# Failed Rocket Momentum Problem (Kleppner 4-4)

• TimSon
In summary, an instrument carrying rocket explodes at the top of its trajectory and breaks into two pieces, one larger and one smaller. The smaller piece, with a mass of m, returns to the launching station while the larger piece, with a mass of 3m, flies away horizontally. Using the equations for momentum conservation and kinematics, it can be determined that the larger piece lands at a distance of (5/3)L away from the launch point. This assumes that the time of flight is the same for all pieces and neglects air resistance and the Earth's curvature.
TimSon

## Homework Statement

An instrument carrying rocket accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is L. The rocket breaks into two pieces that fly apart horizontally. The larger piece has three times the mass of the smaller piece. To the surprise of the scientist in charge, the smaller piece returns to Earth at the launching station. How far away does the larger piece land? Neglect air resistance and effects due to the Earth's curvature.

## Homework Equations

Momentum conserved
Kinematics equations

## The Attempt at a Solution

Let M4 = initial mass
Let 3m = mass that flies away from the launch site.
Let m = mass that flies towards the launch site.

4m * V(M4) = - m * V(m) + 3m * V(3m)

Since velocity at the top is 0:

M4(0) = - m * V(m) + 3m * V(3m)

m * V(m) = 3m * V(3m)

Therefore V(3m) = V(m)/ 3

Since m travels L horizontally,

L = V(m) * t

therefore V(m) = L/ t

If we plug in V(m) = V(3m) * 3

we get V(3m) = L / (t*3) (since velocities are all horizontal, both take the same amount of time to fall down)

Plugging this into the Kinematics Eqn for distance, we get

D = V(3m) * t - .5 * 9.8 * t^2

D = (L/(3*t)) * t - .5 *9.8 * (L/(3*(v(3m)))^2

I don't know how to get rid of the v(3m).

You seem to be confusing vertical velocities and horizontal velocities.

thanks for the hint. it helped me realize that the initial mass's velocity in the x-direction is not zero at the top and then I solved the problem.

May the solution be D=(5/3)L?

did you do by this logic?

Momentum of small piece + Larger piece = total mass * speed before explosion

Yes, and assuming that the time that the total mass, the small piece, and the large piece are flying through the air is the same, as movement in the x direction is concerned.

Is it all right?

## 1. What is the Failed Rocket Momentum Problem?

The Failed Rocket Momentum Problem, also known as Kleppner 4-4, is a physics problem that involves calculating the final velocity of a rocket that has failed to eject its exhaust gases.

## 2. What are the key concepts involved in solving this problem?

The key concepts involved in solving the Failed Rocket Momentum Problem include conservation of momentum, impulse, and the rocket equation. Conservation of momentum states that the total momentum of a system remains constant, while impulse is the change in momentum over time. The rocket equation relates the change in velocity of a rocket to the mass and velocity of the ejected exhaust gases.

## 3. How do you approach solving this problem?

To solve the Failed Rocket Momentum Problem, you must first draw a diagram of the situation and identify the initial and final velocities of the rocket. Then, you can use the conservation of momentum equation to calculate the final velocity. Next, you can use the rocket equation to find the mass and velocity of the ejected exhaust gases. Finally, you can use the impulse equation to calculate the change in momentum and use it to solve for the final velocity of the rocket.

## 4. What are some common mistakes to avoid when solving this problem?

Some common mistakes to avoid when solving the Failed Rocket Momentum Problem include forgetting to include the mass and velocity of the ejected exhaust gases in the momentum and impulse equations, using incorrect units, and not properly setting up the problem with the correct initial and final velocities.

## 5. Why is the Failed Rocket Momentum Problem important in physics?

The Failed Rocket Momentum Problem is important in physics because it demonstrates the principles of conservation of momentum, impulse, and the rocket equation in a real-world scenario. It also helps to understand the behavior of rockets and their propulsion systems, which is crucial in space exploration and other applications of rocket technology.

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