Momentum: Military Test?

Question

In a military test, a 575 kg unmanned spy plane is traveling north at an altitude of 2300 m and a speed of 500 m/s. It is intercepted by a 1360 kg rocket traveling east at 775 m/s. If the rocket and the spy plane become enmeshed in a tangled mess, where, relative to the point of impact, do they hit the ground?

_______ km
_______ angle north of east


Thanks

The angle is correct at 15.26 degrees but none of the distance seem to be correct.

I have also tried, instead of using distance = velocity * time to use d = Vit + 1/2at^2 where Vi = 693.281, t = 21.665, and a = 9.8. When I do that, I get 17.319 km, which is also wrong.

I am having trouble finding the problem. Any suggestions?


Thanks

Answer 1

2300 meters below where they "enmeshed"

Answer 2

Remember momentum is a vector and you are adding two momenta that are at right angles. So total momentum = sqrt((575*500)^2 + (1360*775)^2)) = 1092507 kg-m/s, and the angle is arctan(575*500/1360*775) = 15.26 deg N of E. Velocity is total momentum/total mass = 564.6 m/s; fall time = sqrt(2*2300m/g) = 21.658 s; distance = v*t = 12228 m.
Answer 3 is correct in pointing out that the error in your followup question was to involve horizontal velocity and vertical acceleration in the same equation to solve for fall time. However answer 3 erred in adding the two momenta; the sum is the vector sum, not the simple sum, of the individual momenta, so the velocity and distance are wrong.

Answer 3

I get 15 km 15.26 degrees north of east.

I'll try to explain why in a moment.

EDIT:
To get the angle, we need to figure out the momentum of each object. This is pretty easy, since momentum = mass*velocity.

momentum of plane=
575kg * 500m/s = 287500kg*m/s
momentum of rocket =
1360kg * 775m/s = 1054000kg*m/s

This means that the new object created by the rocket and plane will have an eastward momentum of 1054000kg*m/s and a northward momentum of 2875000kg*m/s. If we set these two vectors at right angles to one another and draw a line between the end points, we can make a triangle. The points of the triangle will be 1) the point of collision 2) 1054000 units east of the collision 3) 2875000 units north of point #2. The new object's direction will be the angle created at the point of collision. Since we know the value of the opposite and adjacent sides, we can use trigonometry to determine the angle:

tangent = opposite / adjacent

tanA = 287500 / 1054000 = 0.2728
Using inverse functioning:
A = 15.26

So we know where the object is headed, but we need to figure out how far it goes. To do this, we need to know how how fast it traveled and how long it traveled.

Determining the object's velocity is easy. Because momentum is always conserved, the total momentum after the collision will be the same as the total momentum before the collision. The total momentum can be determined by adding up the momentum of all the objects in the system:

Total momentum = 287500 + 1054000 = 1341500kg*m/s

To get the new velocity, we just divide the total momentum by the new total weight:

V = 1341500kg*m/s / (1360 + 575)kg = 1341500 / 1935 = 693.28m/s

Figuring out how long it traveled is a little trickier, but not too bad. We can assume that it started falling as soon as it was hit. We can also assume that it stopped moving when it hit the ground. Since the object is accelerating downward at a constant rate (9.81m/s/s due to gravity), we can use a formula to figure out how long it took for it to hit the ground;

D = Vo * T + 0.5 * A * T^2 where:

D is the object's displacement (distance traveled).
D = 2300, since the object was 2300 meters off the ground.
Vo is the original velocity.
Vo = 0, since the object has no downward movement, initially.
A is the rate of acceleration.
A = 9.81. Acceleration due to gravity on Earth is always 9.81m/s/s.
T is the time over which the acceleration occured. This is what we're trying to figure out! So...

2300 = 0 * T + 0.5 * 9.81 * T^2...simplify...
2300 = 4.905 * T^2...solve for T^2...
T^2 = 468.91...solve for T...
T = sqrt468.91 = 21.65

So, our object was in the air for 21.65 seconds. Since we already figured out it's velocity (693.28m/s), determining distance is a snap.

d = v * t, where
d is distance, v is velocity, and t is time.

D = 693.28m/s * 21.65s = 15009.5 meters = 15.0095 kilometers.

Hope this helped!

EDIT #2:
I could be wrong, but I think you're getting confused at the end of your equation. d=v*t is correct to determine how far the rocket traveled, but you have to figure out "t" before you can figure out "d". For this equation, saying d=v*t is the same as saying, "The distance traveled by the object is equal to 693.28m/s multiplied by the amount of time that it takes for the object to hit the ground." You have to figure out how long it took to hit the ground.

You're correct in using (d=Vit + 1/2at^2), but you're using it in the wrong way. We use that equation to determine how long it takes for the object to hit the ground from the height that it was at when the collision occured. Remember, for this equation, we're only concerned about the object's vertical movement, moving DOWNWARD towards the earth. For all intents and purposes, this equation would be the same for any object dropped from the same height.

d = 2300, since that is the altitude from which it began its fall DOWNWARD
Vi = 0, because that object had no DOWNWARD velocity when it began to fall
a=9.81 (or 9.8 if you like), the acceleration DOWNWARD due to gravity

So...
2300 = 0t + 1/2 * 9.8t^2
Since "0t" is equal to 0, we can eliminate it by subtracting 0 from each side...
2300 = 1/2 * 9.8t^2
Multiply both sides by 2...
4600 = 9.8t^2
Divide both sides by 9.8...
469.39 = t^2
Take the square root of both sides
21.67 = t

So, the object traveled for 21.67 seconds before hitting the ground. (This is a little different from the answer above because we used 9.8 for acceleration instead of 9.81.) Since it traveled at a HORIZONTAL velocity of 693.28m/s for 21.67sec, it traveled 15023.38 meters, or 15.023 km.

The important thing is not to get the horizontal velocity confused with the vertical velocity. Remember, speed is just a quantity, but velocity is a vector. It must have both a speed AND a direction.

Feel free to email me through my profile if you want to.

Answer 4

Let v=500m/s(1*i+0*j) – velocity vector of the plane!
Let u=775m/s(0*i+1*j) – velocity vector of missile!
Momentum conservation law: M*v+m*u=(M+m)*w, being M=575kg, m=1360kg, w is their velocity vector after impact to be found.
Falling from height h=2300m they hit our heads in t seconds, i.e. h=g*t^2/2, g=9.8m/s^2.
So t=sqrt(2h/g), while during time t they go on flying with velocity of w // to the ground.
If S is their path vector from impact point, then S=w*t, thus
S=Sx*i+Sy*j=((M*v+m*u)/(M+m))*sqrt(2h/g)
abs(S)=sqrt(Sx*Sx+Sy*Sy), a=atan(Sy/Sx) – don’t be a lazy boy & do calc ur self please!
OK?

Answer 5

11.842 km

15.257 deg N of E