Question

You’re flying your 10,000-kg Scooty Puff Pro spaceship at a constant speed of 20 km/s in the positive x-direction. All of the sudden it malfunctions and many of the thrusters are stuck on. The main thruster is stuck on producing a 800 kN force in the positive x-direction. Two of the small maneuvering thrusters are also stuck on, one producing a 100 kN force in a direction 30° from the positive y-direction towards the negative x-direction, and the other is producing a 150 kN force in a direction 10° from the negative y-direction towards the negative x-direction. After 2 minutes of this the two maneuvering thrusters shut off and the main thruster decreases in magnitude to 200 kN, while still applying the force in the positive x-direction. It takes an additional 6 minutes to fix the problem and turn all of them off. How far and in what direction would you have to travel to get to where you would have been if the malfunction had not occurred.

Answer 1

Mass of the Scooty = 10000 Kg

Speed of the Scooty = 20 km/s = 2010³ m/s in the positive x-direction.

Force in the positive x-direction = 800 kN = 80010³ N

Force in a direction 30° from the positive y-direction towards the negative x-direction = 100 kN = 10010³ N

Force in a direction 10° from the negative y-direction towards the negative x-direction = 150 kN = 15010³ N.

Time t = 2 m = 260 = 120 s

Time = 6 = 660 = 360 s

Solution:

You have to find the net force on the ship, in each situation.

So you will come up with a net force, in a certain direction.

From Newton's second law of motion we know that F = ma, to get the acceleration

F - Force

m - Mass

a - Acceleration

a = F/m = net force / 10000

Then S = V_i t + (1/2)at²

To calculate the magnitude of distance

S - Displacement in meter

V_i - Initial velocity

t - Time

You started with 20 km/s speed in positive x direction, so that will be added to the first calculation.

For each force given, there will be a x component and a y component.

First situation:
x component of the force = 800 KN = 80010³ N

y component of the force = 0

100 kN force in a direction 30° from the positive y-direction towards the negative x-direction

x component of the force = -100cos 30° kN = 1000.86610³ N = 86.610³ N

y component of the force = 100 sin 30° kN = 1000.510³ N = 5010³ N

150 kN force in a direction 10° from the negative y-direction towards the negative x-direction

x component = -150 cos 10° kN = 1500.98510³ N = 147.7210³ N

y component = -150 sin 10° kN = - 1500.173610³ N = - 26.04710³ N

This will give you a net force for the two directions

Once you calculate acceleration, you can apply the following equation

S = V_it + (1/2)at² for each of the two directions, to get the final magnitude for x and y for this section of thrust.

For the next section of thrust, you also need the speed in each direction. That is done using the following equation

V_f = V_i + at

V_f - Final velocity

So after doing all this, you have the x and y position of the craft, and its speed in each direction.

You then repeat this procedure for each of the other two thrust sections, always paying attention to the units as well (we are using Newton, meter and second). So you need to use conversion factors. Now we know the final distance in each direction the ship has gone.

To compare this with the position that the ship would have reached while coasting at its original speed and direction of 20 km/s in the positive x direction for this amount of time.. So we must calculate this, again paying attention to the units. Once you find this, then you need to find the vector that would take you back from the malfunction position to this resulting coasting position.

To find this magnitude, you subtract the malfunction x from the coasting x, to get the x distance. The y distance is just whatever the malfunction y is, as the coasting rocket had no y component.

Use the following formula to get the distance

D = (x² + y²)

Use arcsin (ydist / total dist) to get direction.