Elastic Collision: Express Velocity u in Terms of m,M,v

In summary, in elastic collisions, the velocities after the collision are equal to the initial velocities plus the velocities due to the collision.
  • #1
espen180
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2

Homework Statement



When two balls are dropped like shown in situation 1 in the figure, the top ball shoots up after the impact, while the bottom ball loses some of its mechanical energy, as seen in situation 3. By simplifying the situation, we can assume that the bottom ball stops and bounces up before the top ball, so we get an elastic collision between the two, as shown in situations 2 and 4. If the mass of the small ball is [tex]m[/tex] and the mass of the large ball is [tex]M[/tex] and they have the same speed [tex]v[/tex] before the collision and velocities [tex]u[/tex] and [tex]U[/tex] after, what is [tex]u[/tex] is terms of [tex]m,M,v[/tex]?

Problem may be rewritten and simplifyed as:

Two spheres with masses [tex]m[/tex] and [tex]M[/tex] collide head on with the same speed [tex]|v|[/tex] and exit the collision with velocities [tex]u[/tex] and [tex]U[/tex]. The collision is elastic. Express [tex]u[/tex] in terms of [tex]m[/tex], [tex]M[/tex] and [tex]v[/tex]

http://img401.imageshack.us/img401/4555/fysikkkollisjon.png

Homework Equations



Definition of momentum: [tex]p=mv[/tex] (1)

Conservation of momentum: [tex]\Delta \Sigma p=0[/tex] (2)

For elastic collisions:
[tex]v_1+u_1=v_2+u_2[/tex] (3) (v=initial velocity , u=velocity after collision)


The Attempt at a Solution



The problem is one-dimentional, so I have omitted the usual vector notation and instead defined a positive direction, to the right in situation 4 in the figure.
Inserting values into the equations, we get
(2)[tex]Mv-mv=MU+mu[/tex]

[tex]U=\frac{Mv-mv-mu}{M}[/tex]

(3)[tex]u-v=v+U[/tex]

substituting for U:

[tex]u-v=v+v-\frac{m}{M}(v-u)[/tex]

[tex]u+\frac{m}{M}u=3v-\frac{m}{M}v[/tex]

[tex]u(1+\frac{m}{M})=v(3-\frac{m}{M})[/tex]

[tex]u=v\frac{3-\frac{m}{M}}{1+\frac{m}{M}}[/tex]

So I have a solution. Is it reasonable? I think so, on the grounds that [tex]u=v[/tex] when [tex]m=M[/tex].

What do you think?

Thanks for any help.
 
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  • #2
One step is wrong.
u - v = v + v - m/M*(v + u )
 
  • #3
Thanks for pointing that out.
So it will be like this then:

[tex]u-v=v+v-\frac{m}{M}(v+u)[/tex]

[tex]u-\frac{m}{M}u=3v-\frac{m}{M}v[/tex]

[tex]u(1-\frac{m}{M})=v(3-\frac{m}{M})[/tex]

[tex]u=v\frac{3-\frac{m}{M}}{1-\frac{m}{M}}[/tex]

But now, if m=M, u is infinately large. How do I explain that?
 
  • #4
Careful with your signs on the second line :wink:
 
  • #5
Oh yeah! The first line was wrong, but the others were correct, so I had the right expression from the start, right?
 
  • #6
espen180 said:
Oh yeah! The first line was wrong, but the others were correct, so I had the right expression from the start, right?
Indeed you did :smile:
 
  • #7
Troublesome sign errors. :rolleyes:

Thanks for the help!
 

FAQ: Elastic Collision: Express Velocity u in Terms of m,M,v

What is an elastic collision?

An elastic collision is a type of collision in which both kinetic energy and momentum are conserved. This means that the total energy and total momentum of the system before and after the collision are equal.

How is velocity u expressed in terms of m, M, and v in an elastic collision?

In an elastic collision, the velocity of one object after the collision (u) can be expressed as: u = ((m - M)v + 2Mv) / (m + M), where m is the mass of the first object, M is the mass of the second object, and v is the initial velocity of the first object.

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

In an elastic collision, both kinetic energy and momentum are conserved, while in an inelastic collision, only momentum is conserved. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

Can an elastic collision occur between more than two objects?

Yes, an elastic collision can occur between any number of objects, as long as the total kinetic energy and momentum of the system are conserved.

Are all collisions in the real world perfectly elastic?

No, most collisions in the real world are not perfectly elastic. In most cases, some kinetic energy is lost due to factors such as friction and deformation of the objects involved. However, some collisions, such as those between gas molecules, can be considered almost perfectly elastic.

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