How Far Will a Bullet Travel When a Block Is Free to Slide?

[dowjonez]

so a bullet of mass m is fired into a blockof mass M sitting on a frictionless table. In the first case the block has a wall directly behind it. The bullet travels a distance D into the block

Now the bullet is fired again but the wall is removed, how far will the bullet travel now when the block is free to slide along the table. assuming mechanical energy is conserved in the system

I totally get this question but I am not sure how to arrive at the final answer given by the book which is

x = D*m / (M + m)



so first of all momentum of the system will be conserved

mV1 = (M + m)V2
v2 = mv1 / (M+m)

Secondly there is no additional energy being added to the system
so
E = KEbullet - work of bullet done onto block - KEbulletandblock = 0

1/2mv1^2 = F*D + 1/2(M+m)v2^2

plugging in v2


1/2mv1^2 = F*D + 1/2(M+m)((mv1/(M+m))^2

now i think the next step would be to equate force with mass and acceleration
so


mv1^2 = 2*m*a*D + (M+m)((mv1/(M+m))^2

dividing both sides by m would leave

v1^2 = 2*a*D + 1/m*(M+m)((mv1/M+m)^2

so


2aD = v1^2 - 1/m*(M+m)((mv1/M+m)^2


D = (1/2a)* v1^2 - 1/m*(M+m)((mv1/M+m)^2


thats as far as i can get
like that should be a suitable answer but my teacher always wants it the same as the back of that book
so does anyone know know that reduces to

x = Dm / (M+m)


any help would be grealty appreciated

 

[Chi Meson]

so first of all momentum of the system will be conserved
mV1 = (M + m)V2
v2 = mv1 / (M+m)

right so far

Secondly there is no additional energy being added to the system
so
E = KEbullet - work of bullet done onto block - KEbulletandblock = 0
1/2mv1^2 = F*D + 1/2(M+m)v2^2

this is an incorrect assumption. True no additional energy eters the system, but plenty of energy will leave the system as heat and sound.
At thsi point you need to use the work energy theorem to find a term for the force that the block applies to the bullet. This force will be the same for the second situation (an optimistic assumption, but it works out; I get the "correct answer.")
Do the work energy theorem again for the work done on the block to change its KE from zero to the final velocity. Use F from the first part times "x".
This is obviously an approximation; I disagree wil the validity of this question and the "correct answer" because it is based on the bullet doing work on the "bullet+block"; really the bullet does work only on the block. I personally think the answer is
(DMm)/(m+M)^2