Elastic collision, conceptual

In summary, two particles collide. Particle 1 has initial velocity v, directed to the right, and particle 2 is initially stationary. If the collision is elastic, the final velocities v1 and v2 of particles 1 and 2 are 2v and v, respectively.
  • #1
ph123
41
0
Let two particles collide. Particle 1 has initial velocity v, directed to the right, and particle 2 is initially stationary. Now assume that the mass of particle 1 is 2m, while the mass of particle 2 remains m. If the collision is elastic, what are the final velocities v1 and v2 of particles 1 and 2?

Express answer in terms of the initial velocity, v.


My conservation of energy equation for the two particles is (I already canceled out the masses):

v1^1 + 0.5v2^2 = v^2

My conservation of momentum equation is (masses already cancelled):

2v1 + v2 = 2v

Thus,

v2 = 2v-2v1

v1 = (2v-v2)/2

I tried plugging these values for v1 and v2 into the conservation of energy formula to try and isolate v for each variable. For v1, I ran into difficulty because of the middle term of the binomial equation, -8vv1.

v1^1 + 0.5v2^2 = v^2
v1^2 + 0.5[(2v-2v1)(2v-2v1)] = v^2
v1^2 + 0.5[4v^2 - 8vv1 + 4v1^2] = v^2
-v^2 = v1^2 - 4vv1 + 2v1^2
3v1^2 - 4vv1 = -v^2

From here I spent like 10 lines trying to rearrange terms to separate v from that middle term (now -4vv1), but am unable to do so. Any ideas? Thanks!
 
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  • #2
Recognize that you have a quadratic equation for v1. Solve it!
 
  • #3
Well, there are two different terms in the quadratic. How can you solve for two terms?
 
  • #4
Actually, there are three terms if you put it into standard form:
[tex]ax^2 + bx + c = 0[/tex]

Can you solve that for x? If so, you can solve for v1 in terms of v.
 
  • #5
  • #6
You lost me. I assume you mean:

3v1^2 - 4vv1 + v^2 = 0

[-b +/- sqrt(b^2 - 4ac)]/2a = v1

[4v +/- sqrt([-4v]^2 - 4(3v)(v^2))]/2(3v)

[4v +/- sqrt(16v^2 - 12v^3)]/6v

But now I'm left with

(2/3)v +/- [sqrt(16v^2 - 12v^3)/6v]

I can't combine terms in the square root because they have different exponents, so this doesn't seem like it makes much sense.
 
  • #7
ph123 said:
You lost me. I assume you mean:

3v1^2 - 4vv1 + v^2 = 0

[-b +/- sqrt(b^2 - 4ac)]/2a = v1

[4v +/- sqrt([-4v]^2 - 4(3v)(v^2))]/2(3v)

[4v +/- sqrt(16v^2 - 12v^3)]/6v

But now I'm left with

(2/3)v +/- [sqrt(16v^2 - 12v^3)/6v]

I can't combine terms in the square root because they have different exponents, so this doesn't seem like it makes much sense.


Those "v"s in red shouldn't be there. There is no v in the coefficient of your "a" term. Fix that up and it will be much easier to work with.
By the way, you don't necessarily have to use the quadratic equation. You could just factor it.
 
  • #8
Are you certain that all the information is given? Because the way I see it, you have three variables (v, v1, v2) and only two equations (conservation of momentum and conservation of KE).
 
  • #9
two unknowns right? since the question asks for Vf's in terms of Vi.
 
  • #10
all the info is given. i derived v1 and v2 from conservation of momentum. i have to do each individually, so really there are only two unknowns with two equations.

i see what you mean, hage, about the extra v's i put in. ill try that.
 
  • #11
that works. i got (1/3)v = v1. Thanks everyone for all of your help
 

What is an elastic collision?

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

How is momentum conserved in an elastic collision?

In an elastic collision, momentum is conserved. This means that the total momentum of the system before the collision is equal to the total momentum after the collision. This is due to the fact that there is no net external force acting on the system.

What is the difference between an elastic collision and an inelastic collision?

In an inelastic collision, some kinetic energy is lost due to the objects sticking together or deforming. In an elastic collision, there is no loss of kinetic energy and the objects bounce off each other with no change in their shape or structure.

What factors affect the outcome of an elastic collision?

The outcome of an elastic collision can be affected by the mass and velocity of the objects involved. A higher mass or velocity can result in a greater change in momentum and potentially a more dramatic collision.

What are some real-life examples of elastic collisions?

Examples of elastic collisions can include billiard balls colliding on a pool table, balls bouncing off each other in a game of basketball, or a tennis ball hitting a racket and bouncing off. In each of these cases, there is no loss of kinetic energy and the objects involved maintain their shape and structure after the collision.

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