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

 

[quicknote]

Hello, it seems like you are on the right track in your solution. Let me walk you through the steps to arrive at the final answer.

First, let's consider the conservation of momentum in the system. The initial momentum of the bullet is mv1 and after it gets embedded in the block, the final momentum is (M+m)v2. Since momentum is conserved, we can set these two equal to each other:

mv1 = (M+m)v2

Solving for v2, we get:

v2 = mv1 / (M+m)

Next, let's consider the conservation of energy in the system. Initially, the bullet has kinetic energy of 1/2mv1^2 and after it gets embedded in the block, it has kinetic energy of 1/2(M+m)v2^2. We also have to take into account the work done by the bullet onto the block, which is equal to the force (F) multiplied by the distance traveled (D). Therefore, we can write the conservation of energy equation as:

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

Substituting the value of v2 we found earlier, we get:

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

Now, we need to solve for the force (F). We know that force is equal to mass (M+m) multiplied by acceleration (a). If we assume that the acceleration is constant, we can write:

F = (M+m)a

Substituting this value of F into the conservation of energy equation, we get:

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

Dividing both sides by m, we get:

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

Now, let's rearrange this equation to solve for the acceleration (a):

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

Since we know that the distance traveled by the bullet (D) is the same in both cases, we can equate the two values of acceleration (a) and solve for the distance traveled by the bullet in the second case