Velocities after totally elastic collision

In summary, the problem involves two marbles with different masses and velocities colliding in a one-dimensional space. The conservation of momentum and energy equations can be used to solve for the final velocities of the marbles after the collision. If the collision is inelastic, some kinetic energy will be converted into other forms of energy, while in an elastic collision, both momentum and kinetic energy are conserved.
  • #1
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I found this problem on a website. Is it right? (BTW, this is not homework, but I am just curious if I am thinking right about it.)
A marble with a mass of 2 grams moves to the left with a velocity of 2 m/s when it collides with a 3 gram marble moving in the opposite direction with a velocity of 2 m/s; if the first marble has a velocity of 1.5 m/s to the right after the collision, determine the velocity of the second marble after the collision.

I thought it could be solved using the conservation of momentum:

momentum before = momentum after
p.moving right + p.moving left = p.moving left + p.moving right

3(2) + 2(-2) = 3(v1') + 2(+1.5)

solving for v1' gives -0.3333... I think and ignoring sig.figs.

But then I checked using the conservation of energy. It didn't check.
So I guess one of the velocities is wrong. Let's say the velocity of the first marble after the collision is unknown. Then what would v1' and v2' be?
Someone else on that website said v2' would be 2.333..., but then that doesn't seem right. Why would the larger mass have more energy aft.than before?
I don't have my books and I forgot the formulas. I tried deriving, but I am not sure how to deal with the signs after equating the momentum factors in the difference of two squares. In fact I don't think I have seen this derivation for probably ten years - I couldn't even find it on the web.
Finally, could it be that the collision is not totally elastic? How would that work? I guess then the velocity of the 3 gram after could be 1.5, but not sure. Idk, brain fried. I stayed awake trying to figure it out last night. I guess what I really want to know is what do you get for velocities when it is a totally elastic collision?
 
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  • #2
Google "1d elastic collision equation"
 
  • #3
The difficulty is that you have set up an impossible situation. If given that the first marble has mass 2 g and velocity -2 m/s, the second marble has mass 4 g and velocity 2 m/s, you cannot simply assign an arbitrary velocity to one marble after the collision.

Before the collision, the total momentum is 2(-2)+ 3(2)= 2 g m/s and the total kinetic energy is (1/2)(2)(4)+ (1/2)(3)(4)= 10 kg m^2/s^2. Assigning velocities v1 and v2 after the collision we have the two equations 2(v1)+ 3(v2)= 2 and v1^2+ (3/2)v2^2= 10 to solve for both v1 and v2.
 
  • #4
The original statement of the problem did not specify that the collision is elastic. You correctly used conservation of momentum to find the final velocity of the second marble. As you noticed, kinetic energy is not conserved. So, the collision is indeed inelastic. That's ok and, in fact, that's usually the case for everyday objects.

If you start with the same setup but impose the condition that the collision is elastic, then you are not free to specify the final velocity of either marble. The final velocities are determined by conservation of momentum and conservation of kinetic energy. You can use the equations for a 1d elastic collision to find the final velocities of both marbles (or just set up the conservation equations yourself and work through the algebra).

If you are bothered by why kinetic energy in not conserved in an inelastic collision, it's because some of the kinetic energy of the marbles is converted into other forms of energy such as heat and sound. If you keep track of all the changes in all of the different forms of energy, it would turn out that the total energy is conserved.
 
  • #5


I would say that your approach using the conservation of momentum is correct. The equation you used is a valid way to solve for the velocities after a totally elastic collision. However, it is possible that there are some errors in the calculations or assumptions made, which could explain why the result did not match the conservation of energy. It is always important to double check your work and make sure all variables are accounted for in the equations.

Regarding the velocities of the marbles after the collision, it is correct that the velocity of the first marble (2 grams) would be -0.3333... m/s and the velocity of the second marble (3 grams) would be 2.333... m/s. This may seem counterintuitive, as the larger mass would have a higher velocity, but this is due to the conservation of momentum, which takes into account both the mass and velocity of the objects.

If the collision is not totally elastic, then the velocities after the collision would be different. In a totally elastic collision, the kinetic energy is conserved, meaning that the total energy before the collision is equal to the total energy after the collision. In a partially elastic collision, some of the kinetic energy is lost as heat or sound, resulting in a lower total energy after the collision. This could explain why the conservation of energy did not hold in your calculations.

In summary, your approach to solving the problem using the conservation of momentum is correct, but it is important to double check your work and make sure all variables are accounted for. Additionally, if the collision is not totally elastic, the velocities after the collision will be different. I hope this helps clarify things for you.
 

1. What is a totally elastic collision?

A totally elastic collision is a type of collision between two objects where the total kinetic energy remains the same before and after the collision. This means that the objects involved do not lose any energy to heat, sound, or deformation during the collision.

2. How do you calculate the velocities after a totally elastic collision?

The velocities after a totally elastic collision can be calculated using the conservation of momentum and conservation of kinetic energy equations. These equations take into account the masses and velocities of the objects before and after the collision.

3. Are there any real-life examples of totally elastic collisions?

Yes, there are many real-life examples of totally elastic collisions. Some examples include billiard balls colliding on a pool table, rubber balls bouncing off each other, and atoms colliding in a gas.

4. How is a totally elastic collision different from an inelastic collision?

In a totally elastic collision, the total kinetic energy of the system remains the same before and after the collision. In contrast, in an inelastic collision, some of the kinetic energy is lost to other forms of energy, such as heat or sound.

5. Can a totally elastic collision occur in a vacuum?

Yes, a totally elastic collision can occur in a vacuum. The presence or absence of air or other particles does not affect the conservation of momentum and kinetic energy principles, which govern the behavior of totally elastic collisions.

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