Coservation of energy and impulse

In summary, a stream of elastic glass beads with a mass of 0.46 g each is released at a rate of 108 per second from a horizontal tube. The beads fall 0.54 m to a balance pan and bounce back to their original height. To keep the pointer at zero, a mass of 10.227 g must be placed in the other pan of the balance to balance the force of 0.323 N created by the falling beads. To solve for this mass, the conservation of energy is used to find the velocity of the beads, which is then used in the momentum calculation.
  • #1
gongshow29
10
0

Homework Statement


A stream of elastic glass beads, each with a mass of 0.46 g, comes out of a horizontal tube at a rate of 108 per second. The beads fall a distance of 0.54 m to a balance pan and bounce back to their original height. How much mass must be placed in the other pan of the balance to keep the pointer at zero?

Homework Equations


1/2mv^2

The Attempt at a Solution


What I attempted was solving for the velocity of the ball as it hits the pan, then I multiplied by the mass, but I am not sure of how to account the rate.
 
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  • #2
Rate is necessary as force is momentum change per unit time. You need not think in terms of energy rather think of force acting on the pan as momenta lost by the beads PER UNIT TIME.
 
  • #3
The force on the pan will be 2*n*d/dt(m*v)
n being the number of beads hitting the pan per unit time.
v being the VERTICAL velocity of a bead as it hits the pan.
 
  • #4
n is the rate here so don't you think d/dt(mv) is misleading?
 
  • #5
Yes it is. It should be 2*n*m*v?
 
  • #6
Yes.
 
  • #7
Would I just use the conservation of energy to solve for the velocity then? Or momentum?
 
  • #8
Use conservation of energy to find the velocity. Then use the velocity in the momentum calculation.
 
  • #9
so I would use the following setup mgh=1/2mv^2 correct?
I calculated my velocity with that setup
(0.46)(9.81)(0.54)=(1/2)(0.46)v^2
ended up with 3.254...
plugged that into the formula suggested which was 2(n)(m)(v):
2(108)(0.46)(3.254)=323g, but that is giving me the incorrect answer?
Where am I going wrong in my calculations?
 
Last edited:
  • #10
0.46g is in g; should be in kg.
 
  • #11
Ok, I switched out 0.46g into 0.00046kg, did the same calculation ended up with .010227kg, and my answer wants it in grams, so I multiplied it by 1000, for 10.227g, and its still incorrect. I am really confused now.
 
  • #12


Ok, first of all, you should get a value of 0.323, but that is in Newtons.
Find the mass that is needed to balance that force.
 
  • #13
Ok that makes sense, appreciate everyone's help.
 

FAQ: Coservation of energy and impulse

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, it can only be transferred from one form to another. This means that the total amount of energy in a closed system remains constant over time.

2. How is the conservation of energy related to impulse?

The conservation of energy is related to impulse through the principle of conservation of momentum. Impulse is the change in momentum of an object, and momentum is directly related to an object's mass and velocity. When there is no external force acting on a system, the total energy and momentum of the system will remain constant.

3. What is an example of conservation of energy and impulse in everyday life?

A classic example of conservation of energy and impulse is a pendulum. When the pendulum is released, it swings back and forth with a constant total energy and momentum. The potential energy at the top of the swing is converted into kinetic energy as it moves towards the bottom, and then back into potential energy as it swings back up.

4. How does the conservation of energy and impulse apply to collisions?

In collisions, the total momentum and energy of the system is conserved. This means that in a perfectly elastic collision, where there is no loss of energy, the total energy before and after the collision will be the same. In a perfectly inelastic collision, where there is energy lost to heat or sound, the total energy after the collision will be less than the total energy before.

5. What are the limitations of the law of conservation of energy and impulse?

The law of conservation of energy and impulse has some limitations in certain situations, such as when dealing with systems that involve nuclear reactions or extreme gravitational forces. In these cases, the law may not hold true and other factors, like mass-energy equivalence or general relativity, must be considered.

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