'dy = -35 m I don't know why they give us the distance traveled is 45 m because this appears to be the horizontal distance all of our values are derived by the vertical change becuase it is what betermines the effect gravity will have. This answer is not what the book gives. My answer is a lot closer to the books than what they give. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:10:12 Univ. 7.74 (7.62 in 10th edition). 2 kg pckg, 53.1 deg incline, coeff kin frict .20, 4 m from spring with const 120 N/m. Speed just before hitting spring? How far compressed? How close to init pos on rebound?
......!!!!!!!!...................................
RESPONSE --> Not in this class. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:10:20 ** The forces acting on the package while sliding down the incline, include gravitiational force, normal force and friction; and after encountering the spring the tension force of the spring also acts on the package. The normal force is Fnormal = 2 kg * 9.8 m/s^2 * cos(53.1 deg) = 11.7 N, approx.. This force is equal and opposite to the component of the weight which is perpendicular to the incline. The frictional force is f = .2 * normal force = .2 * 11.7 N = 2.3 N, approx.. The component of the gravitational force down the incline is Fparallel = 2 kg * 9.8 m/s^2 * sin(53.1 deg) = 15.7 N, approx.. Friction acts in the direction opposite motion, up the incline in this case. If we choose the downward direction as positive the net force on the package is therefore 15.7 N - 2.3 N = 13.4 N. So in traveling 4 meters down the incline the work done on the system by the net force is 13.4 N * 4 m = 54 Joules approx. Just before hitting the spring we therefore have .5 m v^2 = KE so v = +-sqrt(2 * KE / m) = +-sqrt(2 * 54 J / (2 kg) ) = +- 7.4 m/s. If we ignore the gravitational and frictional forces on the object while the spring is compressed, which we really don't want to do, we would conclude the spring would be compressed until its elastic PE was equal to the 54 J of KE which is lost when the object comes to rest. The result we would get here by setting .5 k x^2 equal to the KE loss is x = sqrt(2 * KE / k) = .9 meters, approx.. However we need to include gravitational and frictional forces. So we let x stand for the distance the spring is compressed. As the object moves the distance x its KE decreases by 54 Joules, its gravitational PE decreases by Fparallel * x, the work done against friction is f * x (where f is the force of friction), and the PE gained by the spring is .5 k x^2. So we have `dKE + `dPE + `dWnoncons = 0 so -54 J - 15.7N * x + .5 * 120 N/m * x^2 + 2.3 N * x = 0 which gives us the quadratic equation 60 N/m * x^2 - 13.4 N * x - 54 N m = 0. (note that if x is in meters every term has units N * m). Suppressing the units and solving for x using the quadratic formula we have x = ( - (-13.4) +- sqrt(13.4^2 - 4 * 60 (-54) ) / ( 2 * 60) = 1.03 or -.8 meaning 1.07 m or -.8 m (see previous note on units). We reject the negative result since the object will continue to move in the direction down the incline, and conclude that the spring would compress over 1 m as opposed to the .9 m obtained if gravitational and frictional forces are not accounted for during the compression. This makes sense because we expect the weight of the object (more precisely the weight component parallel to the incline) to compress the spring more than it would otherwise compress. Another way of seeing this is that the additional gravitational PE loss as well as the KE loss converts into elastic PE. If the object then rebounds the spring PE will be lost to gravitational PE and to work against friction. If the object goes distance xMax back up the incline from the spring's compressed position we will have`dPE = -.5 k x^2 + Fparallel * xMax, `dKE = 0 (velocity is zero at max compression as well as as max displacement up the incline) and `dWnoncons = f * xMax. We obtain `dPE + `dKE + `dWnoncons = 0 so -.5 k x^2 + Fparallel * xMax + 0 + 33 N * xMax = 0 or -.5 * 120 N/m * (1.07 m)^2 + 15.7 N * xMax + 2.3 N * xMax = 0 We obtain 18 N * xMax = 72 N m, approx., so that xMax = 72 N m / (18 N) = 4 meters, approx.. This is only 2.93 meters beyond the position of the object when the spring was compressed. Note that the object started out 4 meters beyond this position. **
......!!!!!!!!...................................
RESPONSE --> Not in this class. self critique assessment: 3
.................................................
?\???}???J??????? assignment #019 019. `query 19 Physics I 10-21-2007
......!!!!!!!!...................................
19:18:15 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
......!!!!!!!!...................................
RESPONSE --> The x component of the vector can be found by taking the magnitude of the vector times the cos of the angle. The y component of the vector can be found by taking the magnitude of the vector times the sin of the angle. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:19:04 ** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **
......!!!!!!!!...................................
RESPONSE --> This is the same thing I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:22:58 Explain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components.
......!!!!!!!!...................................
RESPONSE --> A vector is made up of the x and y components of the force. That is why if you have the x and y components using the pythogreantheorum you can find the magnitude of the vector. In other words if you tie a rope to an object you want to move and apply force along the componets of a vector equal to their forces it will have the same effect as pulling in the direction of the vector by its force. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:23:20 ** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. **
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:25:12 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
......!!!!!!!!...................................
RESPONSE --> If we caculate the vectors at the given point we can use pythogorean theorum to find its magnitude and we can use the arctan of y/x t find the angle of the vector. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:25:49 ** Using initial conditions and the equations of motion we can determine the x and y velocities vx and vy at a given point, using the usual procedures for projectiles. The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is arctan(vy / vx), plus 180 deg if x is negative. **
......!!!!!!!!...................................
RESPONSE --> Close to the same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:30:16 Explain how we can calculate the initial velocities of a projectile in the horizontal and vertical directions given the magnitude and direction of the initial velocity.
......!!!!!!!!...................................
RESPONSE --> it sounds like we are just breaking the vector down intos its componets by using the direction which is given and the magnitude of the vector. confidence assessment: 2 Not sure about what you where given and wanted to find in this question.
.................................................
......!!!!!!!!...................................
19:31:37 ** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude and the angle with respect to the positive x axis. Initial vel in the y direction is v sin(theta). **
......!!!!!!!!...................................
RESPONSE --> This is the same thing that we have been saying. I understand this concept. Given values it would have been easier to determine what you were looking for. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:32:06 Univ. 8.58 (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force?
......!!!!!!!!...................................
RESPONSE --> not in this class. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:32:16 ** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.002 s) = -240 Newtons, approx. **
......!!!!!!!!...................................
RESPONSE --> not in this class. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:33:08 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I believe that I have a much better understanding of vectors now. Before i was having trouble determining when to use sin or cos. self critique assessment: 3
.................................................
course PHY 201 ???yz[?????R????T~assignment #018
......!!!!!!!!...................................
03:48:27 `q001. Note that this assignment contains 7 questions. . The Pythagorean Theorem: the hypotenuse of a right triangle has a length c such that c^2 = a^2 + b^2, where a and b are the lengths of the legs of the triangle. Sketch a right triangle on a set of coordinate axes by first locating the point (7, 13). Then sketch a straight line from the origin of the coordinate system to this point to form the hypotenuse of the triangle. Continue by sketching a line straight down from (7, 13) to the x axis to form one leg of the triangle, then form the other leg by continuing straight along the x axis back to the origin. How long are these two legs? How long therefore is the hypotenuse?
......!!!!!!!!...................................
RESPONSE --> The horizontal leg will be 7 units while the vertical leg will be 13 units. So the hypotenuse will be sqrt(7^2 + 13^2) = sqrt ( 218) = 14.76 Units. confidence assessment: 3
.................................................
......!!!!!!!!...................................
03:48:59 The leg from (7, 13) to the x axis drops from the point (7, 13) to the point (7,0) and so has length 13. The second leg runs from (7,0) to the origin, a distance of 7 units. The legs therefore have lengths a = 13 and b = 7, so that the hypotenuse c satisfies c^2 = a^2 + b^2 and we have c = `sqrt(a^2 + b^2) = `sqrt( 13^2 + 7^2 ) = `sqrt( 216) = 14.7, approximately.
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
03:51:24 `q002. Sketch a right triangle on a set of coordinate axes whose leg along the x axis has length 12 and whose hypotenuse has length 15. What must be the length of the second leg of the triangle?
......!!!!!!!!...................................
RESPONSE --> Using the formula c^2 = a^2+b^2 then c=15 a=12 b^2 = c^2 -a^2 b = sqrt (15^2 - 12^2) b = sqrt (81) b = 9 confidence assessment: 3
.................................................
......!!!!!!!!...................................
03:51:36 Let c stand for the length of the hypotenuse and a for the length of the known side, with b standing for the length of the unknown side. Then since c^2 = a^2 + b^2, b^2 = c^2 - a^2 and b = `sqrt(c^2 - a^2) = `sqrt( 15^2 - 12^2) = `sqrt(225 - 144) = `sqrt(81) = 9.
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
04:02:39 `q003. If a line of length L is directed from the origin of an x-y coordinate system at an angle `theta with the positive x axis, then the x and y coordinates of the point where the line ends will be y = L * sin(`theta) and x = L * cos(`theta). Sketch a line of length 10, directed from the origin at an angle of 37 degrees with the positive x axis. Without doing any calculations estimate the x and y coordinates of this point. Give your results and explain how you obtained your estimates.
......!!!!!!!!...................................
RESPONSE --> (7.9,6) I strated with an origin point on graph paper I printed off. I found the distance of 10 squares to be 6 cm on the paper. I then lined up on what I was using for the x axis and marked an angle of 37 degrees. I then drew a line that started at the origin and went 6 cm(10 units on graph paper) toward the angle mark I made. I then counted the x and y position of the point as close as I could. confidence assessment: 3
.................................................
......!!!!!!!!...................................
04:03:26 The line will run closer to the x axis then to the y axis, since the line is directed at an angle below 45 degrees. It won't run a whole lot closer but it will run significantly closer, which will make the x coordinate greater than the y coordinate. Since the line itself has length 10, it will run less than 10 units along either the x or the y axis. It turns out that the x coordinate is very close to 8 and the y coordinate is very close to 6. Your estimates should have been reasonably close to these values.
......!!!!!!!!...................................
RESPONSE --> Mine were. self critique assessment: 3
.................................................
......!!!!!!!!...................................
04:06:46 `q004. Using your calculator you can calculate sin(37 deg) and cos(37 deg). First be sure your calculator is in degree mode (this is the default mode for most calculators so if you don't know what mode your calculator is in, it is probably in degree mode). Then using the sin and cos buttons on your calculator you can find sin(37 deg) and cos(37 deg). What are these values and what should therefore be the x and y coordinates of the line directed from the origin at 37 degrees from the x axis?
......!!!!!!!!...................................
RESPONSE --> sin(37 deg) = 0.7986 cos (37 deg) = 0.601 These are the x and y cordinates of a line that is 1 unit long along this angle. The values for a line 10 units long should be 10 times this value. confidence assessment: 3
.................................................
......!!!!!!!!...................................
04:07:45 sin(37 deg) should give you a result very close but not exactly equal to .6. cos(37 deg) should give you a result very close but not exactly equal to .8. Since the x coordinate is L cos(`theta), then for L = 10 and `theta = 37 deg we have x coordinate 10 * cos(37 deg) = 10 * .8 = 8 (your result should be slightly different than this approximate value). Since the y coordinate is L sin(`theta), then for L = 10 and `theta = 37 deg we have y coordinate 10 * sin(37 deg) = 10 * .6 = 6 (your result should be slightly different than this approximate value).
......!!!!!!!!...................................
RESPONSE --> Same thing I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
04:10:27 `q005. Show that the x and y coordinates you obtained in the preceding problem in fact give the legs of a triangle whose hypotenuse is 10.
......!!!!!!!!...................................
RESPONSE --> if a=7.99 b= 6.02 c^2=a^2+b^2 c^2 = 7.99^2+6.02^2 c^2 = 100.0805 c = sqrt (100.0805) c = 10.00 confidence assessment: 3
.................................................
......!!!!!!!!...................................
04:10:54 If c stands for the hypotenuse of the triangle, then if a = 8 and b = 6 are its legs we have c = `sqrt(a^2 + b^2) = `sqrt(8^2 + 6^2) = `sqrt(100) = 10. The same will hold for the more precise values you obtained in the preceding problem.
......!!!!!!!!...................................
RESPONSE --> Same thing I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
04:22:51 `q006. A vector drawn on a coordinate system is generally depicted as a line segment of a specified length directed from the origin at a specified angle with the positive x axis. The vector is traditionally drawn with an arrow on the end away from the origin. In the preceding series of problems the line segment has length 10 and was directed at 37 degrees from the positive x axis. That line segment could have been drawn with an arrow on its end, pointing away from the origin. The components of a vector consist of a vector called the x component drawn from the origin along the x axis from the origin to x coordinate L cos(`theta), and a vector called the y component drawn from the origin along the y axis to y coordinate L sin(`theta). What are the x and y components of a vector directed at an angle of 120 degrees, as measured counterclockwise from the positive x axis, and having length 30 units?
......!!!!!!!!...................................
RESPONSE --> x vector will be 30 cos (120 deg) = -15 y vector will be 30 sin (120 deg) = 26 That will put it in the second quadrant where it needs to be. (-15,26) confidence assessment: 3
.................................................
......!!!!!!!!...................................
04:23:12 The x component of this vector is vector along the x axis, from the origin to x = 30 cos(120 degrees) = -15. The y component is a vector along the y axis, from the origin to y = 30 sin(120 degrees) = 26, approx.. Note that the x component is to the left and the y component upward. This is appropriate since the 120 degree angle, has measured counterclockwise from the positive x axis, takes the vector into the second quadrant, which directs it to the left and upward.
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
04:29:59 `q007. The angle of a vector as measured counterclockwise from the positive x axis is easily determined if the components of the vector are known. The angle is simply arctan( y component / x component ) provided the x component is positive. If the x component is negative the angle is arctan( y component / x component ) + 180 deg. If the x component is positive and the y component negative, arctan( y component / x component ) will be a negative angle, and in this case we generally add 360 degrees in order to obtain an angle between 0 and 360 degrees. The arctan, or inverse tangent tan^-1, is usually on a calculator button marked tan^-1. Find the angle and length of each of the following vectors as measured counterclockwise from the positive x axis: A vector with x component 8.7 and y component 5. A vector with x component -2.5 and y component 4.3. A vector with x component 10 and y component -17.3.
......!!!!!!!!...................................
RESPONSE --> Arctan (5/8.7) = 29.89 deg Point (8.7,5) First Quadrant Arctan (4.3/-2.5) = -59.8+180 = 120.17 Point (-2.5,4.3) Second Quadrant Arctan (-17.3/10) = -60 +360 = 300 Point (10,-17.3) Forth Quadrant confidence assessment: 3
.................................................
......!!!!!!!!...................................
04:31:58 A vector with x component 8.7 and y component 5 makes an angle of arctan (5/8.7) = 30 degrees with the x axis. Since the x component is positive, this angle need not be modified. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(8.7^2 + 5^2) = 10. A vector with x component -2.5 and y component 4.3 makes an angle of arctan (4.3 / -2.5) + 180 deg = -60 deg + 180 deg = 120 degrees with the x axis. Since the x component is negative, we had to add the 180 degrees. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt((-2.5)^2 + 4.3^2) = 5. A vector with x component 10 and y component -17.3 makes an angle of arctan (-17.3 / 10) = -60 degrees with the x axis. Since the angle is negative, we add 360 deg to get 300 deg. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(10^2 + (-17.3)^2) = 20.
......!!!!!!!!...................................
RESPONSE --> I didn't find the length of the vectors but I understand how they are easily found using the pythagorean theorem. self critique assessment: 3
.................................................
?H??????????assignment #019 019. Vector quantities Physics II 10-21-2007
......!!!!!!!!...................................
16:23:16 `q001. Note that this assignment contains 5 questions. . If you move 3 miles directly east then 4 miles directly north, how far do end up from your starting point and what angle would a straight line from your starting point to your ending point make relative to the eastward direction?
......!!!!!!!!...................................
RESPONSE --> If you thind of the cordinates as being on the cordinat plane the point will be (3,4) the angle of this line will be the arctan of 4/3 = 53 degrees. To find out how far you are from the starting poing use the pythogoream theorm c^2 = a^2 + b^2 c = sqrt ( 3^2 + 4^2) c = sgrt ( 25) =5 miles from the starting point. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:23:40 If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees.
......!!!!!!!!...................................
RESPONSE --> Same thing I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:28:57 `q002. When analyzing the force is acting on an automobile as it coasts down the five degree incline, we use and x-y coordinate system with the x axis directed up the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000 Newton weight of the car is represented by a vector acting straight downward. This puts the weight vector at an angle of 265 degrees as measured counterclockwise from the positive x axis. What are the x and y components of this weight vector? Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline? ** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees. If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. ** It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component.
......!!!!!!!!...................................
RESPONSE --> The x component of the vector will be 1500 cos (265) = -130 Newtons The y component of the vector will be 1500 sin (265 deg) = -149 Newtons. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:32:27 The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline.
......!!!!!!!!...................................
RESPONSE --> I did the problem correctly but didn't have enough zeros on the value multiply times the sin and cos. In the con I put 1500 while with the sin i only put 150 on my calculator. self critique assessment: 3 I made a mistake in my calculations but understand where I went wrong.
.................................................
......!!!!!!!!...................................
16:37:16 `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force?
......!!!!!!!!...................................
RESPONSE --> The angle will be arctan (-400/300) = -53 +360 = 307 degrees starting at the x-axis and going clock wise. Within the 4th quadrant. The force exerted will be sqrt ( -400^2+300^2) The force exerted will be +-500 Newtons along the angle given above. It will be positive. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:37:56 Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees.
......!!!!!!!!...................................
RESPONSE --> Same thing that I said. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:48:50 `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force and what will be its direction?
......!!!!!!!!...................................
RESPONSE --> The x vector of the first will be 173.205 N = 200 N cos (30) The y vector of the first angle will be 200 N Sin (30) = 100 N The x vector of the second will be 300 N cos 150 deg = -258.81 N The y vector of the second will be 300 N sin 150 deg = 150 N If we then add the x vectors we get 173.20N -258.81 N = -85.61 N If we then add the y vectors we get 100 N + 150 N = 250 N So the x vector will have a magnitude of -85.61 N while the y vector has a magnitude of 250 N the angle of the line will be arctan (250N/-85.61 N) = -71.1 + 180 = 109 deg If you sketch the 2 vectors you should see that the line that connects the 2 is going to be in the 3rd quadrant direction from the origin. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:50:51 My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This means that the action of my force is completely equivalent to the action of two forces, one of 173 Newtons in the x direction and one of 100 Newtons in the y direction. Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action of your force is completely equivalent the action of two forces, one of -260 Newtons in the x direction and one of 150 Newtons in the y direction. In the x direction and we therefore have forces of 173 Newtons and -260 Newtons, which add up to a total x force of -87 Newtons. In the y direction we have forces of 100 Newtons and 150 Newtons, which add up to a total y force of 250 Newtons. The total force therefore has x component -87 Newtons and y component 250 Newtons. We easily find the magnitude and direction of this force using the Pythagorean Theorem and the arctan. The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons, approximately. The angle at which the force is directed, as measured counterclockwise from the positive x axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg.
......!!!!!!!!...................................
RESPONSE --> I forgot to give the magnitude of the angle but that was just an oversight on my part. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:56:45 The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately.
......!!!!!!!!...................................
RESPONSE --> I hit a wrong key before I could answer. I saw that the magnitude of the vector that is made up of the 2 component vectors is what you were looking for. I knew that we would have to find the x and y components of each vector and then add them together to find the vectors of the resulting vecor. Then using the pythogrean theroum we could find the magnitude of the vector and using the arctan on y/x vectors we could find the angle of the line. There is no point in me doing the math since i can already see all the math you did, but I do understand this concept completly. self critique assessment: 3
.................................................