How Fast Must the Hammer Move to Ring the Bell at the Carnival?

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Homework Help Overview

The problem involves determining the speed at which a hammer must strike a target to send a metal piece upward to ring a bell at a carnival. The scenario includes a 9.00 kg hammer and a 0.400 kg metal piece, with the bell located 5.00 m above the ground. The problem states that 25% of the hammer's kinetic energy is used to lift the metal piece.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the energy required to raise the metal piece and the expression for kinetic energy (KE). There is an exploration of how to incorporate the 25% energy usage into the equations. Questions arise regarding the variables involved and the interpretation of initial and final heights.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the relationships between the kinetic and potential energy. Some guidance has been provided regarding the algebraic manipulation of the equations, but there is still uncertainty about the definitions of variables and the setup of the problem.

Contextual Notes

Participants are working under the constraints of the problem statement, including the specific masses and the height to which the metal piece must be raised. There is a focus on ensuring the correct interpretation of the kinetic energy and its application in the context of the problem.

Parzival
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Homework Statement


At a carnival, you can try to ring a bell by striking a target with a 9.00 kg hammer. In response, a 0.400 kg metal piece is sent upward toward the bell, which is 5.00 m above. Suppose that 25.0% of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?


Homework Equations



W = final mechanical energy - initial mechanical energy

1/2m*final velocity^2 + mg*finalheight = 1/2m*initial velocity^2 + mg*initial height


Conservation of mechanical energy



The Attempt at a Solution


I tried setting up an equation using the second given equation, but unfortunately there are too many variables.
 
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Start simple, how much energy will it take to raise 0.4 kg mass to a height of 5 m?

What is the expression for KE?
 
rock.freak667 said:
Start simple, how much energy will it take to raise 0.4 kg mass to a height of 5 m?

What is the expression for KE?

KE = 1/2mv^2;

or KE = mg(hf - h0)
 
Parzival said:
KE = 1/2mv^2;

or KE = mg(hf - h0)

Right but they said that 25% of the hammer's KE is used to do the work, so how should you incorporate this into the equation?
 
rock.freak667 said:
Right but they said that 25% of the hammer's KE is used to do the work, so how should you incorporate this into the equation?

KE = 1/8mv^2

or KE = 1/4mg(hf-h0)
 
Parzival said:
KE = 1/8mv^2

Correct.

So now you have 1/8mv2=Mg(hf-h0)

You know m, M,g, hf and h0, solve for v.
 
rock.freak667 said:
Correct.

So now you have 1/8mv2=Mg(hf-h0)

You know m, M,g, hf and h0, solve for v.

So: v^2 = 8mMg(hf - h0)

v = sqrt(8mMg(hf-h0))

but what is h0? Do I assume it is 0 m?

And there are two masses: the 0.400 kg one and the 9.00 kg one.
 
Parzival said:
So: v^2 = 8mMg(hf - h0)

Recheck your algebra on this one, you will need to divide by something.


Parzival said:
but what is h0? Do I assume it is 0 m?

Yes it is zero.


Parzival said:
And there are two masses: the 0.400 kg one and the 9.00 kg one.

Right, which mass has the kinetic energy? Which mass will have the potential energy gain?
 
rock.freak667 said:
Recheck your algebra on this one, you will need to divide by something.




Yes it is zero.




Right, which mass has the kinetic energy? Which mass will have the potential energy gain?

Sorry, I failed miserably. The equation is

sqrt(8Mg(hf-h0)/ m)

Let me guess. Plug in the numbers; then, multiply this by four to get the original KE, and then solve for the final velocity.
 
  • #10
Parzival said:
Sorry, I failed miserably. The equation is

sqrt(8Mg(hf-h0)/ m)

Let me guess. Plug in the numbers; then, multiply this by four to get the original KE, and then solve for the final velocity.

Well you already have the formula for the final velocity v.

v = \sqrt{\frac{8Mg(h_f -h_0)}{m}}

So just input all the numbers.
 
  • #11
Sorry, I forget. What does capital M represent? I know lowercase m represents the mass, right?
 
  • #12
Parzival said:
Sorry, I forget. What does capital M represent? I know lowercase m represents the mass, right?

Both are mass. One is just the mass of the hammer and the other is the mass of the piece of metal.
 

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