Conservation of energy of bullet into sandbag

In summary: I think for c) you should talk about energy rather than momentum.You need to:-explore the law of conservation of energy-discover all forms of energy that were involved in this question-and finally answer the question of why the system did not lose any energy, or why the total energy stayed constant.In summary, the given question involves a ballistic pendulum consisting of a bullet and a bag of sand, with the bullet being fired into the sandbag at 840 m/s. By calculating the mechanical energy before and after the collision, it is determined that 5103.73 J is converted into thermal energy. This example demonstrates the application of the first law of thermodynamics, as energy both enters and leaves the system. Additionally
  • #1
pharm89
32
0

Homework Statement



Hi I am taking a self study physics course and not sure about the following question. Any help would be much appreciated. Thanks

1. A 15.0 g bullet is fired at 840 m/s into a ballistic pendulum consisting of a 24.0 kg bag of sand. If the bag rises to a height of 80.0 cm, how much energy is converted into thermal energy in the bag?

Homework Equations





The Attempt at a Solution



1. Given: m = 24.0 kg
h= 80 cm or 0.08 m
v= 840 m/s

Required: thermal energy

Analysis: Determine total mechanical energy as the bag of sand rises and falls.

Bag of sand at the top
mgh
(24.0 kg)(9.8 m/s^2)(0.08 m )
=18.81 J

At the bottom
Ek + Eg
1/2(24.0 kg ) (0.08m)
=0.96
Total mechanical energy = 18.81 J - 0.96 = 17.85 J

I am kind of lost with this question and not sure of the right procedure to follow.
Thanks for the assistance.

 
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  • #2
Think of this problem as having two parts:
(1) The collision of bullet with sandbag, which is perfectly inelastic.
(2) The rising of the bullet + sandbag, after the collision.

Hint: Different quantities are conserved in each part.
 
  • #3
I wrote a reply and deleted it because I thought I had misunderstood the question, but I don't think I misunderstood anything that matters, so here's my reply again (a bit better than before):

kinetic energy of bullet = increase of gravitational potential energy of bag with bullet inside + heat

You can calculate the potential energy increase by calculating the work it would take to lift the bag and a bullet 80 cm straight up. The contribution from the mass of the bullet may not be significant, but you shouldn't just assume that it isn't.

And 80 cm is 0.8 m, not 0.08 m.
 
  • #4
Doc Al said:
Think of this problem as having two parts:
(1) The collision of bullet with sandbag, which is perfectly inelastic.
(2) The rising of the bullet + sandbag, after the collision.

Hint: Different quantities are conserved in each part.

Thanks for the help. Am I understanding the question any better.
I tried the problem again and did calculations for both parts.

1. the collision of the bullet and the sandbag
Mechanical Energy - Ek + eg
= Eh since h = 0
1/2 (0.015 kg) (840 m/s)
=1/2 (12.6)
= 6.3 J

2. The rising of the bullet and the sandbag
Mechanical Energy = Ek + Eg
=mgh, since v=0
=(0.015kg (bullet) + 24 kg (sandbag))(9.8 m/s^2)(80.0 cm)
=282.24 J
Therefore, total change in mechanical energy = 282.29 J-6.3 J = 274 J
??:))
Thanks
Pharm 89
 
  • #5
It looks to me like you understand that you need to compare the energy before the collision of bullet with sandbag (which can be calculated directly) to the energy after the collision (which can be calculated from the height the sandbag rises). That's the right approach, but you made a few mistakes.

1. the collision of the bullet and the sandbag
Mechanical Energy - Ek + eg
= Eh since h = 0
1/2 (0.015 kg) (840 m/s)
=1/2 (12.6)
= 6.3 J
Here you attempted to calculate the energy of the bullet: Good! But your formula for kinetic energy is incorrect--the speed must be squared. Do this step over.

2. The rising of the bullet and the sandbag
Mechanical Energy = Ek + Eg
=mgh, since v=0
=(0.015kg (bullet) + 24 kg (sandbag))(9.8 m/s^2)(80.0 cm)
=282.24 J
Here you attempted to calculate the potential energy of the sandbag + bullet: Good. But you made a few errors: (1) express the height in meters, not cm; (2) recheck your arithmetic.

Sanity check: The final energy can't be greater than the initial energy--that should have tipped you off that something was wrong somewhere. :wink:
 
  • #6
Doc Al said:
It looks to me like you understand that you need to compare the energy before the collision of bullet with sandbag (which can be calculated directly) to the energy after the collision (which can be calculated from the height the sandbag rises). That's the right approach, but you made a few mistakes.


Here you attempted to calculate the energy of the bullet: Good! But your formula for kinetic energy is incorrect--the speed must be squared. Do this step over.


Here you attempted to calculate the potential energy of the sandbag + bullet: Good. But you made a few errors: (1) express the height in meters, not cm; (2) recheck your arithmetic.

Sanity check: The final energy can't be greater than the initial energy--that should have tipped you off that something was wrong somewhere. :wink:

True, of course that would not be right:))) Thanks for the tips.
The collision of the bullet and the sandbag
1/2(0.015 kg) (840m/s)^2
= 1/2 (0.015kg) (705600)
=5292 J
The rising of the bullet and sandbag
Emechanical = mgh
=(0.015 kg (bullet) + 24 kg (sandbag))(9.8m/s^2)(0.8m)
=(24.015kg)(7.84)
=188.27 J
therefore, 5292 j - 188.27 = 5103.73 J is converted into thermal energy.
 
  • #7
Now you've got it. :approve:
 
  • #8
Doc Al said:
Now you've got it. :approve:

Thanks for the help. The question makes much more sense now.

There are two other parts to that question which are theory based.
They are:
b) Explain 1 application of the first law of thermodynamics in this example.

My answer: Energy both enters and leaves this system. The thermal energy leaves this system as heat to the surroundings. The system does work on its surroundings, that is, energy leaves the system and is converted to other forms outside the system.

(c) Explain the law of conservation of energy as it applies to this example.

My answer: a resource is needed (energy) to increase the height of the pendulum. The bag of sand absorbs the bullets momentum and converts it into upward movement, therefore the total momentum of a system is conserved.

Am i on the right track? Thanks :0
 
  • #9
I think for c) you should talk about energy rather than momentum.
You need to:
-explore the law of conversation of energy
-discover all forms of energy that were involved in this question
-and finally answer to the question that why the system did not lose any energy?, in other words why did the total energy stay constant.
 
  • #10
rootX said:
I think for c) you should talk about energy rather than momentum.
You need to:
-explore the law of conversation of energy
-discover all forms of energy that were involved in this question
-and finally answer to the question that why the system did not lose any energy?, in other words why did the total energy stay constant.

Thanks..I guess I need to break down the question into different parts.

-exploring the law of conservation of energy
This law states that the total amount of energy in the universe remains constant.

discover all forms of energy in the question:
-Thermal energy - converted in the sandbag
-sound energy - when the bullet hits the sandbag
-kinetic energy - the time up until and after the bullet hits the sandbag
-potential energy - when the sandbag rises to a height of 80.0 cm

Why did the total energy stay constant
-The system changes forms of energy however before and after the bullet hits the sandbag the total amount of energy remains constant.

Are these correct interpretations. Thanks for the help.
Pharm 89
 

Related to Conservation of energy of bullet into sandbag

1. How does conservation of energy apply to a bullet entering a sandbag?

When a bullet enters a sandbag, the kinetic energy of the bullet is converted into other forms of energy, such as heat and sound. However, the total amount of energy remains the same, following the principle of conservation of energy.

2. Does the mass of the bullet affect the conservation of energy?

Yes, the mass of the bullet does affect the conservation of energy. According to the law of conservation of energy, the total energy before and after the collision remains the same. Therefore, the more massive the bullet, the more energy it will transfer to the sandbag upon impact.

3. How does the speed of the bullet impact the conservation of energy?

The speed of the bullet also plays a role in the conservation of energy. The kinetic energy of an object is directly proportional to its mass and the square of its velocity. This means that a faster-moving bullet will have more kinetic energy and will transfer more energy to the sandbag upon impact.

4. What other factors influence the conservation of energy in this scenario?

In addition to the mass and speed of the bullet, other factors that can impact the conservation of energy in this scenario include the composition and density of the sandbag, the angle and trajectory of the bullet, and any external forces acting on the system.

5. Is conservation of energy the only principle at play in this scenario?

No, conservation of energy is not the only principle at play in this scenario. The conservation of momentum also plays a role, as the momentum of the bullet is transferred to the sandbag upon impact. Additionally, the law of inertia applies, stating that an object in motion will remain in motion unless acted upon by an external force.

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