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Linear momentum

  • Thread starter louie3006
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  • #1
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Homework Statement


a simple pendulum consist of 1.50Kg mass connect to a cord without any mass or friction. initially the pendulum is vertically positioned when a 2.00Kg mass collides with it, causeing the pendulum to displace vertically upward 1.25m. after the collision, the 2.00Kg mass travels along the frictionless horizontal surface until it meets a 30.0˚ incline with coefficient of kinetic friction of .400. if the mass travels a maximum distance of 1.125m up the incline, determine the initial velocity that the 2.00Kg mass strikes the pendulum with.


Homework Equations


Pο = P
mVο+ MVο = MV +mv


The Attempt at a Solution

 

Answers and Replies

  • #2
Kurdt
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Where is your attempt?
 
  • #3
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I don't know where to start?
 
  • #4
Hootenanny
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I don't know where to start?
I think that conservation of energy might me a more appropriate concept to use here.

HINT: How much energy does the 2.00kg mass transfer to the 1.50kg mass?
 
  • #5
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so here is what i tried, but I don't know if I'm right i just followed an example in the book that is somehow similar to this problem.
m1=2.00kg
m2= 1.50kg
height to which the block and pendulum swings hf=1.25
(m1+m2)Vf= m1Vο + 0
Vο = (m1+m2/m1)Vf
Vο= (m1+m2/m1) √ghf
Vο = (2.00kg + 1.50kg / 2.00kg)√2(9.80m/s^2)(1.25m)
Vο = 8.66M/s?


http://s44.photobucket.com/albums/f6/kingLA/?action=view&current=04132009360-1.jpg
 
Last edited:
  • #6
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I think that conservation of energy might me a more appropriate concept to use here.

HINT: How much energy does the 2.00kg mass transfer to the 1.50kg mass?
PE=mgh= (2.00kg)(9.80m/s^2)(1.25m) =24.5J
 
  • #7
Hootenanny
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PE=mgh= (2.00kg)(9.80m/s^2)(1.25m) =24.5J
Does the 2kg mass travelling 1.25m upwards?
 
  • #8
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well, what happens is that the 2.00Kg hits the 1.50Kg mass which cause it to swing and then keeps moving straight forward till it reaches 30 degree incline and travels a distance of 1.125m
I posted a link that has the diagram up above if you like to look at it.
 
  • #9
Hootenanny
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well, what happens is that the 2.00Kg hits the 1.50Kg mass which cause it to swing and then keeps moving straight forward till it reaches 30 degree incline and travels a distance of 1.125m.
So if the 1.50kg mass swings upwards, why have you used 2.00kg in your calculation of the potential energy?
 
  • #10
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I was wondering if anyone ever figured this one out completely? I think that if you get the initial velocity of the 2.00kg mass at the base of the incline then that can be used as the final velocity after it strikes the 1.50 mass, and so can be used in this equation
MVinitial + mVinitial = MVfinal + mVfinal
Is this correct thinking?
 

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