# Elastic collision of two asteroids

• Arejang
In summary: A and B are at rest at the beginning. I'm getting a bit lost here. Can you walk me through it again?In summary, two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing blow. Asteroid A, which was initially traveling at 40.0 m/s, is deflected 30.0\circ from its original direction, while asteroid B travels at 45.0 \circ to the original direction of A. After the collision, asteroid A's kinetic energy is conserved, while asteroid B's momentum is conserved.
Arejang

## Homework Statement

Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing blow. Asteroid A, which was initially traveling at 40.0 m/s, is deflected 30.0$$\circ$$ from its original direction, while asteroid B travels at 45.0 $$\circ$$ to the original direction of A

Find the speed of asteroid A and asteroid B after the collision.

What fraction of the original kinetic energy of asteroid A dissipates during this collision?

## Homework Equations

Conservation of Kinetic energy
$$1/2m_{a}v_{a1x}^{2}+1/2m_{b}v_{b1x}^{2}=1/2m_{a}v_{a2x}^{2}+1/2m_{b}v_{b2x}^{2}$$

Conservation of momentum
$$m_{a}v_{a1x}+m_{b}v_{b1x}=m_{a}v_{a2x}+m_{b}v_{b2x}$$

## The Attempt at a Solution

I really want to assume that asteroid B is initially at rest, but since it's not stated; I'm not sure I should do so. But for this problem, I feel like it may be the only way to solve this problem so that's what I will assume, please correct me if I'm wrong. Anyway, it seems like you would have to solve for one of the final velocities through substitution. Since the masses of the two asteroids are the same, we can factor them out of both equations, leaving us with only the velocities to worry about.

So our Kinetic Energy Conservation formula looks like this:

$$1/2v_{a1x}^{2}+1/2v_{b1x}^{2}=1/2v_{a2x}^{2}+1/2_{b2x}^{2}$$

I then proceeded to solve for $$v_{a2x}$$ and got

$$v_{a2x}=\sqrt{v_{b2x}^{2}-v_{a1x}^{2}}$$

This expression of $$v_{a2x}$$ I subbed into the momentum conservation formula, getting:

$$v_{a1x}=v_{b2x}+\sqrt{v_{b2x}^{2}-v_{a1x}^{2}$$

It seems logical that I solve for the velocity of asteroid B, then plug that in the kinetics formula then solve for A, but I'm not certain how to integrate the angle of the collisions into each formula. In fact, I'm actually not sure if I even approached this problem correctly. But please let me know where I need to go with this.

Oh and to Doc. Al if you're reading this, it turned out my previous problem was correct and that the online problem had the wrong answer input as the correct one. Consequently, everyone got full credit for the problem.

Bump! Can anyone help me please?

I think you'd have to assume asteroid B is at rest initially for it to be possible since, in a nutshell, you'd absolutely have to know the total energy of the system initially, and that would be mighty hard to do without enough info

For my mental image I picture the asteroid A initially traveling down the positive x axis, and colliding with B at the origin, then A is traveling 30 degrees "north" of the x-axis and B is 45 degrees "south" of the x axis

So the initial momentum of the system is just the mass of A times its initial velocity, it's all in the +x direction

Afterwards, remember that the x-component of the momentum needs to be conserved, and the y components need to cancel.

The angles come into play because you use trig to find the x and y components of the velocity

The kinetic energy dissipated question is a pseudo-trick question. coughcoughelasticcolissiondefinitioncough

EDIT: Well I should be careful with that. In an elastic collision no energy is dissipated into other forms like heat or anything. The kinetic energy of asteroid A may, and I think will have to be, different than it was initially. But it wasn't "dissipated" >_>

The kinetic energy of the SYSTEM is the same though, so uh, interpret that how you will ^_^

Last edited:
blochwave said:
The kinetic energy dissipated question is a pseudo-trick question. coughcoughelasticcolissiondefinitioncough

LOL...hee hee...:rofl:

So the initial momentum of the system is just the mass of A times its initial velocity, it's all in the +x direction

So how would I be able to utilize a mass that's not given?

Well they're the same so there should be cancellation

In the x-direction:

The momentum before is going to be mass times velocity in the x direction(which with the way I drew it is just its speed) and the momentum after is going to be the mass times velocity in the x direction of A and B, those velocity components can be found using trig and those angles. All the masses cancel

Same thing in the y direction except the initial momentum is 0

## 1. What is an elastic collision?

An elastic collision is a type of collision in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

## 2. How do two asteroids exhibit an elastic collision?

In an elastic collision of two asteroids, both asteroids will experience a change in velocity and direction, but the total kinetic energy of the system will remain constant.

## 3. What factors determine the outcome of an elastic collision between two asteroids?

The outcome of an elastic collision between two asteroids is determined by the mass, velocity, and angle of approach of each asteroid. These factors will determine the change in velocity and direction of each asteroid after the collision.

## 4. What is the significance of studying elastic collisions of two asteroids?

Studying elastic collisions of two asteroids can provide valuable insight into the dynamics of celestial bodies, as well as help us understand how objects in space interact with each other. This knowledge can be applied to better understand the formation and evolution of our solar system.

## 5. Can an elastic collision between two asteroids result in a completely elastic collision?

No, an elastic collision between two asteroids can never be completely elastic. Some kinetic energy will always be lost due to factors such as gravitational attraction and non-perfectly elastic materials. However, in some cases, the loss of kinetic energy may be negligible and can be considered as a nearly elastic collision.

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