Totally elastic and momentum collision equation help Thanks

In summary, the equation you use to solve for the speed of the clay before impact is 1/2(m1+m2)V2^2 = friction(m1+m2)gL.
  • #1
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Homework Statement



A 12kg wad of clay is hurled horizontally at a 100g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50m before coming to rest. If the coefficient of friction between the bloack and the surface is .650, what is the SPEED of the clay immediately before impact?

Homework Equations



This is a perfectly inelastic collision correct?

My book tells me in the answer section to use the following formula, but I am not sure where it came from or why to use this. Am I to use this for equations related to inelastic collisions and friction involved?

1/2(m1+m2)V2^2 = friction(m1+m2)gL

How did it come up with that? I guess I am just having trouble figuring out how I was suppose to come up with that formula.

Any clarification?



The Attempt at a Solution

 
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  • #2
nukeman said:
This is a perfectly inelastic collision correct?
Yes.

My book tells me in the answer section to use the following formula, but I am not sure where it came from or why to use this. Am I to use this for equations related to inelastic collisions and friction involved?

1/2(m1+m2)V2^2 = friction(m1+m2)gL

How did it come up with that? I guess I am just having trouble figuring out how I was suppose to come up with that formula.
It's telling you that the kinetic energy the system had after the collision must equal the work done by friction. Note that V2 is the speed of the 'clay+block' after the collision. You'll use that to figure out the speed of the clay before the collision.
 
  • #3
nukeman said:
1/2(m1+m2)V2^2 = friction(m1+m2)gL

Do you think that above equation is to be used before impact or after impact?

Now i notice that another poster beat me.
 
  • #4
Ok great.

When looking at this problem... What is the first and second things I should be trying to figure out. Mainly, when looking at this problem specifically, what would I be looking to do first?

I want to systematically break this equation down from start!
 
  • #5
grzz said:
Do you think that above equation is to be used before impact or after impact?

Now i notice that another poster beat me.

I really am not sure... ?
 
  • #6
nukeman said:
When looking at this problem... What is the first and second things I should be trying to figure out. Mainly, when looking at this problem specifically, what would I be looking to do first?

I want to systematically break this equation down from start!
Break the problem into two stages:
(1) The collision itself. What's conserved?
(2) The movement against friction after the collision.

The equation you have deals with part #2.

What equation can you come up with to deal with #1?
 
  • #7
I thought for step one, I would use:

Vf = m1V1i + m2V2i / (m1 + m2)

?
 
  • #8
nukeman said:
I thought for step one, I would use:

Vf = m1V1i + m2V2i / (m1 + m2)

?
Good. (Careful with parentheses!)

What does Vf correspond to?

What about V1i and V2i?
 
  • #9
Doc Al said:
Good. (Careful with parentheses!)

What does Vf correspond to?

What about V1i and V2i?

Vf would be velocity of the system?

v1i would be for the clay and v2i would be wood ?
 
  • #10
nukeman said:
Vf would be velocity of the system?
Yes, immediately after the collision.

v1i would be for the clay and v2i would be wood ?
Right.
 
  • #11
Ok, so first thing I do when looking at this is figure out the Vf of the system after the collision.

With the formula I supplied for the first step Vf = m1V1i + m2V2i / (m1 + m2) , how do I go from there, to here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL
 
  • #12
Doc? Can you explain the above reply to me if you can, or anyone else? :)
 
  • #13
nukeman said:
Ok, so first thing I do when looking at this is figure out the Vf of the system after the collision.
Right. You'll do that by solving the second part first, since you have all the needed data. (It's often the case that its best to work a problem 'backwards'.)

With the formula I supplied for the first step Vf = m1V1i + m2V2i / (m1 + m2) , how do I go from there, to here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL
You'll solve the second step first.

How will you relate the two steps? Hint: How does Vf in the first formula relate to V2 in the second? (Generally it's a good idea to use the same symbols for the same quantities. Otherwise things get confusing.)

In your first formula, Vf = (m1V1i + m2V2i) / (m1 + m2), what is V2?
 
  • #14
Oh ok Doc, I am starting to understand it a little better. Just having trouble coming to a conclusion on what exact formula to use. Like what stpes I take to get the right formula...

You ask what is V2 ? Well, that's the Velocity of the wood, which is zero, correct? What significance is that may I ask?

Again, MUCH thanks for the help!

EDIT: I guess what I am missing is how I would come to the formula to get the speed of the system, as detailed here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL

I have never seen that before!
 
Last edited:
  • #15
nukeman said:
You ask what is V2 ? Well, that's the Velocity of the wood, which is zero, correct? What significance is that may I ask?
Yes, in your conservation of momentum equation (which describes the collision), V2 stands for the initial speed of the block, which is zero. You'll need that equation to solve for the initial speed of the bullet. (You'll use this equation last.)

EDIT: I guess what I am missing is how I would come to the formula to get the speed of the system, as detailed here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL
That's just an energy equation stating that the change in KE of the system after the collision is equal to the work done by friction. What does V2 stand for here?
 

1. What is the equation for a totally elastic collision?

The equation for a totally elastic collision is m1v1i + m2v2i = m1v1f + m2v2f, where m represents mass and v represents velocity.

2. How is momentum conserved in a totally elastic collision?

In a totally elastic collision, momentum is conserved because the total momentum before the collision is equal to the total momentum after the collision. This is represented by the equation p1i + p2i = p1f + p2f, where p represents momentum.

3. Can you use the totally elastic collision equation to solve for unknown variables?

Yes, you can use the totally elastic collision equation to solve for unknown variables. However, you will need to know at least three of the variables (mass, initial velocity, or final velocity) in order to solve for the remaining unknown variable.

4. What is the difference between a totally elastic collision and an inelastic collision?

In a totally elastic collision, kinetic energy is conserved, meaning that the objects involved in the collision maintain the same amount of kinetic energy before and after the collision. In an inelastic collision, some kinetic energy is lost, usually in the form of heat or sound.

5. How is the coefficient of restitution related to a totally elastic collision?

The coefficient of restitution is a measure of the elasticity of a collision. In a totally elastic collision, the coefficient of restitution would be 1, indicating that all kinetic energy is conserved. A lower coefficient of restitution would indicate that some kinetic energy is lost, making the collision less elastic.

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