Linear Momentum: Rocket Explosion

[smtcenterchic]

Homework Statement

A 975-kg two-stage rocket is traveling at a speed of 5.80 x 10³ m/s with respect to Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a relative speed (relative to each other) of 2.20 x 10³ m/s along the original line of motion. What is the speed and direction of each section (relative to earth) after the explosion?

Homework Equations

$$m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b$$
where m = 975 kg, v = 5.80 x 10³ m/s relative to earth, and m_a = m_b = 487.5 kg

The Attempt at a Solution

At first I thought that special relativity and the addition of velocities could be used to find the speed of the rockets relative to earth. Then I realized that linear momentum is in chapter 7, while relativity was in chapter 27. Therefore, that idea didn't make sense. One student suggested the following:

$$m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b$$
(975)(5.80 x 10³) = (487.5)(2.20 x 10³)(v_a) + (487.5)(v_b)

His logic was that one could then set v_a and v_b equal to each other, since you had accounted for the difference in velocities relative to each other. I didn't think this made sense either, but I'm at a loss for any other way to approach the problem. Any and all help is welcomed and appreciated.

 

[anathema]

Dear student,

Thank you for your post. You are correct in thinking that using special relativity and the addition of velocities is not the appropriate approach for this problem. Instead, we can use the conservation of momentum to solve for the final velocities of each section of the rocket.

As you correctly noted, we can set up the equation m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b, where m = 975 kg, v = 5.80 x 103 m/s relative to earth, and ma = mb = 487.5 kg. This equation represents the conservation of momentum, which states that the total momentum before the explosion is equal to the total momentum after the explosion.

We know that the initial momentum of the rocket before the explosion is 975 kg * 5.80 x 103 m/s = 5.655 x 106 kg*m/s. After the explosion, the two sections of the rocket will have equal but opposite momentums, since they are moving in opposite directions. This means that each section will have a momentum of 2.8275 x 106 kg*m/s.

Now, we can plug this value into our equation and solve for the final velocities. We get:

5.655 x 106 kg*m/s = (487.5 kg)(va) + (487.5 kg)(vb)

Solving for va and vb, we get:

va = 11.6 m/s and vb = -11.6 m/s

This means that one section of the rocket will be moving at a speed of 11.6 m/s in the same direction as the original rocket, and the other section will be moving at a speed of 11.6 m/s in the opposite direction.

I hope this helps clear up any confusion. Let me know if you have any further questions. Good luck with your studies!