Elastic Collision Formula: Solving for Final Velocity and Mass Ratios

In summary, the conversation is about a perfectly elastic collision between two objects, one with mass m1 and initial velocity v1i, and the other with mass m2 and initial velocity v2i. The question is how v1i should relate to v2i in order for m1 to stop (v1f=0). The relevant equations used are KE = .5mv^2 and P = mv. The solution involves using the conservation of energy and momentum equations, and leads to the conclusion that v1i = (m2(1 - 2v2i))/(m1). There may be a typo in the final equation as 1-velocity is dimensionally invalid.
  • #1
Puddles
12
0
Homework Statement

An object of mass m1 traveling with velocity v1i has a perfectly elastic collision in which it rear ends and object of mass m2 (m2>>m1) traveling with velocity v2i. How must the velocity v1i relate to v2i if the mass m1 is to stop in its tracks (v1f=0)? What happens if velocity v1i is greater than this? If it is smaller?

Relevant equations
KE = .5mv^2
P = mv

The attempt at a solution
Cons Energy
.5m1v1i^2 + .5m2v2i^2 = .5m2v2f^2

V2f = sq.rt(( m1v1i^2 + m2v2i^2 )/(m2))

Cons Momentum
m1v1i + m2v2i = m2v2f

V2f = ( m1v1i + m2v2i )/(m2)

Set equal to each other, but my answer keeps getting more complex? It's a math error, but I'm not sure what it is…

I get to here:(m1^2v1i^2)+(2m1v1im2v2i)+(m2^2v2i^2) = (m1v1i^2)+(m2v2i^2)

Can anyone help me continue to work this out? I'm frustrated because this is a simple problem but I can't get it.
 
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  • #2
Okay, I think I found my error, I've worked out that v1i = (m2(1 - 2v2i))/(m1), how can I plug this back into check it? I'm struggling to find a way to do so but I know there must be a way…
 
  • #3
Puddles said:
Okay, I think I found my error, I've worked out that v1i = (m2(1 - 2v2i))/(m1), how can I plug this back into check it? I'm struggling to find a way to do so but I know there must be a way…
Is there a typo there? 1-velocity is dimensionally invalid.
 

1. What is a simple elastic collision?

A simple elastic collision is a type of collision between two objects where kinetic energy is conserved and the objects bounce off each other without losing any energy. This type of collision only occurs in idealized situations where there is no friction or loss of energy due to deformation of the objects.

2. How is momentum conserved in a simple elastic collision?

In a simple elastic collision, momentum is conserved because the total momentum of the two objects before the collision is equal to the total momentum after the collision. This means that the total mass and velocity of the objects must remain the same before and after the collision.

3. What is the formula for calculating the final velocities in a simple elastic collision?

The formula for calculating the final velocities in a simple elastic collision is vf1 = ((m1 - m2) / (m1 + m2)) * vi1 and vf2 = ((2 * m1) / (m1 + m2)) * vi1, where vi1 and vi2 are the initial velocities of the two objects and m1 and m2 are their masses.

4. Can a simple elastic collision occur between objects of different masses?

Yes, a simple elastic collision can occur between objects of different masses as long as the total momentum and kinetic energy are conserved. However, the final velocities of the objects will depend on their masses and initial velocities.

5. Why is a simple elastic collision considered an idealized situation?

A simple elastic collision is considered an idealized situation because it does not take into account factors such as friction and energy loss due to deformation of the objects. In real-world situations, these factors will affect the outcome of a collision and the conservation of momentum and kinetic energy may not hold true.

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