00:01
Okay, today we're going to be talking a little bit about conservation of energy.
00:07
So in this problem, we are given the mass of the bullet, the mass of the block, and the height in which the bullet and block system reaches after the collision.
00:20
So we have these things over to the right here.
00:26
And so the first part of the question asks us to find an expression of the space.
00:35
Be of the block bullet system immediately after the collision in terms of defined quantities in g.
00:41
So because of the conservation of energy, we know that the energy of the system right after the collision is going to be the same as the potential energy when it reaches a certain height of 0 .95.
00:56
So we can express that here.
01:01
The kinetic energy right after collision equals the potential energy after it reaches a defined height.
01:09
So if we write that out, we have one half mv squared equals m gh.
01:21
So the masses of the systems are going to be the same in each of these equations so we can cancel those out.
01:29
And then we just need to find an expression for the speed.
01:33
And so we can switch everything around so that we get velocity squared equals 2gh and then finally velocity equals square root to gh so that's part a part b now asks us to find the value of the actual speed of this system in meters per second right off the collision so we can just plug the numbers we have into this equation we've found.
02:07
So part b is velocity equals a square root of 2 times 9 .8 times 0 .95.
02:19
9 .8 is in meters per second square and 9 .95 is in meters.
02:24
So we get our final value to equal about 4 .32 meters per second.
02:38
And so these are our two answers for the first part.
02:42
Of this problem, parts a and b.
02:47
Okay.
02:48
Now we can move on to part c.
02:58
Okay.
03:00
So for this part, ask us to enter an expression for the initial speed of the bullet in terms of defined quantities and cheat.
03:08
And so we can use the conservation of energy to do a very similar thing.
03:13
So we know that the speed of the bullet, or the kinetic energy of the bullet, we'll call it kb, will be equal to the kinetic energy of the system right after the collision, we'll call that ks, which is equal to the potential energy at the end when it reaches a height.
03:34
So using that knowledge, we can determine an expression for the, sorry, we can determine an expression for the initial speed of the bullet.
03:54
So we can take the kinetic energy of the bullet, which will be one half, mass of the bullet, and times velocity squared equals the mass of the system which is the mass of the bullet plus the mass of the block times gravity times height so now we just need to switch this equation around to find the an expression for the velocity of the bullet so doing that really quickly and get e squared equals 2m plus large m then add that square root in so we can get two great so there we have part c i'll just box that part d wants us to find the actual value of that speed so what we're going to do is we're just going to plug those numbers we have in to this equation two times zero point zero seven five kilograms times times no not times excuse me plus 0 .93 kilograms times 9 .8 meters per second squared times 0 .95 meters all over 0 .93 kilograms and then putting that into a calculator we get the final velocity to be equal to about 48 .24 meters per second and that's our answer for part d.
06:32
The next part of the problem in part e wants us to find the initial kinetic energy of the bullet in jules.
06:41
So we can use the initial speed of the bullet and the mass of the bullet to find this...