00:01
Neglecting non -conservative forces such as friction, then mechanical energy is conserved.
00:05
If we also neglect external forces, then momentum is conserved.
00:09
We will apply these two concepts as we will deal with the following problem.
00:13
Here we have a 1 .28 kilogram block that is attached to a horizontal spring with spring constant of 2 ,750 newton per meter.
00:21
The block is initially at rest on a frictionless surface, and then it is being hit by a 10 .6 gram bullet.
00:28
The bullet sticks or embeds itself on the block.
00:32
As a consequence, the block would compress the spring by an amount of 11 .9 centimeter.
00:39
We wish to find the speed by which the bullet hits the block.
00:44
So that is denoted with vb in our problem.
00:47
So first is that that assume that there are no non -conservative forces such as, well, it's already given here that the surface is frictionless.
00:55
So we can then assume that momentum is conserved as there are no non -conservative forces.
01:01
So looking at scenario or looking at situation 2 and 3, the kinetic energy of the bullet with the block in it at the instant, right immediately after the block embeds itself, is going to be converted into elastic potential energy as the block bullet would be compressing the spring.
01:22
So we say then that the, from scenario to, 2 to 3 where conservation of mechanical energy is upheld, so the kinetic energy at 2 becomes potential energy at 3.
01:38
We solve for kinetic energy as the product of 1 half times the mass, where in this case the mass will be the total mass of the block and the bullet times the speed times the square of the speed, whereas the elastic potential energy is equal to 1 half times the spring constant, times the amount of compression of the spring x so we can cancel the one -half here and so our equation becomes m plus capital m v squared equals k x squared solving for v here gives us square root of k x squared divided by m plus m we have to have everything in s i so that your mass for the bullet needs to be in kilograms just as the mass of the bullet is so just as the mass of the block is so we have to divide the mass of the bullet by 1000 noting that 1 ,000 grams is 1 kilogram so that means 10 .6 divided by 1 ,000 will is going to be equal to 0 ,000 0 .0106.
02:51
And also your x needs to be in meters.
02:55
So we'll divide this by 100, noting that one meter is 100 centimeters.
03:01
So this becomes 0 .119.
03:05
So let's plug those values in our equation to get v.
03:10
Your k is 2 .750...