Perfectly Elastic Collision (Easy, just confusing)

In summary, the problem is to determine the final velocities, v1 and v2, of two particles in a perfectly elastic collision, with particle 1 having a mass of 2m and an initial velocity directed to the right, and particle 2 having a mass of m and being at rest. This can be solved using the principles of conservation of momentum and energy, resulting in two equations with two variables. The final velocities can be expressed in the form of (constant)v, indicating the ratio of the initial velocity that each particle was left with. However, the solution may involve square roots and appear more complex than expected.
  • #1
ForrestFire
4
0

Homework Statement



Let two particles of collide in a perfectly elastic collision. Particle 1 has a mass of 2m, while particle 2 has a mass of m. Particle 1 has initial velocity v, directed to the right, while particle 2 is at rest. What are the final velocities, v1 and v2, of particle 1 and particle 2?

Homework Equations



Conservation of Momentum

m1v1+m2v2=m1v3+m2v4

Conservation of Energy

(1/2)mv1^2+(1/2)mv2^2=(1/2)mv3^2+(1/2)mv4^2

The Attempt at a Solution



I'm not sure what to do. I've been messing around with stuff, but I can't seem to get the right answer, and I'm running out of answers on my online homework. I know you have to solve in one equation, and then substitute to the other in order to get down to a workable amount of variables, but I'm not entirely sure how. I"m pretty sure the answer has to be in the form of (constant)v, to show the ratio of the initial velocity that each object was left with, and I just don't know how to whittle it down to that.
 
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  • #2
v1 and v2 are already given. You have two equations in two variables. Solve it. To make it simpler you can also use the equation from restitution instead of energy. :wink:
 
  • #3
See, I thought it would be straightforward, but I'm getting a very weird answer, with a lot of square roots and such. My answer for V4, which is the velocity of the second object after the collision, was:

sqrt(v)*sqrt(sqrt(v^2-2)+v)

When it should be some nice number like 1/4v.

I don't understand what I'm doing :-\
 

Related to Perfectly Elastic Collision (Easy, just confusing)

What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In other words, there is no loss of energy during the collision.

Can you give an example of a perfectly elastic collision?

One example of a perfectly elastic collision is when two billiard balls collide on a pool table. The total kinetic energy of the two balls before the collision is equal to the total kinetic energy after the collision, as there is no loss of energy during the collision.

How is a perfectly elastic collision different from an inelastic collision?

In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. This means that the total kinetic energy after the collision is less than the total kinetic energy before the collision. In a perfectly elastic collision, there is no loss of kinetic energy.

What are the conditions for a perfectly elastic collision to occur?

In order for a collision to be perfectly elastic, the objects involved must be elastic, meaning they can be compressed or stretched without permanently deforming. The collision must also occur without any external forces acting on the objects, such as friction or air resistance.

Why is the concept of perfectly elastic collisions important in physics?

The concept of perfectly elastic collisions is important in physics because it helps us understand the conservation of energy and momentum in collisions. It also allows us to make predictions about the behavior of objects in collisions, which can be applied in various fields, such as engineering and sports.

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