Conservation of energy of bullet into sandbag

1. Apr 28, 2007

pharm89

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 28, 2007

Staff: Mentor

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. Apr 28, 2007

Fredrik

Staff Emeritus
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. May 5, 2007

pharm89

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. May 5, 2007

Staff: Mentor

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.

6. May 5, 2007

pharm89

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. May 6, 2007

Staff: Mentor

Now you've got it.

8. May 6, 2007

pharm89

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. May 6, 2007

rootX

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. May 22, 2007

pharm89

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