Completely inelastic collision

In summary, the energy in a collision is lost in two different ways: 1/2 of it is lost in the first collision and 1/6 of it is lost in the second.
  • #1
Order
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This is a problem I have thought about for a long time without any progress. Please give me a hint.

Homework Statement



Cars B and C are at rest with their brakes off. Car A plows into B at high speed, pushing B into C. If the collisions are completely inelastic, what fraction of the initial energy is dissipated in car C? Initially the cars are identical.

Homework Equations



[tex]p_{i}=p_{f}[/tex]

The Attempt at a Solution



The momentum is conserved and since it is a completely inelastic collision the cars are stuck to each other with velocity v, different from the initial velocity v0. [tex]mv_{0}=3mv[/tex] and [tex]v=\frac{v_{0}}{3}[/tex] and for car C
[tex]\frac{E_{C}}{E_{0}}=\frac{(v_{0}/3)^{2}}{v_{0}^{2}}=\frac{1}{9}.[/tex] Unfortunately the answer in my answer sheet is 1/6. What is wrong?
 
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  • #2
Order said:
This is a problem I have thought about for a long time without any progress. Please give me a hint.


The momentum is conserved and since it is a completely inelastic collision the cars are stuck to each other with velocity v, different from the initial velocity v0. [tex]mv_{0}=3mv[/tex] and [tex]v=\frac{v_{0}}{3}[/tex] and for car C
[tex]\frac{E_{C}}{E_{0}}=\frac{(v_{0}/3)^{2}}{v_{0}^{2}}=\frac{1}{9}.[/tex] Unfortunately the answer in my answer sheet is 1/6. What is wrong?

The final mass is 3m moving at speed v0/3.

[tex]E_f = \frac{1}{2}3m\left(\frac{v_0}{3}\right)^2[/tex]

So 2/3 of the energy is lost in the collisions. But there are two collisions:
The first is between A and B. What is the energy lost in that collision? The rest is lost in the second.AM
 
  • #3
Ok, so 1/2 of the energy is lost in the first collision and 1/6 in the second with a total loss of 2/3. My problem was that I did not understand the word dissipate. Maybe transmit was the word I confused it with? (My english is not so good, so I don't know.) So I guess my calculation was correct but not asked for. Thanks anyway Andrew.
 
  • #4
Order said:
Ok, so 1/2 of the energy is lost in the first collision and 1/6 in the second with a total loss of 2/3. My problem was that I did not understand the word dissipate. Maybe transmit was the word I confused it with? (My english is not so good, so I don't know.) So I guess my calculation was correct but not asked for. Thanks anyway Andrew.
I can see why you may have been confused. Dissipated can mean dispersed. Part of the initial kinetic energy is dispersed from one to three cars. But in this context dissipated refers to the loss of kinetic energy.

AM
 
  • #5


As a scientist, it is important to first understand the concept of completely inelastic collisions. In this type of collision, the objects involved stick together after impact and move with a common final velocity. In contrast, in a partially inelastic collision, the objects may stick together momentarily but then separate.

In this problem, we can assume that cars B and C are of equal mass and car A has a much larger mass. When car A collides with car B, it transfers all of its initial kinetic energy to car B, causing it to move with a velocity of v0/2. However, since the collision is completely inelastic, car B then collides with car C and transfers all of its energy to car C. This means that car C is now moving with a velocity of v0/2 as well.

To calculate the fraction of initial energy dissipated in car C, we need to consider the initial kinetic energy of car A (which is the same as the total initial kinetic energy of all three cars) and the final kinetic energy of car C.

Initial kinetic energy = 1/2 mv0^2

Final kinetic energy of car C = 1/2 (1/2m)(v0/2)^2 = 1/8 mv0^2

Therefore, the fraction of initial energy dissipated in car C is (1/8 mv0^2) / (1/2 mv0^2) = 1/4.

It seems like there may be a mistake in your calculation for the final velocity of car C. Instead of v0/3, it should be v0/2 as explained above. I hope this hint helps you solve the problem.
 

Related to Completely inelastic collision

1. What is a completely inelastic collision?

A completely inelastic collision is a type of collision where the two objects involved stick together after the collision and move as one mass. This results in a decrease in the total kinetic energy of the system.

2. How is momentum conserved in a completely inelastic collision?

In a completely inelastic collision, momentum is conserved because the total momentum before the collision is equal to the total momentum after the collision. This means that the combined mass of the objects and their velocities before the collision must equal the combined mass and velocity after the collision.

3. What is the difference between an elastic and a completely inelastic collision?

In an elastic collision, the objects involved bounce off each other and move apart after the collision. This results in no loss of kinetic energy and the total momentum of the system is conserved. In a completely inelastic collision, the objects stick together and move as one, resulting in a decrease in kinetic energy and the total momentum is still conserved.

4. What factors affect the outcome of a completely inelastic collision?

The factors that affect the outcome of a completely inelastic collision include the mass and velocity of the objects involved in the collision. The greater the mass and velocity, the greater the change in kinetic energy and momentum after the collision.

5. Can a completely inelastic collision occur in real life?

Yes, completely inelastic collisions can occur in real life. Examples include when a car collides with a wall, when a bullet hits a target, or when two objects stick together upon impact. In these scenarios, kinetic energy is lost and the objects involved move together as one mass after the collision.

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