00:01
Here in this given problem mass of the bullet which is being fired at the block, hanging block, its mass is m is equal to 9 .53 grams.
00:21
Its speed, initial speed before colliding with the block vi that is 270 meter per second, then mass of the block that is m is equal to 1 .85 kilogram.
00:50
Considering a perfectly inelastic collision, initial momentum of the bullet m into vi that will be equal to final momentum of the bullet along with the block m plus m multiplied by vf.
01:29
So, this vf will be given by m times vi divided by m plus m means this is 9 .53 multiplied by this is 9 .53 gram or 9 .53 multiplied by 10 to the power minus 3 kilogram multiplied by speed, initial speed of the bullet 270 meter per second divided by 9 .53 gram or 9 .53 into 10 power minus 3 kilogram again plus 1 .85 kilogram.
02:06
So, this vf is calculated to be equal to 1 .38 meter per second speed of the block along with the bullet.
02:19
Now, in the diagram with the light string this is the block attached, this is the block and the bullet going to be embedded in this block.
02:50
Then after the bullet is embedded into the block, the block will rise up...