How Does Hammer Impact Affect Pole Penetration and Energy Dissipation?

In summary: The problem states that the collision is perfectly inelastic. What does that tell you about the masses after...In summary, the collision is perfectly inelastic, so the resistance to the penetration is 84.9 N.
  • #1
notsoclever
8
0
Hello everyone, here's the problem

Homework Statement


A pole with mass m2 = 2 Kg is planted on the ground with a consecutive hammer hits. The hammer ha a mass m1 = 4 kg and is dropped from 1 meter upon the pole. With a single hit the pole penetrate the ground of 2 cm.
Find:
a) the total resistence R to the penetration, supposing constant that R is costant for each hit and the collision is perfectly inelastic.
b)The dissipation of energy after the collision

Homework Equations


Momentutm
Conservation of energy

The Attempt at a Solution


I've found the velocity of the hammer just before the collision using the conservation of energy, that is 4.4 m/s.
Using the momentum consrvation and knowing that it is an perfectly inelastic collision I've found that the velocity of the sistem after the hammer hit is 2.9 m/s.
I think till now there is anything to say (If I made any mistakes please tell me).
After that I've though to find the acceleration of the pole knowing that the penetration is 0.02 meters and that the final velocity is 0, using this formula [tex]a=(V_0)^2/2S[/tex] where [tex]V_0 = 2.9[/tex].
Knowing that the [tex]F=ma[/tex] I've thougt to find the resistance R but the result is wrong.
Can you help me please? (The result of R should be 84.9 N)
 
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  • #2
notsoclever said:
Hello everyone, here's the problem

Homework Statement


A pole with mass m2 = 2 Kg is planted on the ground with a consecutive hammer hits. The hammer ha a mass m1 = 4 kg and is dropped from 1 meter upon the pole. With a single hit the pole penetrate the ground of 2 cm.
Find:
a) the total resistence R to the penetration, supposing constant that R is constant for each hit and the collision is perfectly inelastic.
b)The dissipation of energy after the collision

Homework Equations


Momentutm
Conservation of energy

The Attempt at a Solution


I've found the velocity of the hammer just before the collision using the conservation of energy, that is 4.4 m/s.
Using the momentum consrvation and knowing that it is an perfectly inelastic collision I've found that the velocity of the system after the hammer hit is 2.9 m/s.
I think till now there is anything to say (If I made any mistakes please tell me).
After that I've though to find the acceleration of the pole knowing that the penetration is 0.02 meters and that the final velocity is 0, using this formula [tex]a=(V_0)^2/2S[/tex] where [tex]V_0 = 2.9[/tex].
Knowing that the [tex]F=ma[/tex] I've thougt to find the resistance R but the result is wrong.
Can you help me please? (The result of R should be 84.9 N)
That looks like the correct approach.

You don't show your results for the last two parts.

What did you get for acceleration?

What did you use for the mass in the last step? What did you get for the resisting force?
 
  • #3
To find the acceleration I've used the equations for decelerated motion:
[tex]v = v_0 - at[/tex] and [tex]s=v_0t - \frac{1}{2}a t^2[/tex]
expressing the time t from the first equation, the second equation becomes:
[tex]s=\frac{v_0^2}{2a}[/tex]
with a inverse function the acceleration is equal to
[tex]a=\frac{v_0^2}{2s}[/tex]
The value computed is 217.78 m/s^2
Using the second principle of dynamics [tex]F=ma[/tex] and using only the mass of the pole, the force is about F = 434 N
I thought the resistence should be equal to this force but as I told the result is 84.9 N.

Do you think I did any other mistakes?
 
  • #4
I suspect that the given answer is incorrect. I can't see how a force of about 85 N could stop the masses in such a short distance. The force of gravity alone would eat up nearly 60 N of it, leaving only about 25 N to slow them. Your acceleration of 217.78 m/s2 is about 22g's, so expect something around 22 x 60 N as a ballpark figure.
 
  • #5
Well, that value for the acceleration is the value I got, but I'm not completely sure It's correct... what do you think about my resoning?
 
  • #6
notsoclever said:
Well, that value for the acceleration is the value I got, but I'm not completely sure It's correct... what do you think about my resoning?
Your acceleration value looks reasonable to me. I think you should draw a free body diagram for the situation after the collision, while the masses are decelerating. As I mentioned, the force due to gravity plays a role in how much force needs to be applied in order to accomplish the required deceleration.
 
  • #7
notsoclever said:
F=ma and using only the mass of the pole,
Why only the mass of the pole?
 
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  • #8
Because I thought that the penetration is due only to the mass of the pole... What do you think about that?
 
  • #9
notsoclever said:
Because I thought that the penetration is due only to the mass of the pole... What do you think about that?
The problem states that the collision is perfectly inelastic. What does that tell you about the masses after collision?
 
  • #10
notsoclever said:
Because I thought that the penetration is due only to the mass of the pole... What do you think about that?
What will happen to the hammer after impact? Does it disappear? Move more slowly than the pole?
 
  • #11
The masses after the collision move with the same velocity, but I don't know the direction.
I supposed the direction was opposite one to each other and I decide to use just the mass of the pole to compute the force.
 
  • #12
I think you need to review the details of what constitutes a perfectly inelastic collision. Look it up.
 
  • #13
gneill said:
I think you need to review the details of what constitutes a perfectly inelastic collision. Look it up.
Ok! You're right! The objects stick together after a perfectly inelastic collision. So the mass to consider is 6 kg istead of 2.
This leads to a greater force of 1302 N... I'm going to draw a free body diagram and post it as soon as I can because I'm not figuring it out :-(
 
  • #14
notsoclever said:
The masses after the collision move with the same velocity, but I don't know the direction.
I supposed the direction was opposite one to each other and I decide to use just the mass of the pole to compute the force.
That would violate the conservation of momentum equation you used. Momentum is a vector, so the direction of travel matters.
 

Related to How Does Hammer Impact Affect Pole Penetration and Energy Dissipation?

What is a collision?

A collision is an event in which two or more objects come into contact with each other. It can be either an elastic collision, where there is no loss of energy, or an inelastic collision, where some energy is lost.

What is kinetic energy?

Kinetic energy is the energy possessed by an object due to its motion. It is calculated as half of the mass of the object multiplied by its velocity squared.

How is energy conserved in a collision?

In a closed system, the total energy before a collision is equal to the total energy after the collision. This means that energy is conserved in a collision, even if it is transferred from one object to another.

What factors affect the outcome of a collision?

The outcome of a collision is affected by factors such as the masses and velocities of the objects involved, as well as the type of collision (elastic or inelastic).

What is the difference between elastic and inelastic collisions?

In an elastic collision, there is no loss of kinetic energy, and the objects bounce off each other. In an inelastic collision, some kinetic energy is lost, and the objects stick together after the collision.

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