course Phy 121 זyӗٚyxwCassignment #015
......!!!!!!!!...................................
14:49:56 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
......!!!!!!!!...................................
RESPONSE --> First of all, we find the acceleration of the problem acceleration = 10N / 2 kg = 5 m/s^2 for 3 seconds...... 5 m/s^2 * 3 = 15 m/s Impulse & Momentum....10N * 3 sec = 30 kg m/s So then after we find the quantites that are missing, we can plug in these numbers above to find the change in velocity 30 kg m/s / 2 kg = 15 m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:51:56 The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second.
......!!!!!!!!...................................
RESPONSE --> I competely understand the problem and I arrived at the answer by using the methodology of using what was provided. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:00:36 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
......!!!!!!!!...................................
RESPONSE --> I think that here the momentum was unchanged. The change in impulse was 30 kg m/s and the time interval was 3 seconds One quantity was before velocity and one is after the momentum is added. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:02:07 Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second. Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second. The two quantities m * `dv and Fnet * `dt are identical.
......!!!!!!!!...................................
RESPONSE --> Well, I guess I mixed up the quantity in the first that was being asked. But yes, the two quantites are the same. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:12:15 `q003. The quantity m * v is called the momentum of the object. The quantity Fnet * `dt is called the impulse of the net force. The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
......!!!!!!!!...................................
RESPONSE --> 2000 N * 1.5 seconds = 3000 kg m/s is the impulse of the force. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:12:58 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:17:34 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
......!!!!!!!!...................................
RESPONSE --> 3000 kg m/s / 1200 kg = 2.5 m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:18:30 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:30:05 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
......!!!!!!!!...................................
RESPONSE --> When I tried to solve this problem, I got 10 m/s as the change in velocity using the imp-mom theorem. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:34:07 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
......!!!!!!!!...................................
RESPONSE --> I missed the problem because I didn't plug in the right numbers in the proper place. self critique assessment: 2
.................................................
ڳƖզn| assignment #016 016. Projectiles Physics II 11-05-2008
......!!!!!!!!...................................
17:43:16 11-05-2008 17:43:16 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:50:02 11-05-2008 17:50:02 The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:50:28 11-05-2008 17:50:28 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:51:43 11-05-2008 17:51:43 As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:57:05 11-05-2008 17:57:05 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
18:03:20 11-05-2008 18:03:20 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
18:11:25 11-05-2008 18:11:25 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
18:14:48 11-05-2008 18:14:48 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
......!!!!!!!!...................................
NOTES ------->
.................................................ζ®泂
assignment #016 016. Projectiles Physics II 11-05-2008
......!!!!!!!!...................................
19:28:06 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
......!!!!!!!!...................................
RESPONSE --> to find out how far the object dropped, we first need to find out how much it travels under the influence of gravity displacement = 1/2 a t^2 t^2 = sqrt of .408 seconds so the object takes 0.638 seconds to fall 2 meters confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:44:18 The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second.
......!!!!!!!!...................................
RESPONSE --> i didn't round the seconds i understood the problem self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:44:51 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
......!!!!!!!!...................................
RESPONSE --> 12 meters * .64 seconds = 7.68 meters confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:45:44 As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
19:58:10 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
......!!!!!!!!...................................
RESPONSE --> as before we use the uniform acc. motion formulas to solve for the quantities that are not provided t^2 = sqrt (2* 1.5 m) / 9.8 m/s^2 t^2 = .31 t = .553 seconds so the object will travel 6 meters, we measure that to the .553 seconds and get...... 6 meters/sec * .54 seconds = 3.24 meters being traveled by the object confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:58:31 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:07:39 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
......!!!!!!!!...................................
RESPONSE --> we use the displacement formulas again and solve for t = sqrt (2* 4m) / 9.8m/s^2 t = 0.9 seconds horizontal displacement 32 meters / .9 sec = 35 m/s final velocity 9.8 m/s^2 * .9 sec = 8.8 m/s since the object would be going in a downward position, it would be safe to say that the horizontal velocity does not change confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:09:32 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
"