Conservation of momentum and conservation of kinetic energy

In summary: The center of mass does not move. The velocities of the objects relative to the center of mass are the same before and after the collision.
  • #1
fredrogers3
40
0

Homework Statement


Here is the problem:

http://www.chegg.com/homework-help/questions-and-answers/7-blocks-collide-elastically-frictionless-surface-write-conservation-momentum-relation-thi-q237700

Homework Equations


There is conservation of momentum involved along with conservation of kinetic energy.
Thus,
m1v1+m2v2=m1v1 final - m2v2 final

The Attempt at a Solution


When I do all the calculation and solve the quadratic for part c, I come up with a final velocity of -1 for the heavier block. The question I have is that is this a sign error? It would make more sense that the collision would result in the blocks going in the direction that the heavier object was going. Also, I got 3 for the other block (positive).

Homework Statement


Homework Equations


The Attempt at a Solution

 
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  • #2
fredrogers3 said:

Homework Statement


Here is the problem:

http://www.chegg.com/homework-help/questions-and-answers/7-blocks-collide-elastically-frictionless-surface-write-conservation-momentum-relation-thi-q237700

Homework Equations


There is conservation of momentum involved along with conservation of kinetic energy.
Thus,
m1v1+m2v2=m1v1 final - m2v2 final

The Attempt at a Solution


When I do all the calculation and solve the quadratic for part c, I come up with a final velocity of -1 for the heavier block. The question I have is that is this a sign error? It would make more sense that the collision would result in the blocks going in the direction that the heavier object was going. Also, I got 3 for the other block (positive).
(It wouldn't have been all that difficult to do some cutting & pasting to post the problem in your post.)

7. Two blocks collide elastically on a frictionless surface.
(a) Write down the conservation of momentum relation for this problem.
(b) Write down the conservation of energy for this problem.
(c) Solve the set of simultaneous equations above for the final velocities of the blocks
after the collision given v1i= 2.0 m/sec, v2i= −2.0 m/sec, m1 = 0.50 kg, and
m2 = 0.30 kg? (zero credit for just using canned formulas given in Halliday &
Resnick—I expect to see your own algebra). (1.0, 3.0)​

I'll be back later. You might receive help from elsewhere by then.
 
  • #3
To continue:

Here we have both objects approaching each other with the same speed, 2 m/s.

Consider the situation in which the two object also have the same mass.

In that case they would both move away from each other at 2 m/s, the one on the left would move at -2 m/s, the other moving at +2 m/s.

Now, let's continues to repeat this collision using objects from the right (the side with initial velocity of -2 m/s) which have less and less mass. A small decrease in mass will make only a small difference in the outcome. The object on the left will still rebound back to the left, so its velocity will be negative, but its speed will be less than 2 m/s . Continue to decrease the mass of the object coming from the right, and at some point, the object from the right will have zero velocity after the collision. (Do you know what ratio of masses produces this result?) For any object approaching from the left with even less mass, the object coming from the left will continue with positive velocity after the collision, although at reduced speed.

For the situation you have to solve, the masses are not different enough for object 1 to continue on to the right. It does have negative velocity after the collision.


You know, you can easily check your result by seeing if momentum and energy are both conserved.


By the way, if you know how to work with the center of mass frame of reference, elastic collisions are easy to analyze.
 

1. What is the law of conservation of momentum?

The law of conservation of momentum states that the total momentum of a closed system remains constant, even in the presence of external forces. This means that in any interaction between two objects, the total momentum before the interaction is equal to the total momentum after the interaction.

2. How does the law of conservation of momentum apply to collisions?

In collisions, the total momentum of the system remains constant. This means that the sum of the momentums of the objects before the collision is equal to the sum of the momentums after the collision. This is true for both elastic and inelastic collisions.

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

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy before the collision is equal to the total momentum and total kinetic energy after the collision. In an inelastic collision, only momentum is conserved, while some kinetic energy is lost in the form of heat, sound, or deformation.

4. How is conservation of kinetic energy related to conservation of momentum?

Conservation of kinetic energy is a special case of conservation of momentum. In a closed system, if there are no external forces acting on the objects, the total kinetic energy of the system will also remain constant. This is because kinetic energy is directly proportional to velocity, which is a component of momentum.

5. Can the law of conservation of momentum and kinetic energy be violated?

No, the law of conservation of momentum and kinetic energy are fundamental laws of physics and have been proven to hold true in all observed interactions. However, in extreme cases such as nuclear reactions or at the quantum level, these laws may need to be modified to account for other factors.

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