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Laws of motion problem

  • Thread starter coldblood
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  • #1
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Hi friends, The problem is from Newton's Laws.

The problem is as follows:
https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash4/1006364_1417581381802301_666606151_n.jpg

IInd law states,
Force, F = dP/dt
=> F = d(mv)/ dt

Out come,

If m is constant, v is variable, F = m.[d(v)/dt] => F = m.a

If v is constant, m is variable, F = v.[d(m)/dt] => F = v.[rate of change of mass]

If both m and v are variable, F = m.[d(v)/dt] + v.[d(m)/dt]

Hence the answer of the question should be Option (B). But the book states that answer is option (C) is correct. How is it so.

Please friends help me in solving this issue.

Thank you all in advance. I would appreciate the help.
 

Answers and Replies

  • #2
[itex]\vec{v}[/itex] is a variable. It is the instantaneous velocity at all times. So [itex]\vec{a}[/itex]=[itex]\frac{d\vec{v}}{dt}[/itex]. [itex]\lambda[/itex] is the constant rate at which sand is leaking out. So the amount of mass that would be lost after a period t would be [itex]\lambda[/itex]t.
 
  • #3
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[itex]\vec{v}[/itex] is a variable. It is the instantaneous velocity at all times. So [itex]\vec{a}[/itex]=[itex]\frac{d\vec{v}}{dt}[/itex]. [itex]\lambda[/itex] is the constant rate at which sand is leaking out. So the amount of mass that would be lost after a period t would be [itex]\lambda[/itex]t.
So why not Option (B) is correct?
 
  • #4
In your last step where you assume that both m and v are variable you should actually use partial differentiation, not ordinary differentiation.
 
  • #5
arildno
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Rewrite your original post with non-psychotic letters, please.
 
  • #6
In applying the product rule the first mass will also be m not mo as in answer B.
 
  • #7
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I did that problem over and over and concluded that option C is correct and please, I half had a seizure reading the 1st post :/
 
  • #8
133
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Thank you all friends. The problem has been cleared. A apologize for the bad fonts.
 

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