Finding the final velocity using momentum and KE

In summary, the conversation discusses the equations for momentum and kinetic energy before and after an elastic collision between two masses with different initial velocities. It is noted that the equations can be used to determine the final velocities of the masses, but the direction of the final velocities cannot be determined until the equations are solved. The concept of choosing a direction as positive and then determining if the final velocity is negative or positive is also discussed.
  • #1
lonewolf219
186
2
Before the collision, there are two masses. Mass 1 is at rest and mass 2 is moving. No friction. It is an elastic collision and mass 2 is moving in positive x direction.

After the collision, is it correct that mass 1 moves in positive x direction but mass 2 rebounds and moves in negative x direction? If so, is this the equation for momentum:

m[itex]_{2}[/itex]v[itex]_{2i}[/itex]=m[itex]_{1}[/itex]v[itex]_{1f}[/itex]-m[itex]_{2}[/itex]v[itex]_{2f}[/itex]

And the equation for KE:

[itex]_{2}[/itex](v[itex]_{2i}[/itex])^2=m[itex]_{1}[/itex](v[itex]_{1f}[/itex])^2+m[itex]_{2}[/itex](v[itex]_{2f}[/itex])^2

And if solving for v[itex]_{1f}[/itex] :

v[itex]_{1f}[/itex]=[itex]\sqrt{\frac{m_{2}(v_{2i})^2}{m_{1}}}[/itex]

And the velocity of this mass should be in the positive x direction.

Does everything look correct?
 
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  • #2
lonewolf219 said:
is this the equation for momentum:

m[itex]_{2}[/itex]v[itex]_{2i}[/itex]=m[itex]_{1}[/itex]v[itex]_{1f}[/itex]-m[itex]_{2}[/itex]v[itex]_{2f}[/itex]
Does the RHS represent total momentum after collision? Doesn't look like it.
 
  • #3
Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
 
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  • #4
lonewolf219 said:
Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
What do the equations tell you? You shouldn't need to be taking any square roots, thereby introducing ambiguity in sign. If you handle the conservation equations correctly you can obtain a simple ratio between relative velocities. Combined with the mass ratio you can then determine all velocities in terms of the initial velocity of mass 2.
 
  • #5
Thanks for pointing out the proper method. So since mass 1 has half the mass of mass 2, mass 2 does not rebound and change directions? What if the masses were switched? If mass 1 were moving and collided with the more massive mass 2 at rest, would mass 1 change direction in this case?
 
  • #6
lonewolf219 said:
Thanks for pointing out the proper method. So since mass 1 has half the mass of mass 2, mass 2 does not rebound and change directions? What if the masses were switched? If mass 1 were moving and collided with the more massive mass 2 at rest, would mass 1 change direction in this case?
Is that what the equations tell you? Please post your working.
 
  • #7
lonewolf219 said:
Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
It is not a matter of "changing direction" or "not changing direction". Total momentum is always [itex]m_1v_1+ m_2v_2[/itex]. Whether or not a mass changes direction is signaled by the sign of its v. If mass 2 is twice mass 1 and has initial velocity [itex]v_i[/itex] then "conservation of momentum" is [itex]2mv_i= mv_1+ 2mv_2[/itex] where [itex]v_1[/itex] is the velocity after the collision of mass 1 and [itex]v_2[/itex] is the velocity after the collision of mass 2.
"Conservation of kinetic energy" is [itex](1/2)2mv_i^2= (1/2)mv_1^2+ (1/2)(2m)v_2^2[/itex]. Of course,, you can divide both equations by "m" eliminating that. You will have [itex]v_1+ 2v_2= 2v_i[/itex] and [itex]v_1^2+ 2v_2^2= 2v_i^2[/itex]. You could then, for example, solve the first equation for [itex]v_1= 2v_i- 2v_2[/itex] and replace [itex]v_1[/itex] in the second equation by that to get a quadratic equation for [itex]v_2[/itex].
 
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  • #8
Thanks Haruspex and HallsofIvy, I appreciate your help!

Haruspex has asked me more than once "Is that what the equation tells you?" Now I realize what you are saying (thanks HallsofIvy for your explanation)... you are saying that we do not know if the mass will rebound or not until we solve the equation. So, we can't choose a direction for the final velocity before we have solved for it. But if the final velocities are in opposite directions, then we would conclude that the moving mass rebounded. Is this right, Haruspex?
 
  • #9
lonewolf219 said:
Thanks Haruspex and HallsofIvy, I appreciate your help!

Haruspex has asked me more than once "Is that what the equation tells you?" Now I realize what you are saying (thanks HallsofIvy for your explanation)... you are saying that we do not know if the mass will rebound or not until we solve the equation. So, we can't choose a direction for the final velocity before we have solved for it. But if the final velocities are in opposite directions, then we would conclude that the moving mass rebounded. Is this right, Haruspex?
Yes. It often happens when trying to find a vector (force, velocity, whatever) that you don't know in advance which direction it will be in. So you just choose a direction as positive, and if the equations give you a negative answer then you know it was the other way.
 
  • #10
Awesome! Thanks for your help, Haruspex... This concept is very important!
 

1. How do you calculate final velocity using momentum and kinetic energy?

To calculate the final velocity using momentum and kinetic energy, you can use the formula: Vf = √(2KE/m), where Vf is the final velocity, KE is the kinetic energy, and m is the mass of the object. This formula is based on the conservation of momentum and energy principles.

2. What is the relationship between momentum and kinetic energy?

Momentum and kinetic energy are both measures of an object's motion, but they measure different aspects of it. Momentum is a measure of an object's mass and velocity, while kinetic energy is a measure of an object's mass and speed. In other words, momentum is related to an object's motion in a certain direction, while kinetic energy is related to its overall motion.

3. Can an object have a high momentum but low kinetic energy?

Yes, an object can have a high momentum but low kinetic energy. This is possible when the object has a high mass and low velocity. For example, a large truck moving at a slow speed can have a higher momentum than a small car moving at a fast speed.

4. How does the mass of an object affect its final velocity when using momentum and kinetic energy?

The mass of an object does affect its final velocity when using momentum and kinetic energy. The larger the mass of the object, the more momentum and kinetic energy it will have. This means that the final velocity will be lower for a larger mass compared to a smaller mass.

5. What are the units of measurement for momentum and kinetic energy?

The units of measurement for momentum are kilogram-meters per second (kg·m/s), while the units for kinetic energy are joules (J). These units can also be expressed in other forms, such as Newton-seconds (N·s) for momentum and kilogram-meters squared per second squared (kg·m^2/s^2) for kinetic energy.

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