Shovel Nyan, Much appreciated. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
The string has a horizontal force working on the bullet during the swing. I think that’s why horizontal momentum is not conserved. I hope I understand it correctly.
Aww this is great, I love this problem. So, okay; lets say that instead of pendulum, the bullet hits a block attached to a spring. We can use the same set up by replacing mgh with 1/2 kx^2 (spring potential energy) while the external force causing the loss of momentum is the frictional force, right? *finger's crossed*
Jossy, Correct, mostly. Momentum is conserved when the bullet digs into the block. The bullet stops due to friction between the wood and the bullet. Of course, mechanical energy is not conserved here (you put a big pit into the wood and this took energy), but momentum is conserved. Now the block is moving and starts to compress the spring. At this point you conserve energy, so the kinetic energy of the block turns into potential energy of the spring once it's completely compressed. For this part, assume no friction. So you are correct that you can replace mgh with 1/2 kx^2 and you'll get the right answer, but I wanted to comment on your language. Cheers, Dr. A p.s. You may uncross your fingers now.
I don't know about this, as I believe some energy from the bullet will be converted to sound and heat, so therefore it wouldn't completely be conservation law of energy. That is why DR. A split it into two parts because the initial part where the bullet and block are separate will cause an inelastic collision, so momentum should instead be used. But I might be wrong.
Kinetic energy is not conserved during the inelastic collision because mechanical energy is lost when the bullet penetrates the wood. momentum is not conserved during the blocks movement upward because the force of gravity is acting on it. So the velocity of the bullet cannot be directly determined through by equating these end points. We instead use the point after impact because once the collision has ended the kinetic energy of the system is conserved. Once we know the velocity of that block, we can use the conservation of momentum to find the velocity of the bullet because momentum is conserved during the collision. I too am learning
well, if you set the first set of equations to solve for V1, then you don't have enough information to solve for V1. V2 would be an unknown value. However if you go the other way around like what he did, through second set of equations (E conserved), you have another equation for V2 which you could plug in the values and solve for V1. I mean hopefully I was answering your question. This is what I thought. Im learning here too :)
Searam Park is correct. It all depends on what you're given to start off with. With the first equation, you can solve for either V1 or V2, and likewise plugin for whichever you're trying to solve for in the second equation. In this example, he chose to solve for the initial velocity of the bullet, very much like most homework and test questions will ask for and hence "ballistic".
What if the Block was not selected, the engineer or physicist did not run the numbers and select the proper length, thickness, hardness, mass of the wood engineered to match the parameters of the velocity and mass of the bullet to absorb it, and it just disintegrates? Lol...
hey Matt, this has been very helpful.
I think the smart glass works for me.
also, it's very cool when the students participate.
cheers.
This is very detailed and understandable, musch respect for you prof
Shovel Nyan,
Much appreciated.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Thank you for explaining this so well! Glad to have you as a teacher.
exhilir,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
Amazing explanation man🔥😃
BEST PHYSICS LECTURES EVER !!!
Thank you doctor strange.
The string has a horizontal force working on the bullet during the swing. I think that’s why horizontal momentum is not conserved. I hope I understand it correctly.
the best tutor ever........ much respect........
Thank you. I'm glad to hear you're finding these useful.
Keep up with the physics.
Cheers,
Dr. A
Aww this is great, I love this problem. So, okay; lets say that instead of pendulum, the bullet hits a block attached to a spring. We can use the same set up by replacing mgh with 1/2 kx^2 (spring potential energy) while the external force causing the loss of momentum is the frictional force, right? *finger's crossed*
Jossy,
Correct, mostly.
Momentum is conserved when the bullet digs into the block. The bullet stops due to friction between the wood and the bullet. Of course, mechanical energy is not conserved here (you put a big pit into the wood and this took energy), but momentum is conserved.
Now the block is moving and starts to compress the spring. At this point you conserve energy, so the kinetic energy of the block turns into potential energy of the spring once it's completely compressed. For this part, assume no friction.
So you are correct that you can replace mgh with 1/2 kx^2 and you'll get the right answer, but I wanted to comment on your language.
Cheers,
Dr. A
p.s. You may uncross your fingers now.
Great video! 👍
Things make so much more sense years later...
An easier way would be to set the initial kinetic energy of the bullet itself equal to the final gravitational potential energy of the bullet + block.
I don't know about this, as I believe some energy from the bullet will be converted to sound and heat, so therefore it wouldn't completely be conservation law of energy. That is why DR. A split it into two parts because the initial part where the bullet and block are separate will cause an inelastic collision, so momentum should instead be used. But I might be wrong.
When the bullet enters the block would it be a perfectly inelastic collision not an inelastic collision?
Bullet entering block is certainly an "inelastic" collision because it has to push wood fibers out of the way, which takes energy.
Cheers,
Dr. A
What I'm wondering is, why doesn't he solve for V1 just using the momentum equation directly instead of solving for V2 ?? (Someone please answer)
Kinetic energy is not conserved during the inelastic collision because mechanical energy is lost when the bullet penetrates the wood. momentum is not conserved during the blocks movement upward because the force of gravity is acting on it. So the velocity of the bullet cannot be directly determined through by equating these end points. We instead use the point after impact because once the collision has ended the kinetic energy of the system is conserved. Once we know the velocity of that block, we can use the conservation of momentum to find the velocity of the bullet because momentum is conserved during the collision. I too am learning
well, if you set the first set of equations to solve for V1, then you don't have enough information to solve for V1. V2 would be an unknown value. However if you go the other way around like what he did, through second set of equations (E conserved), you have another equation for V2 which you could plug in the values and solve for V1. I mean hopefully I was answering your question. This is what I thought. Im learning here too :)
Searam Park is correct. It all depends on what you're given to start off with. With the first equation, you can solve for either V1 or V2, and likewise plugin for whichever you're trying to solve for in the second equation. In this example, he chose to solve for the initial velocity of the bullet, very much like most homework and test questions will ask for and hence "ballistic".
how does he write backwards so well
He doesn't. He cheats. Answer here: www.learning.glass
Cheers,
Dr. A
What if the Block was not selected, the engineer or physicist did not run the numbers and select the proper length, thickness, hardness, mass of the wood engineered to match the parameters of the velocity and mass of the bullet to absorb it, and it just disintegrates? Lol...
I think then it's tough to do the same measurement.
Cheers,
Dr. A
thank u very much this was very helpful
Thank-you sir for very nice explanation
Student of India😂😂
Why does energy conserves?
Do you write backwards???????
Can anyone help me on this experiment? Attach contact info
Are you writing backwards?
Nope. Mirror image of the writing.
Cheers,
Dr. A
Am I the only one blown away by how he is able to write everything backwards so easily??
No, but sadly not true either. See:
ua-cam.com/video/CWHMtSNKxYA/v-deo.html
Cheers,
Dr. A
are you happy that you spent 3.3k on a piece of glass
the squeaking of the marker is horrible
Christine,
Agreed. We are fixing it on my new website: www.universityphysics.education
Cheers,
Dr. A