Fully perfect elastic collision

In summary, the conversation revolved around deriving the formula for a perfectly elastic collision and the validity of the equations used. The expert summarizer notes that the final equation found is a simple and efficient way to derive the new velocities in a perfectly elastic collision. The conversation also touches on the possibility of no velocity change in an elastic collision and the use of Cramer's rule to solve equations quickly.
  • #1
madah12
326
1

Homework Statement


ok so I was trying to figure out how to derive the formula for true elastic collision and I came up with the following but I want to make sure I didn't make a mistake.

Homework Equations


(K1+K2)before=(K1+K2)after
1/2 m1 v1^2 + 1/2 m2 u1^2 = 1/2 m1 v2^2 + 1/2 m2 u2^2

m1v1 + m2u1=m1v2 + m2 u2

The Attempt at a Solution


(1/2 m1 v1^2 + 1/2 m2 u1^2 = 1/2 m1 v2^2 + 1/2 m2 u2^2)*2

m1v1^2+m2 u1^2=m1v2^2+m2 u2^2

m1(v1^2-v2^2)=m2(u2^2-u1^2) *(1)

m1v1 +m2u1=m1v2 + m2 u2

m1(v1-v2)=m2(u2-u1) *(2)
(1)/(2)
=
v1+v2=u1+u2
u2=v1+v2-u1

m1v1 + m2u1=m1v2 + m2 (v1+v2-u1)

m1v1 + m2u1=m1v2 + m2 v1+m2 v2 - m2 u1

2m2u1 +m1v1 -m2v1 =v2 (m1+m2)

v2= (2m2u1 +v1(m1-m2)) / (m1+m2)

EDIT: the point of this is that my friend told me that the derivation will take more than 3 pages so I am thinking there must be something incorrect here or I ate some lines lol.
 
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  • #2
This is a very nice and simple way to derive the new velocities in a perfectly elastic collision. Just a note: The equations 1 and 2 allow the possibility that the velocities do not change, so nothing happens (v1=v2, u1=u2 is also a possible solution). Congratulation if you found this out by yourself!

ehild
 
  • #3
ok but isn't it possible in an elastic collision that after the collision the objects velocities don't change? so the equation is still valid right...?
 
  • #4
madah12 said:
EDIT: the point of this is that my friend told me that the derivation will take more than 3 pages so I am thinking there must be something incorrect here or I ate some lines lol.
Heh, reminds me of when I was an undergrad. My friends would turn in three or four pages of algebra per problem while I turned in one. Often after I first solved a problem, I'd go back and try figure out how to reduce it to the fewest number of steps so I could follow the logic. Doing this helped develop better algebra skills, plus I learned tricks and techniques that allowed me to solve problems more efficiently later on. (Plus when I was writing up a problem to turn in, I was lazy and wanted to write down just enough so the grader could follow what I was doing.)

Once you reach the point where you have the two equations

[tex]\begin{align*}
m_1 v_1 + m_2 u_1 &= m_1 v_2 + m_2 u_2 \\
v_1 - u_1 &= -v_2 + u_2
\end{align*}
[/tex]

you can solve them quickly using Cramer's rule. For example, for v2, you'd get

[tex]v_2 = \frac{\begin{vmatrix}m_1 v_1 + m_2 u_1 & m_2 \\ v_1-u_1 & 1 \end{vmatrix}}{\begin{vmatrix}m_1 & m_2 \\ -1 & 1\end{vmatrix}} = \frac{m_1 v_1 + m_2 u_1 - m_2(v_1-u_1)}{m_1 + m_2} = \frac{(m_1-m_2)v_1 + 2 m_2 u_2}{m_1+m_2}[/tex]

It doesn't make much of a difference for this problem, but occasionally, it can help you avoid some tedious algebra.
madah12 said:
ok but isn't it possible in an elastic collision that after the collision the objects velocities don't change? so the equation is still valid right...?
If the objects collide, their velocities will change, and when you divided equation (1) by equation (2), you assumed this, otherwise you'd be dividing by 0.

The original equations, however, admit the possibility that the objects don't collide and their velocities remain unchanged, but that's not the case you're interested in anyway.
 
  • #5


Your attempt at deriving the formula for a perfectly elastic collision is correct. However, it is important to note that this formula is only valid for a perfectly elastic collision, where there is no loss of kinetic energy during the collision. In reality, most collisions are not perfectly elastic and some energy is lost in the form of heat, sound, or deformation.

To verify that your derivation is correct, you can substitute in some values for the masses and velocities and see if the equation holds true. You can also try to derive the formula using other methods, such as conservation of momentum and conservation of kinetic energy, to further confirm your results.

Overall, your understanding and approach to deriving this formula is sound. Keep up the good work!
 

1. What is a fully perfect elastic collision?

A fully perfect elastic collision is a type of collision between two objects where there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision is the same.

2. How does a fully perfect elastic collision differ 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. In a fully perfect elastic collision, all of the kinetic energy is conserved and remains in the form of kinetic energy.

3. Is it possible for a real-life collision to be fully perfect elastic?

No, in reality, there is always some energy lost due to factors such as friction and heat. However, in some cases, collisions can be very close to fully perfect elastic, such as in the collision of subatomic particles.

4. How is the elasticity of a collision determined?

The elasticity of a collision is determined by calculating the coefficient of restitution, which is the ratio of the relative velocity of separation to the relative velocity of approach. A fully perfect elastic collision would have a coefficient of restitution of 1.

5. What are some applications of fully perfect elastic collisions?

Fully perfect elastic collisions are commonly seen in the movement of subatomic particles, such as in particle accelerators. They are also used in sports equipment, such as in the bouncing of a basketball or the rebound of a golf ball on a club. Additionally, understanding fully perfect elastic collisions is important in fields such as engineering and astrophysics.

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