00:01
Hello, in this question we apply the conservation of energy as well as linear momentum.
00:08
We are told that a bullet hits a block at a speed of 400 meters per second.
00:17
The mass in kilograms is 0 .005.
00:22
And the block, also mass of 1 kilogram, compresses against or moves against a spring, the spring constant of 900, compressing the spring constant of 9, the spring through a distance of 5 cm or 0 .05 meters.
00:46
So we first established that the elastic potential energy gained by the spring during compression is equal to the kinetic energy of the block.
00:59
If the block comes to rest, then all the kinetic energy of the block becomes elastic potential energy of the spring.
01:06
So the half mv squared kinetic energy of the block is equal to the connectic energy of the block is is equal to half kx squared formula for elastic potential energy of a spring which is half of 900 0 .05 squared.
01:27
This gives us 1 .125 joules.
01:32
So this is how much kinetic energy the block had and upon coming to rest this becomes the potential energy of the spring.
01:41
So we can find the the speed of the block knowing that its kinetic energy is 1 .125 so making velocity of block the subject now with 2 1 .125 times 2 and that is 1 .5 meters per second now we are to find the the speed at which the bullet images from the block so to do that we apply the conversion of momentum momentum before the bullet embedded in the rock must recall to momentum after the bullet leaves the block so before the bullet embass in the block momentum was momentum of the bullet which is its mass times velocity momentum of the block was zero because the block was stationary afterwards momentum of the block is one kilogram moving at a speed of 1 .5 and momentum of the bullet is 0 .005 moving at the speed of speed of v.
03:07
This is 0 .005...