#$&*
course Phy 121
11/26 11 am
017. collisions
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Question: `q001. Note that this assignment contains 6 questions.
A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.
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Your solution:
Well we know that for the second object, the m is 2 kg, and the v0 = 0, we know that the `dt is .03. I know that we need to use the eqations momentum = m * `dv, and Impulse of Fnet = Fnet * `dt, but I'm not sure how to find the `dv. Would it be the difference in the first object's veocity?
confidence rating #$&*:
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Given Solution:
By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N.
Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.
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Self-critique (if necessary):
I'm not sure if I understand this completely. Does this mean that the momentum of the first object is equal to the momentum of the second object?
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Self-critique rating:3
@&
The impulse-momentum theorem does not allow us to find the momentum of either object. It is stated only in terms of the change in momentum.
All that is asked here is the average force required to change the velocity of the 10 kg object.
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Question: `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.
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Your solution:
Well following what was correct in the first problem, I would say it is +667 Newtons.
Therefore we can find the vf.
Fnet * `dt = m * (vf - v0)
[(Fnet * `dt) / m] + v0 = vf
vf = [(667 Newtons * .03 s) / 2 kg] + 0 = 20 kg m/s / 2 kg = 10 m/s
confidence rating #$&*:
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Given Solution:
Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second.
This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object.
A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s.
Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?
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Your solution:
I would say it is greater than since the one object wasn't moving and that made up for the slowing down of the other object.
confidence rating #$&*:
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Given Solution:
The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules.
The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules.
Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?
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Your solution:
I think that the momentum afterwards is greater than the momentum beforehand since the `dv increased for the second object, even considering the decrease in `dv of the first object.
confidence rating #$&*:
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Given Solution:
The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision.
The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second.
The total momentum after collision is therefore equal to the total momentum before collision.
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Self-critique (if necessary):
I got that wrong, I did not expect it to stay the same at all. This is very interesting.
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Self-critique rating:3
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Question: `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?
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Your solution:
Since the Fnet and `dt doesn't change, which makes up the impulse, which is equal to the momentum, neither can the momentum change.
@&
The impulse is not equal to the momentum, but rather to the change in momentum.
*@
confidence rating #$&*:
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Given Solution:
Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum.
Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.
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Self-critique (if necessary):OK
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Self-critique rating:OK
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q006. A 2 kg mass moving at 6 m/s collides with a 3 kg mass, after which the 2 kg mass is moving at 4 m/s.
By how much did its momentum change?
By how much did the momentum of the 3 kg mass change?
How do the forces experienced by the two masses during the collision compare?
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Your solution:
Momentum before = 2 kg * 6 m/s = 12 kg m/s Momentum after = 2 kg * 4 m/s = 8 kg m/s, therefore it's momentum change by 4 kg m/s.
@&
Good, but this would be - 4 kg m/s.
*@
We know that the total momentum should still equal 12 kg m/s after the collision, making the change in momentum overall 0.
From this we can determine how much the momentum changed in the 3 kg mass
Total momentum = [momentum after for 2 kg] + [momentum after for 3 kg]
Therefore momentum aterwards for 3 kg = [Total momentum] - [momentum after for 2 kg] = 12 kg m/s - 8 kg m/s = 4 kg m/s, I assume that the original momentum for this object was 0, making the change 4 kg m/s also.
@&
Your thinking is good. More specifically:
The change in momentum of the first object is -4 kg m/s.
The total momentum remains constant.
Therefore the momentum of the second object must change by +4 kg m/s.
*@
From this we can determine the v of the 3 kg object.
v = momentum / m = 4 kg m/s / 3 kg = 4/3 m/s
@&
You can't determine the velocity of the second object, since you don't know its initial velocity. All you can determine is the change in its velocity.
4/3 m/s is the change in its velocity.
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THe momentum forces for the two masses are equal.
confidence rating #$&*:
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Self-critique (if necessary):
Not sure if I did this correctly.
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Self-critique rating:2
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q006. A 2 kg mass moving at 6 m/s collides with a 3 kg mass, after which the 2 kg mass is moving at 4 m/s.
By how much did its momentum change?
By how much did the momentum of the 3 kg mass change?
How do the forces experienced by the two masses during the collision compare?
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Your solution:
Momentum before = 2 kg * 6 m/s = 12 kg m/s Momentum after = 2 kg * 4 m/s = 8 kg m/s, therefore it's momentum change by 4 kg m/s.
@&
Good, but this would be - 4 kg m/s.
*@
We know that the total momentum should still equal 12 kg m/s after the collision, making the change in momentum overall 0.
From this we can determine how much the momentum changed in the 3 kg mass
Total momentum = [momentum after for 2 kg] + [momentum after for 3 kg]
Therefore momentum aterwards for 3 kg = [Total momentum] - [momentum after for 2 kg] = 12 kg m/s - 8 kg m/s = 4 kg m/s, I assume that the original momentum for this object was 0, making the change 4 kg m/s also.
@&
Your thinking is good. More specifically:
The change in momentum of the first object is -4 kg m/s.
The total momentum remains constant.
Therefore the momentum of the second object must change by +4 kg m/s.
*@
From this we can determine the v of the 3 kg object.
v = momentum / m = 4 kg m/s / 3 kg = 4/3 m/s
@&
You can't determine the velocity of the second object, since you don't know its initial velocity. All you can determine is the change in its velocity.
4/3 m/s is the change in its velocity.
*@
THe momentum forces for the two masses are equal.
confidence rating #$&*:
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Self-critique (if necessary):
Not sure if I did this correctly.
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Self-critique rating:2
#*&!
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q006. A 2 kg mass moving at 6 m/s collides with a 3 kg mass, after which the 2 kg mass is moving at 4 m/s.
By how much did its momentum change?
By how much did the momentum of the 3 kg mass change?
How do the forces experienced by the two masses during the collision compare?
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Your solution:
Momentum before = 2 kg * 6 m/s = 12 kg m/s Momentum after = 2 kg * 4 m/s = 8 kg m/s, therefore it's momentum change by 4 kg m/s.
@&
Good, but this would be - 4 kg m/s.
*@
We know that the total momentum should still equal 12 kg m/s after the collision, making the change in momentum overall 0.
From this we can determine how much the momentum changed in the 3 kg mass
Total momentum = [momentum after for 2 kg] + [momentum after for 3 kg]
Therefore momentum aterwards for 3 kg = [Total momentum] - [momentum after for 2 kg] = 12 kg m/s - 8 kg m/s = 4 kg m/s, I assume that the original momentum for this object was 0, making the change 4 kg m/s also.
@&
Your thinking is good. More specifically:
The change in momentum of the first object is -4 kg m/s.
The total momentum remains constant.
Therefore the momentum of the second object must change by +4 kg m/s.
*@
From this we can determine the v of the 3 kg object.
v = momentum / m = 4 kg m/s / 3 kg = 4/3 m/s
@&
You can't determine the velocity of the second object, since you don't know its initial velocity. All you can determine is the change in its velocity.
4/3 m/s is the change in its velocity.
*@
THe momentum forces for the two masses are equal.
confidence rating #$&*:
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Self-critique (if necessary):
Not sure if I did this correctly.
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Self-critique rating:2
#*&!#*&!
&#Good work. See my notes and let me know if you have questions. &#