How is Thermal Energy Converted in a Ballistic Pendulum?

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A 15g bullet traveling at 840m/s strikes a ballistic pendulum with a 24kg sandbag, causing the bag to rise 80cm. The initial mechanical energy of the bullet is calculated to be approximately 5292J, while the potential energy at maximum height is about 188.3J. The difference, around 5103.7J, represents the thermal energy converted during the collision. The first law of thermodynamics can be applied by relating the internal energy of the sandbag to the work done on it by the bullet. Additionally, the kinetic energy of the bullet transforms into heat and gravitational potential energy as the system reaches its maximum height.
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a) A 15g bullet is fired at 840m/s into a ballistic pendulum consisting of 24 kg bag of sand. |\
| \
| \ { }
| { == }
| { }
15 g ----- /\
=== | 24 kg | | 80 cm
840m/s | sand | |
----- \/

If the baf rises to a height of 80 cm, much enregy is converted into thermal energy?

b)Explain one application of the first law of thermodynamics in this example.

c) Expalin the law of conservation of energy as it applies to this example



ok so i tried this and i rlly don't have a clue i know that Energy mechanical = Epotential + Ekinetic

=mv^2/2 + mg(h)

now that's all i know how do i figure out wat is converted into thermal energy? and is thermal energy mechanical energy?
 
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That is the correct description of mechanical energy. Also, thermal energy is not a type of mechanical energy. Consider the following with this problem. You have an inelastic collision here. What do you know about conservation of MECHANICAL energy in an inelastic collision?
 
a decrese in Ek will result in a increase of Ep? Change in Em = 0? I am very lost
 
The term mechanical energy refers to the sum of the kinetic and potential energies only.

If you calculate the mechanical energy of the bullet and bag before the collision, then calculate the mechanical energy of the bullet and bag at their max height, the difference between these 2 values will be the energy lost to heat and other forms.
 
the potential energy is 0?
 
ohhh ok that makes perfect sense so find the Em at the beggening and find the Em at its max hieght then subract the 2 , and why didnt i think of that?
 
oh also wat is the mass in this question? the mass of the bag and bullet combined? or just the bag or just the bullet?
 
also wat is my speed in the Em as the bag rises? would i use 840 again as i did when the bag was stationary?
 
ohhh right my speeed would be zero because at the max hieght there is no speed
 
  • #10
ok i get around 8500J for my begginign energy and about 20J for my ending energy does that make sense that 8480J would be thermal energy?
 
  • #11
The mass you use depends on what part of the problem you are at. When you are calculating the initial mechanical energy, it is all contained in the bullet, so you would use the bullet's mass. While calculating the mechanical energy at the final height of the system, it is the bullet AND the bag that have moved to that height, so the mass used would be their combined mass. Does this answer your question on the mass?
 
  • #12
grade11 said:
ok i get around 8500J for my begginign energy and about 20J for my ending energy does that make sense that 8480J would be thermal energy?

My answer isn't that high. I am getting a lower number for my initial mechanical energy, and a higher number for my final mechanical energy. You should loose a lot of energy to thermal energy, but it's not that much. Check your numbers and units again. Don't worry, I think your on the right track here.:smile:
 
  • #13
also wat about b and c?
 
  • #14
ohh ok so the mass for the initial is just 0.015 kg ok soo there is not hieght so that's adding no potential energy so i got 0.015(840)^2/2 ok so that's 5300J while my Em will be 18.80J so that's thermal energy of like 5288 j?
 
  • #15
that sound right?
 
  • #16
srry i got to keep posting these to see if u responded or not...
 
  • #17
grade11 said:
ohh ok so the mass for the initial is just 0.015 kg ok soo there is not hieght so that's adding no potential energy so i got 0.015(840)^2/2 ok so that's 5300J while my Em will be 18.80J so that's thermal energy of like 5288 j?

I get initial Ek = (1/2)(0.015)(840^2) = 5292J
I get Em= mgh= (24.015kg)(9.8)(0.8m)= 188.3J

So thermal = 5292-188.3 = 5103.7J
 
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  • #18
ohhh right i was doing 80 centimatres was 0.08 metres hahahaha whooops simple math too ok how about b and c?? anyone can help me out?
 
  • #19
grade11 said:
ohhh right i was doing 80 centimatres was 0.08 metres hahahaha whooops simple math too ok how about b and c?? anyone can help me out?

well for b) remember that thermal energy = internal energy...

can you relate the internal energy of the sandbag to the work done by the sandbag on the bullet use the first law of thermodynamics?

c) well here just explain how the kinetic energy in the bullet transforms to heat and gravitational potential energy...
 
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