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Conservation of momentum (with conservation of energy)

 
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Nov5-07, 12:45 PM   #1
 

Conservation of momentum (with conservation of energy)


1. The problem statement, all variables and given/known data
A 17.00kg sphere is hanging from a hook by a thin wire 3.60m long, it is free to swing in a complete circle. Suddenly it is struck horizontally by a 6.00kg dart that embeds itself in the sphere. What is the minimum initial speed of the dart so the combination makes a complete circular loop after the collision?

2. Relevant equations
[tex]F_c\mbox{top of loop} = T+m_c g\\
E_k=frac{mv^2}{2}\\
E_g=mg\Delta h[\tex]


3. The attempt at a solution
So basically what i did was figured out the velocity at the top of the loop needed to produce 0 tension since the object must make a full loop and not just get to the top.

i did this by setting the centripetal force equal to the force of gravity and the tension
since we know tension is equal to 0, the centripetal force is then equal to the gravitational force acting on the object. the masses cancel and this gives....
[tex]frac{mv^2}{r}=mg\\v=sqrt{gr}[\tex]

we know both r and g so we can then calculate the velocity at the top of the circle needed for 0 tension in the line to be 5.943m/s. Using this, we can then calculate the total energy of the particle at that point (knowing the kinetic and gravitational potential energy)
[tex]E_total = frac{mv^2}{2} + m g \delta h[\tex] (energy at the top of the loop)

this total energy if my calculations are right is equal to 2436.87 Joules. Since theres no way the system could have lost energy over the swing, i presumed that at the bottom the energy was the same(without the gravitational potential energy)... this means that the total energy must be equal to only the kinetic energy of the combined particle and dart at the bottom. so you set kinetic energy equal to 2436.87Joules and solve for the velocity of the particle at the bottom.

i know know the speed at which the particle must be traveling at the bottom. so in order to find the speed at which the dart must hit that particle (and then embed itself in) becomes a simple conservation of momentum question. where the mass of the dart and its velocity (the unknown) is the only momentum in the system prior to impact, and the only momentum after impact is the combined mass of the dart and sphere multiplied by the initial velocity needed to overcome the loop. i get an answer of 55.8m/s for the dart... but its wrong. where and how did i go wrong? sorry for the length of this post.
 
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Nov5-07, 04:35 PM   #2
 
bump :(... anybody?
 
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