00:01
Hi, here in this given problem, mass of the bullet, which is fire to the wood block, that is m, small m is equal to 5 .00 gram.
00:20
And mass of the block, wood block, let it be capital m, and this is 1 .00 kilogram.
00:34
Length of the string l is equal to 2 .00 meter and vertical height raised by the center of mass of the block as a result of collision of bullet in it that height is given as 0 .38 centimeter okay one more thing which is given to us that is initial speed of the bullet before hitting the block and it does not remain embedded within the block it comes out through the block v -not that is given as 450 meter per second and we have to find its final speed when it will come out of the block on other side so, first of all, for the block using energy conservation, its kinetic energy when bullet hits it, that is half m vb squared, vb for the velocity of the block, is equal to m gh.
02:16
So canceling this m, we get an expression for vb, and that is square.
02:22
Root of 2g h.
02:24
So plugging in all the known values here, this vb will be given by 2g means 2 times of 9 .8 multiplied by h 0 .38 centimeter or into 10 to par minus 2 meter.
02:39
And finally it is calculated to be equal to 0 .271 meter per second.
02:47
This will be the speed of the block after bullet passes through it...