# A bullet pushing a knot out of a wood block

• emily081715
In summary, to find the initial velocity of a bullet that is shot into a wooden block, we can use the equation v=v0+at, where v is the final velocity (in this case, 10 m/s), v0 is the initial velocity (what we're trying to find), a is the acceleration (-4.9 × 105 m/s2 in this case), and t is the time it takes for the bullet to travel through the block (1.0 ms). After rearranging and solving for v0, we get an initial velocity of 500 m/s, which is more realistic and correct.
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
pi=pf
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

emily081715 said:
what simple step i have missed
emily081715 said:
1.0 ms to travel through the block,
Does the momentum of the block change?

Bystander said:
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

emily081715 said:
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.

Last edited:
PeroK said:
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

emily081715 said:
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.

PeroK said:
That's not a bad idea.
how though? f=ma?

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 like Serena said:
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

emily081715 said:
i am now confused as to what approach to take

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

I like Serena said:
How about ##v=v_0 + at## (since we have constant deceleration)?
I like Serena said:
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

emily081715 said:
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?

I like Serena said:
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 said:
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

I like Serena

## 1. How does a bullet push a knot out of a wood block?

When a bullet is fired into a wood block, it carries a large amount of kinetic energy. This energy is transferred to the wood block upon impact, causing the molecules in the wood to vibrate and move. As the bullet passes through the wood and reaches the knot, the molecules in the knot are pushed out of the way, allowing the bullet to continue through the block.

## 2. Can a bullet push a knot out of any type of wood?

Yes, a bullet can push a knot out of any type of wood as long as the wood is not too dense or hard. The density and hardness of the wood will affect the amount of energy required to push the knot out, but in most cases, a bullet with enough velocity and energy can successfully push a knot out of the wood.

## 3. How fast does a bullet need to be going to push a knot out of a wood block?

The speed of the bullet needed to push a knot out of a wood block depends on the size and density of the knot, as well as the type of wood. In general, a bullet traveling at a speed of at least 1000 feet per second should be able to push a knot out of most types of wood.

## 4. Is the knot pushed out of the wood block completely or partially?

In most cases, the knot will be pushed completely out of the wood block by the bullet. However, depending on the size and density of the knot, it is possible for the knot to be partially pushed out or only displaced within the wood block.

## 5. Are there any safety concerns when conducting experiments with a bullet and wood block?

Yes, there are safety concerns when conducting experiments with a bullet and wood block. It is important to always follow proper safety precautions, such as wearing protective gear and ensuring a safe distance between yourself and the wood block. It is also important to conduct these experiments in a controlled and supervised environment to avoid any accidents.

• Introductory Physics Homework Help
Replies
11
Views
7K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
25
Views
6K