Momentum and Energy Loss

In summary, when a block in motion collides with a stationary object in a perfectly inelastic collision, the fraction of final kinetic energy over initial kinetic energy can be represented by M2/M1+M2. This equation holds true even if M2 is initially moving, but the calculation may be more complex. The reason for using this specific ratio is because it has a simple mathematical form.
  • #1
MotoPayton
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http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

When a block in motion(M1) collides with a a stationary object (M2) in a perfectly inelastic collision, the fraction of KE final/KE initial can be modeled as M2/M1+M2.

My question is if M2 has to be stationary for this equation to work.

If both objects are moving before the collide in a perfectly inelastic collision does the ratio of lost kinetic energy still hold?
 
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  • #2
MotoPayton said:
My question is if M2 has to be stationary for this equation to work.
Yes.
If both objects are moving before the collide in a perfectly inelastic collision does the ratio of lost kinetic energy still hold?
It's a bit more complicated when both blocks are initially moving. But it's easy enough to work it out.
 
  • #3
I ran the math and it makes sense now.

The only reason we use the M1/M1+M2 is because it is the only ratio of energies that has a simple mathematical form.
 
  • #4
MotoPayton said:
The only reason we use the M1/M1+M2 is because it is the only ratio of energies that has a simple mathematical form.
Yes. It's a special case that is particularly easy to deal with.
 
  • #5


The equation for calculating the fraction of kinetic energy lost in a perfectly inelastic collision, where one object is initially at rest, assumes that the second object is also at rest. This is because the equation is based on the conservation of momentum and energy, which only apply when there is no external force acting on the system.

If both objects are moving before the collision, the equation would need to be modified to take into account the initial momentum of both objects. This would result in a different value for the fraction of kinetic energy lost.

However, in most real-world situations, it is difficult to have two objects collide in a perfectly inelastic manner where both are already in motion. Usually, one object is at rest or moving very slowly compared to the other, so the equation can still be used effectively.

It is important to note that the equation for calculating the fraction of kinetic energy lost in a perfectly inelastic collision is an idealized model and may not accurately represent all real-world situations. Other factors, such as friction and deformation of the objects, can also contribute to energy loss in a collision.
 

1. How is momentum defined and why is it important?

Momentum is defined as the product of an object's mass and velocity. It is important because it describes the amount of motion an object has and is conserved in a closed system, meaning it remains constant unless acted upon by an external force.

2. What is the difference between elastic and inelastic collisions?

Elastic collisions are ones in which the total kinetic energy of the system is conserved, while inelastic collisions involve a loss of kinetic energy due to the deformation of objects involved.

3. How do energy losses occur in a system?

Energy losses in a system can occur due to friction, air resistance, and other dissipative forces. These forces convert kinetic energy into other forms of energy, such as heat or sound.

4. How does the law of conservation of energy apply to momentum and energy loss?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. This applies to momentum and energy loss because although the total momentum and energy of a system may change, the total amount remains the same.

5. How can momentum and energy loss be minimized?

Momentum and energy loss can be minimized by reducing the effects of dissipative forces, using materials with high elasticity, and increasing the efficiency of processes involved in the system. In some cases, adding external forces can also help to minimize energy loss.

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