# Projectile Motion Problem with Momentum

• SprintsMcgee
In summary: So tangent inverse of -500.3/139.5 is the angle which is -74.4 degrees.In summary, two objects, a 575 kg unmanned spy plane and a 1280 kg rocket, collide in a military test. The resulting tangled mess has a velocity of 519.38 m/s and an angle of -74.4 degrees east of north. The final velocity was found using the momentum equations, and the angle was found using the tangent inverse of the y and x components of the velocity.
SprintsMcgee
Problem: In a military test, a 575 kg unmanned spy plane is traveling north at an altitude of 2700m and a speed of 450m/s . It is intercepted by a 1280kg rocket traveling east at 725m/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? Give the direction as an angle east of north.

The relevant equations I used were the momentum equations in the x and y direction which are Px = m1v1 = (m1 + m2)vfx and Py = m2v2 = (m1 + m2)vfy. Then I used the equations to find the mag of Vf and the ø of the equation.

I was able to get the ø to be 74.4 in the east of North direction and then I got the Vf = 519.38. I know that that's the Vi for a projectile motion equation but I cannot figure out what I am doing wrong. I keep getting answers close to 14 km. What mistakes am I making!?

i am currently learning this in class and could double check you if i can see your wrok.

Px = m1v1 = (m1+m2)vfx
= 575(450) = (575+1280)vf => vf = 139.5
Py = m2v2 = (m1+m2)vfy
= -1280(725) = (575+1280)vfy => vf = -500.3

Ptot = √(vfy^2 +vfx^2) = √(139.5^2 + 500.3^2) = 519.38

omega = arctan( y/x) = arctan(-500.3/139.5) = -74.4 degrees.

Thats all I got for now.

We haven't gotten that far yet unfortunately. We just finished up doing problems like two cars crash into each other head on. When i did my work i got an answer that didn't make sense either.

Your answer of fourteen km may be right because the forward momentum of the tangled mess will quickly dispipate and it will fall straight down rather quickly. Wish I could be more help but I am only in high school level physics.

SprintsMcgee said:
omega = arctan( y/x) = arctan(-500.3/139.5) = -74.4 degrees.

What is omega and arctan mean? Opposite/Adjacent?

its omega = tan^-1 (y/x) the y component and the x component of the velocity.

## 1. What is projectile motion?

Projectile motion is a type of motion in which an object is thrown, launched, or dropped and moves along a curved path under the influence of gravity.

## 2. How does momentum affect projectile motion?

Momentum is the product of an object's mass and velocity, and it determines the object's resistance to changes in its motion. In projectile motion, momentum affects the distance and direction the object will travel, as well as how it responds to external forces, such as air resistance.

## 3. How is the trajectory of a projectile calculated?

The trajectory of a projectile can be calculated using the equations of motion, which take into account the initial velocity, angle of launch, and acceleration due to gravity. These equations can be solved for the horizontal and vertical components of displacement, velocity, and time, allowing for the calculation of the projectile's trajectory.

## 4. What factors can affect the trajectory of a projectile?

The trajectory of a projectile can be affected by several factors, including the initial velocity, angle of launch, air resistance, and the presence of external forces. The mass and shape of the projectile can also have an impact on its trajectory.

## 5. How is the conservation of momentum applied to projectile motion problems?

The law of conservation of momentum states that the total momentum of a closed system remains constant. In projectile motion problems, this means that the initial momentum of the object will be equal to the final momentum, taking into account any external forces acting on the object. This principle can be used to solve for unknown variables in projectile motion problems.

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