Conservation of momentum and conservation of kinetic energy

  • #1

Homework Statement


Here is the problem:

http://www.chegg.com/homework-help/...te-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

 

Answers and Replies

  • #2
SammyS
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Homework Statement


Here is the problem:

http://www.chegg.com/homework-help/...te-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
SammyS
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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.
 

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