A bullet pushing a knot out of a wood block

[emily081715]

Homework Statement

You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal surface (Figure 1) . After hitting dead center on a hard knot that runs through the block horizontally, the bullet pushes out the knot. It takes the bullet 1.0 ms to travel through the block, and as it does so, it experiences an x component of acceleration of -4.9 × 105 m/s2. After the bullet pushes the knot out, the knot and bullet together have an x component of velocity of +10 m/s. The knot carries 10% of the original inertia of the block.
What is the initial velocity of the bullet?

Homework Equations

Conservation of momentum equation
p_i=p_f
m1v1i+m2v2i=m1v1f+m2v2f

The Attempt at a Solution

v1f= (0.2)(10)+0.005(10)/ 0.005
=410m/s
the 0.2 came from the 10% of 2kg wooden block (knot)
this answer is incorrect but i am unsure where my error is and what simple step i have missed

 

[Bystander]

what simple step i have missed

1.0 ms to travel through the block,

Does the momentum of the block change?

 

[emily081715]

Does the momentum of the block change?

no the momentum of the block would not change since its still at rest, only the knot moves

 

[PeroK]

no the momentum of the block would not change since its still at rest, only the knot moves

Do you know the required answer?

In any case, I suspect the rest of the block does move, and as you don't know its speed you can't use conservation of momentum.

 

[emily081715]

Do you know the required answer?

In any case, I suspect the rest of the block does move, and as you don't know its speed you can't use conservation of momentum.

no i don't know the final answer. if i do not use conservation of momentum, will i be using something with the acceleration

 

[PeroK]

no i don't know the final answer. if i do not use conservation of momentum, will i be using something with the acceleration

That's not a bad idea.

 

[emily081715]

That's not a bad idea.

how though? f=ma?

 

[I like Serena]

Hey Emily! ;)

There's quite some distracting information there.
But let's see...
Initially the bullet has some unknown speed $v_0$.
Then it was decelerated with $-4.9 × 10^5\text{ m/s}^2$ during $1 \text{ ms}$.
And afterwards it has speed $10\text{ m/s}$.
Is that sufficient information?

 

[emily081715]

Hey Emily! ;)

There's quite some distracting information there.
But let's see...
Initially the bullet has some unknown speed $v_0$.
Then it was decelerated with $-4.9 × 10^5\text{ m/s}^2$ during $1 \text{ ms}$.
And afterwards it has speed $10\text{ m/s}$.
Is that sufficient information?

i am now confused as to what approach to take

 

[I like Serena]

i am now confused as to what approach to take

How about $v=v_0 + at$ (since we have constant deceleration)?

 

[emily081715]

How about $v=v_0 + at$ (since we have constant deceleration)?

How about $v=v_0 + at$ (since we have constant deceleration)?

so to rearrange for
vi=vf/at
=10/(−4.9×105 m/s2)(0.001s)
=0.02
i don't think i did this correctly because the bullet would need to travel much faster then that

 

[I like Serena]

so to rearrange for
vi=vf/at
=10/(−4.9×105 m/s2)(0.001s)
=0.02
i don't think i did this correctly because the bullet would need to travel much faster then that

I agree. The bullet must travel much faster than that - at least more than 10 m/s.
Perhaps the rearrangement isn't quite correct?

 

[emily081715]

I agree. The bullet must travel much faster than that - at least more than 10 m/s.
Perhaps the rearrangement isn't quite correct?

you are right my rearrangement was incorrect

 

[emily081715]

you are right my rearrangement was incorrect

i divided when i should've subtracted, that means the vi=500m/s which is more realistic and correct. thanks