Rocket fires its engine, force/acceleration

[amd123]

Homework Statement

A small space probe of mass 235 kg is traveling at 85.0 m/s. It fires its main engine at an angle of 63.00 to its original direction of travel. The engine produces a thrust of 12,000 N and there is enough fuel on board for a 10.0 s burn. What distance will it have traveled 3.00 hrs after the engine shuts down? Assume no mass loss due to firing the engine.

In which direction is the rocket travelling?

Homework Equations

I'm having trouble understanding what the free body diagram for this would look like.

The Attempt at a Solution

http://img851.imageshack.us/img851/1162/66673407.png

 

[PeterO]

Homework Statement

A small space probe of mass 235 kg is traveling at 85.0 m/s. It fires its main engine at an angle of 63.00 to its original direction of travel. The engine produces a thrust of 12,000 N and there is enough fuel on board for a 10.0 s burn. What distance will it have traveled 3.00 hrs after the engine shuts down? Assume no mass loss due to firing the engine.

Homework Equations

I'm having trouble understanding what the free body diagram for this would look like.

The Attempt at a Solution

http://img851.imageshack.us/img851/1162/66673407.png

One thing I think you can ignore if the force of gravity - this is a space probe, so will be out away from any significant body in space.

The force of the engine is the only force acting, and given the time of the burn, you can find the impulse imparted - that will give the change in momentum, and thus the change in velocity.
That change in velocity will be added to the original velocity to get the new velocity. Mainain that for 3 hours and you know how far the probe will have gone.

 

[amd123]

One thing I think you can ignore if the force of gravity - this is a space probe, so will be out away from any significant body in space.

The force of the engine is the only force acting, and given the time of the burn, you can find the impulse imparted - that will give the change in momentum, and thus the change in velocity.
That change in velocity will be added to the original velocity to get the new velocity. Mainain that for 3 hours and you know how far the probe will have gone.

I haven't learned momentum yet, can't solve the problem using that method.
Also, what direction is the rocket traveling in?

Fey or Fe?

 

[PeterO]

I haven't learned momentum yet, can't solve the problem using that method.
Also, what direction is the rocket traveling in?

Fey or Fe?

The direction the probe is traveling in is irrelevant. It is only important that the force from the rocket motor is at an angle of 63 degrees to the direction. If it will make you more comfortable, make the original direction straight up the page.

Impulse - momentum is only a neat summary of V = V_o + at and F = ma.

Use Newton's second law [ F = ma] to find the acceleration, then use V = V_o + at to find the change in velocity.

For that change, assume V_o = 0 as you are only calculating the change in velocity, to be later added to the original velocity.

 

[rwisz]

I would first clarify the direction of the rocket's travel. From the given information, it is not explicitly stated in which direction the rocket is traveling. Assuming that the original direction of travel is in the positive x-direction, the rocket would be traveling in the positive y-direction after firing its engine at an angle of 63.00 degrees.

To solve for the distance traveled 3.00 hours after the engine shuts down, we can use the equation d = v0t + 1/2at^2, where d is the distance traveled, v0 is the initial velocity, t is the time, and a is the acceleration. In this case, the initial velocity, v0, is 85.0 m/s in the positive x-direction and the acceleration, a, is the net force divided by the mass of the rocket, which can be calculated using Newton's second law, F = ma. Since there is no mass loss due to firing the engine, the mass of the rocket remains 235 kg.

Using trigonometry, we can find the components of the thrust force in the x and y directions. The x-component is Tcos(63.00) and the y-component is Tsin(63.00), where T is the thrust force of 12,000 N. Therefore, the net force in the x-direction is Tcos(63.00) - ma and the net force in the y-direction is Tsin(63.00). Plugging these values into F = ma, we can solve for the acceleration in the x-direction, which is 3.13 m/s^2, and the acceleration in the y-direction, which is 11.9 m/s^2.

Now, we can plug these values into the equation d = v0t + 1/2at^2 to solve for the distance traveled after 3.00 hours. In the x-direction, the distance traveled would be 3.00 hours * 85.0 m/s + 1/2 * 3.13 m/s^2 * (3.00 hours)^2 = 763.5 m. In the y-direction, the distance traveled would be 1/2 * 11.9 m/s^2 * (3.00 hours)^2 = 53.6 m. Therefore, the total distance traveled after 3.00 hours would be the square root of (763.5