Like

Report

An engine of the orbital maneuvering system (OMS) on a space shuttle exerts a force of $(26,700 \mathrm{N}) \}$ for 3.90 $\mathrm{s}$ , exhausting a negligible mass of fuel relative to the $95,000-\mathrm{kg}$ mass of the shuttle. (a) What is the impulse of the force for this 3.90 s? (b) What is

the shuttle's change in momentum from this impulse? (c) What is the shutle's change in velocity from this impulse? (d) Why ean't we find the resulting change in the kinetic energy of the shuttle?

(a) $\vec{J}=104130 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} \hat{\mathrm{j}}$

(b) $\Delta \overrightarrow{\mathrm{p}}=104130 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} \hat{\mathrm{j}}$

(c) $\Delta \overrightarrow{\mathrm{v}}=1.096 \mathrm{m} / \mathrm{s} \hat{\mathrm{j}}$

(d) since the change in kinetic energy depends not only on the change in velocity but also on both the initial and final velocity. And since we can't get any of them, therefore we can't get the change in kinetic energy.

You must be signed in to discuss.

Rutgers, The State University of New Jersey

Numerade Educator

Hope College

University of Sheffield

in this question, the orbital maneuvering system off that space shuttle exerts the force off 26,000 and 700 Newtons, pointing upwards that space shuttle has a mass off 95,000 kg on that force is exerted for 3.90 seconds in the first item. Off this question, we have to evaluate what is the impulse produced by that force during this 3.9 seconds? For that, we can use this expression, which tells us that the impulse is given by the force times the interval off time. And let me add that my reference frame for this question is this one. Everything that is pointing upwards is positive India's reference frame. So the impulse is given by the force off 26,000 and 700 Newtons times Delta T, which is 3.90 seconds. This results in an impulse off 104,000 and 130 but noticed that we had at most three significant figures for the time the reform. We must run the final results to treat significant figures. That means that the final result must be approximated to 104,000, and that's it. You can also write that as 1.4 times 10 to the fifth on the units are units off momentum kilograms, meters per second and this is the answer to the first item In the second item. We have to tell what is the variation in the momentum produced by that impulse? Well, the answer is that the variation in the momentum is equal to the impulse, as you can see from this relation so trivially. The answer to the second item is 1.4 times 10 to the fifth kilograms meters per second and that's it. Now we have to evaluate what is the change in the velocity? Well, the changing the velocity can easily be evaluated from the change in the momentum. So take a look at the following. The change in the momentum is equals to the final momentum, minus the initial momentum. The final momentum is given by the mass times. The final velocity on the initial momentum is given by the mass times the initial velocity notice that I'm not including the change in the mass that is produced by the exhaustion off you. Because the question says that the Mass. A few that was used to produce that force is negligible when compared to the mass of the shuttle. So you can consider the mass of the shuttle Toby approximately constant the reform, the changing, the momentum is given by the mass times, the final velocity minus the initial velocity or the mass times the change in the velocity. Therefore, the changing the velocity of the shuttle is given by the change in the momentum divided by its mass. Therefore, it is 1.4 time stand to the fifth, divided by 95,000 kg, and these results in approximately 1.9 m per second. Note that I approximated the final results to three significant figures because we only had three significant figures for the change in the momentum. Now the final item asks us, Why don't we evaluate the variation in the kinetic energy? Well, the reason is the following the variation in the kinetic energy is given by the final kinetic energy, minus the initial kinetic energy. The final kinetic energy is the mass of the Chateau times, its final velocity squared, divided by true. The initial kinetic energy is given by the mass of the shuttle times its initial velocity squared divided by True. Then we can factor out em over to so that the variation in the kinetic energy is given by the final velocity squared minus the initial velocity squared. The reform Delta K is equals to em over troop. We can write that product as VF plus zero times VF minus zero. The reform the variation in the momentum Delta K is given by em over two times VF plus zero times Delta v the reform. As you can see, it depends explicitly on the initial velocity. And we only know the variation in the velocity. So you're not able to evaluate the variation in the kinetic energy because we don't know what waas the initial velocity.

Brazilian Center for Research in Physics