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(adsbygoogle = window.adsbygoogle || []).push({});

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 im 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

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# Bullet in block question

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