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phy 121
Your 'cq_1_24.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_24.2_labelMessages **
If an airplane is flying north, from what direction would the wind be blowing to most effectively cause the aircraft to veer toward the east?
• What sort of path would it follow for the first second or so?
answer/question/discussion: ->->->->->->->->->->->-> :
A wind blowing west would cause an aircraft going north to veer east.
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• If the wind keeps shifting so as to remain perpendicular to the direction of motion what path will the airplane follow?
answer/question/discussion: ->->->->->->->->->->->-> :
If the force continues perpendicular to the plane’s direction of motion, the plane will move in a clockwise or counterclockwise direction.
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&#Good responses. Let me know if you have questions. &#
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phy 121
Your 'cq_1_24.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_24.2_labelMessages **
A steel ball of mass 60 grams, moving at 80 cm / sec, collides with a stationary marble of mass 20 grams. As a result of the collision the steel ball slows to 50 cm / sec and the marble speeds up to 70 cm / sec.
• Is the total momentum of the system after collision the same as the total momentum before?
answer/question/discussion: ->->->->->->->->->->->-> :
The total momentum before colision is equal to the total momentum after. We can verify this by setting the total momentum equal to each other
M1v1+m2v2 = m1v1`+m2v2`
(60g)(80cm/s) + (20g)(0) = (60g)(50cm/s) + (20g)(70cm/s)
4800g*cm/s + 0 = 3000g*cm/s + 1400g*cm/s
4800g*cm/s = 4400g*cm/s
Here the momentum is not the same, although in some previous notes i recall seeing that the total momentum before is suppossed to equal the total momentum after. Im also aware there is a circumstance in which we are required to divide by the total mass of the system, but if this is required here im not sure where it is used.
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• What would the marble velocity have to be in order to exactly conserve momentum, assuming the steel ball's velocities to be accurate?
answer/question/discussion: ->->->->->->->->->->->-> :
If the marble’s velocity after collision was 90cm/s this would give us a momentum of 1800g*cm/s + 3000g*cm/s = 4800g*cm/s. I was able to discover this intuitively but there is certainly a way in which the equations can be set up to achieve specifically this result.
M1v1+m2v2 = m1v1`+m2v2`
If we call v2` X and solve we should be able to get the right answer. Subtracting m1v1` we get
M1v1+m2v2-m1v1` = m2v2`
Divde by m2` to achieve
V2 = m1v1+m2v2-m1v1` / m2
V2 = (4800g*cm/s) - (3000g*cm/s) / (20g)
V2 = 1800 / 20
V2 = 90
So this method is then appropriate for finding the desired answer, and it appears still safe to say that the total momentum before is equal to the total momentum after, although i am still curious as to where dividing by the total mass comes into play.
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*#&!
&#Your work looks very good. Let me know if you have any questions. &#