00:01
Hello students in this question we are given a mass of a bullet which is equals to 4 grams which is moving horizontally with a speed of the bullet equals to 355 meter per second and it approaches the two wooden blocks and so let us see the situation after it strikes both the block so so this is our diagram for the question here we are also given the masses of the block first and this second block now in the first part we have to find a velocity of the second block when the bullet is embedded into it so we will use conservation of energy so firstly the kinetic energy of the bullet initially will be equals to the kinetic energy of the bullet after passing through the block 1 plus the kinetic energy of the block 1 so here it will be half mass will be 0 .0 4 and the velocity is 355 square it will be equal so it is a kinetic energy of the bullets so it will be equal to the kinetic energy of the block 1 after it is moving so it will be equals to half m which is the mass of the block 1 which is 1 .150 kg multiplied by the velocity which is 0 .550 square plus the kinetic energy.
01:59
Here this is the kinetic energy of the block 1 plus the kinetic energy of the bullet.
02:07
Here it will be the kinetic energy of the bullet after passing the block 1 so from here the kinetic energy of the bullet bullet after passing block 1 will be equals to 2 so it will be 251 .88 joules now we have to find a velocity of the second block after the bullet is embedded so we can again use a conservation of energy so the kinetic energy of the bullet will be equal to the kinetic energy of the block 2 after the bullet is embedded in it because the energy will remain conserved half here the kinetic energy of the bullet after passing through the first block is 251.
03:05
0 .88 is equal to half mass will be 1 .53 plus 0 .004 multiplied by v .2.
03:19
So from here we get v equals to 18.
03:22
So it is 18 .1 2 meter per second.
03:30
Now in the b part we have to find the kinetic energy of the system before and after the core legion.
03:36
So the ratio of the kinetic energy of the system initially to the final kinetic energy of the system.
03:45
Initially the kinetic energy was due to the blade which is half, mass of the blade which is 0 .04...