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Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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asst 12 lab turned in?
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I completed asst 12 lab ball down ramp several weeks ago. I thought I turned it in then. When I reviewed my records, I can't prove I turned it in at all.
Before I turn it in for a possible second time, can you tell me if you have a asst 12 lab from me?
Thanks
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I located the lab. The problem might have been the result of the double submission, but that shouldn't have made a difference so I'm really not sure how it got omitted.
Anyway it will be posted following this response.
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Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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set 4 problem 10 KE compare
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Below is just a copy of the problem I'm working. My questions are in the other boxes below.
Set 4 Problem number 7
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Problem
Solution
Generalized Solution
Problem
An object moving at 9 m/s collides with an object moving at -5 m/s. The mass of the first object is 4 kg and the mass of the second is 8 kg.
After the collision the first object is observed to have velocity -9 m/s.
What will be the velocity of the second object after the collision?
How do the kinetic energy totals before collision compare with those after collision?
Solution
Conservation of Momentum, which follows from the fact that the objects exert equal and opposite forces on one another for equal times, assures us that the momentum changes of the objects are equal and opposite.
We can find the momentum change of the first object:
The first object has a change in velocity of -9 m/s - 9 m/s = -18 m/s.
Its change in momentum is therefore
`dp1 = m1 `dv1 = 4 kg * -18 m/s = -72 kg m/s.
The momentum change of the second object is unknown, but its mass is known. Its momentum change being the product of its mass and its change in velocity, it is easy to find its velocity change:
`dp2 = m2 `dv2, so
`dv2 = `dp2 / m2 = -`dp1 / m2 = --72 kg m/s / 8 kg = 9 m/s.
The final velocity v2' of the second object is easily found from its initial velocity and its change in velocity:
v2' = v02 + `dv2 = -5 m/s + 9 m/s = 4 m/s.
We could have solved this problem from the more detailed statement of momentum conservation for two objects of constant mass:
total momentum before collision = total momentum after collision, or
m1 v1 + m2 v2 = m1 v1' + m2 v2'.
We are given m1, v1, v1', m2 and v2. The only thing we don't know is v2'.
Solving for v2' we get v2' = (m1 v1 + m2 v2 - m1 v1') / m2.
Substituting the values of m1, v1, m2, v2 and v1' we find that v2' = 4 m/s.
Generalized Solution
Knowing that in a system of two objects, in which the only forces acting are the forces exerted on the objects by one another, the forces will be equal and opposite and therefore the momentum changes will be equal and opposite, we have the statement `dp2 = - `dp1. In terms of masses and velocity changes this tells us that
m2 `dv2 = m1 `dv1
In the present situation, where we know v2 but don't know `dv2, we can solve this equation for `dv2, obtaining
`dv2 = - (m1 / m2) * `dv1.
Knowing `dv2 we can find the final velocity v2' = v2 + `dv2 of the second object.
Alternatively we can simply solve the conservation of momentum equation
m1 v1 + m2 v2 = m1 v1' + m2 v2' (equivalent to but more detailed than m2 `dv2 = m1 `dv1)
for v2', obtaining
v2' = ( m1 v1 - m1 v1' + m2 v2 ) / m2.
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The problem I'm working above says set 4 problem 7, but in reality it is listed under set 4 problem 10. This problem asks for how does the KE compare before and after the collision. I solved for KE before and after and then realized the problem doesn't talk anymore about the question. Said another way I don't have a solution to compare to.
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I solved for total system KE before as 62J and after the collision total system KE as -98J.
Do you have the KE before vs. after solution? If my numbers are correct, I don't need it.
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The KE before collision is
1/2 * 4 kg * (9 m/s)^2 + 1/2 * 8 kg * (-5 m/s)^2 = 660 Joules, approx..
After collision we get
1/2 * 4 kg * (-9 m/s)^2 + 1/2 * 8 kg * (4 m/s)^2 = 193 Joules.
Check my arithmetic, but these numbers do seem to make sense.
Note: Do remember to offset your questions with multiple question marks so I don't inadvertently miss them, or spend unnecessary time looking for them.
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#$&*
Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
set 4 problem 11 mass
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Below is a copy of the problem I'm working. My questions are in the following boxes.
Set 4 Problem number 11
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Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
Two objects collide and remain stuck together after collision.
One object has mass 13 kg and is moving in the positive direction at 18 m/s and the other has mass 13 kg and and moves at 13 m/s in the negative direction.
What is their total momentum, and what will be the velocity of this system immediately after collision?
Solution
The momentum of the first is ( 18 m/s)( 13 kg)= 234 kg m/s, and that of the second is ( - 13 m/s)( 18 kg) = -234 kg m/s.
The total momentum is therefore 234 kg m/s + -234 kg m/s = 0 kg m/s.
The total mass is easily found to be 31 kg. Since the total momentum after collision is the same as that before collision, we see that after collision we have a mass of 31 kg with momentum 0 kg m/s.
Dividing we obtain velocity ( 0 kg m/s)/( 31 kg) = 0 m/s.
Generalized Solution
The total momentum of a mass m1 moving at velocity v1 and a mass m2 moving at velocity v2 is
pTot = m1 v1 + m2 v2.
By Newton's Third Law and the Impulse-Momentum Theorem this momentum will remain unchanged during collision.
After collision we wil have one object of mass m1 + m2 and momentum m1 v1 + m2 v2. The object will therefore have velocity
velocity = momentum / mass = (m1 v1 + m2 v2) / (m1 + m2).
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The problem states One object has mass 13 kg and is moving in the positive direction at 18 m/s and the other has mass 13 kg and and moves at 13 m/s in the negative direction.
The solution statesThe momentum of the first is ( 18 m/s)( 13 kg)= 234 kg m/s, and that of the second is ( - 13 m/s)( 18 kg) = -234 kg m/s.
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The problem has the second object at 13kg mass and the solution is worked as the second object being 18kg mass.
I beleive I understand either way. I just couldn't check my solution because of the differences.
Thanks.
Self-critique (if necessary):
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Self-critique rating:
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The momentum of the second would be -13 m/s * 13 kg = -169 kg m/s.
The total momentum would therefore be 65 kg m/s.
With total mass being 26 kg, the common velocity of the two masses after collision would be
65 kg m/s / (26 kg) = 2.5 m/s.
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