# A 0.027 kg bullet is firedstraight up at a falling...

A 0.027 kg bullet is firedstraight up at a falling wooden block that has a mass of4.0 kg. The bullet has a speed of710 m/s when it strikes the block. Theblock originally was dropped from rest from the top of a buildingand had been falling for a time t when the collision withthe bullet occured. As a result of the collision, the block (withthe bullet in it) reverses direction, rises, and comes to amomentary halt at the top of the building. Find the timet.
s

1. Speed of block after falling, from rest, a distances    V^2 = 2 * g * s
Momentum of block P = M * ?(2 * g * s)
Bullet must have momentum of 2 * P (ignore mass of bulletin block since it is small)
So m * v = 2 * M * ?(2 * g * s)
s = m^2 * v^2 / (M^2 * 8 * g)     Also s =1/2 * g * t^2
Then t = 1/2 * ?(m^2 * v^2 / (M^2 * g^2)) = m * v/ (2 * M * g) = .24 sec
2. The velocity equation for the block dropped from rest for a timet is v1B = -gt. The mass of the block is m1=4 kg, that of the bullet is m2 = 0.027 Kg, and thevelocity of the bullet before the collision is v2B = 710m/s. Note that after the collision, the velocity of the block (nowwith the bullet in it) is exactly reversed to vf =gt

This is a completely inelastic collision. Applyingconservation of momentum, we have

m1v1B + m2v2B =(m1 + m2)vf

Substitute the time dependency, we have
m2v2B - m1gt = (m1+ m2)gt. Solving this for the time yeilds:
t = (m2v2B)/(2m1 +m2)g = 0.24369 s
3.  4.  5. here bullet and block are moving in opposite directions
take upward direction as positive
initial momentum of bullet = mu =   0.0027*710kgm/s
initial momentum of block = -Mv
the block has fallen for a time t then
initial momentum of block = -Mgt = -4*9.8*t kg m/s
then total initial momentum = mu -Mv =  0.0027*710kgm/s   - 4*9.8*t kg m/s
final momentum of bullet and block =( M+m)v'
let total block and bullet has travelled a distance s to thetop of the building
then v2-v'2= -2gs---1
here v = 0 at the top
but we know that s = (1/2)gt2
because block also travelleled a distance s in downwarddirection initially in time t
therefore from 1 v' = gt = 9.8*t
then initial momentum = final momentum
0.0027*710 kgm/s   - 4*9.8*t kg m/s = ( M+m)v'
=(4+0.027)*9.8*t
solve for t
6. here bullet and block are moving in opposite directions
take upward direction as positive
initial momentum of bullet = mu =   0.0027*710kgm/s
initial momentum of block = -Mv
the block has fallen for a time t then
initial momentum of block = -Mgt = -4*9.8*t kg m/s
then total initial momentum = mu -Mv =  0.0027*710kgm/s   - 4*9.8*t kg m/s
final momentum of bullet and block =( M+m)v'
let total block and bullet has travelled a distance s to thetop of the building
then v2-v'2= -2gs---1
here v = 0 at the top
but we know that s = (1/2)gt2
because block also travelleled a distance s in downwarddirection initially in time t
therefore from 1 v' = gt = 9.8*t
then initial momentum = final momentum
0.0027*710 kgm/s   - 4*9.8*t kg m/s = ( M+m)v'
=(4+0.027)*9.8*t
solve for t
7. here bullet and block are moving in opposite directions
take upward direction as positive
initial momentum of bullet = mu =   0.0027*710kgm/s
initial momentum of block = -Mv
the block has fallen for a time t then
initial momentum of block = -Mgt = -4*9.8*t kg m/s
then total initial momentum = mu -Mv =  0.0027*710kgm/s   - 4*9.8*t kg m/s
final momentum of bullet and block =( M+m)v'
let total block and bullet has travelled a distance s to thetop of the building
then v2-v'2= -2gs---1
here v = 0 at the top
but we know that s = (1/2)gt2
because block also travelleled a distance s in downwarddirection initially in time t
therefore from 1 v' = gt = 9.8*t
then initial momentum = final momentum
0.0027*710 kgm/s   - 4*9.8*t kg m/s = ( M+m)v'
=(4+0.027)*9.8*t
solve for t