Bullet Momentum Force: 3.15x104 N

[Pao44445]

Homework Statement

a bullet 15 g travels 300 m/s then hit the plastic sheet that is thick 2 cm and pass through it. After the crash, the bullet travels 90 m/s ( continue to lose the velocity ) How much the average force acting on the bullet?

Homework Equations

Momentum = Mass x Velocity

The Attempt at a Solution

From the diagram, While the bullet is travelling, it has momentum (15x10^-3)(300) = 4.5 kg(m/s) and after the crash, it has momentum (15x10^-3)(90) = 1.35 kg(m/s)
4.5 - 1.35 = 3.15 kg(m/s)
the change in momentum is equal to Impulse
Impulse = 3.15 kg(m/s)
F(t) = 3.15 kg(m/s)

I found the time with S=ut+1/2at²
2x10^-2 = 300(t) + 1/2[(90-300)/t](t)²
t is about 1x10^-4 s

F(10^-4) = 3.15 kg(m/s)
F = 3.15x10⁴ N

 

[oz93666]

Are you asking a question ?

The only error I can see is that you have force as kg(m/s) when it should be kg(m/s2)

 

[JMatt7]

I found the time with S=ut+1/2at²
2x10^-2 = 300(t) + 1/2[(90-300)/t](t)²
t is about 1x10^-4 s

You're making the assumption that the acceleration is constant during the process, which is unlikely. Try not to assume anything about what happens inside the material, except that all the reaction forces acting on the bullet can be averaged into a single force that slows it down.

I'll suggest the best and cleanest way to solve this problem is through energy. Is the energy of the bullet conserved?

 

[PeroK]

You're making the assumption that the acceleration is constant during the process, which is unlikely. Try not to assume anything about what happens inside the material, except that all the reaction forces acting on the bullet can be averaged into a single force that slows it down.

I'll suggest the best and cleanest way to solve this problem is through energy. Is the energy of the bullet conserved?

These problems that ask for the "average" force are ambiguous: does that mean the average over time or the average over distance? If you want to calculate the average over time, then you do have to assume constant acceleration, as otherwise the time is variable.

Otherwise, you can calculate the average over distance: using energy, as you suggest.

 

[haruspex]

These problems that ask for the "average" force are ambiguous

That's being kind. Seems to me that, unless average over distance is specified, average over time is implied. Average acceleration is well defined as $\frac{\Delta v}{\Delta t}$. Average force should equal average acceleration times mass, so $\frac{\Delta p}{\Delta t}$.
Questions which imply to the student that it is ok in general to calculate average force as $\frac{\Delta E}{\Delta s}$ should be erased from the textbooks.

 

[SammyS]

Furthermore, the sketch implies that the plastic sheet is displaced somewhat as the bullet passes through it. Thus the deceleration occurs over a distance that's greater than the 2cm thickness of the plastic sheet. By the way, that's very thick for what I would refer to as a plastic sheet.

 

[Pao44445]

My teacher said I can actually use energy equation to solve this but since I am learning about momentum :/

 

[haruspex]

My teacher said I can actually use energy equation to solve this but since I am learning about momentum :/

If you have a good relationship with your teacher, you might care to bring section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ to his/her attention. Otherwise, just do as your teacher says but remember that the method is not valid.

 

[Abro baba jhon]

<Moderator's note: Member has been warned to open new threads for new questions and not trying to avoid the template and request on own effort by this backdoor method.>

how do I calculate the Energy absorbed by the fabric.?
since
initial velocity of bullet = Vi = 755 f/s
final velocity of bullet = Vf = 343 f/s
weight (mass) of bullet = m = 1.75 grams

 

[haruspex]

how do I calculate the Energy absorbed by the fabric.?
since
initial velocity of bullet = Vi = 755 f/s
final velocity of bullet = Vf = 343 f/s
weight (mass) of bullet = m = 1.75 grams

Does the bullet's energy change?

 

[Abro baba jhon]

Does the bullet's energy change?

yes.

 

[Abro baba jhon]

yes.

sorry, velocity changed

 

[haruspex]

yes.

So how much energy does it lose and where does the lost energy go?

 

[haruspex]

sorry, velocity changed

... which changes the kinetic energy, right?

 

[Abro baba jhon]

sorry, velocity changed

and now I want to calculate the Energy absorbed

 

[Abro baba jhon]

... which changes the kinetic energy, right?

yes, definitely

 

[Abro baba jhon]

So how much energy does it lose and where does the lost energy go?

the energy of bullet utilized to penetrate the fabric

 

[haruspex]

the energy of bullet utilized to penetrate the fabric

Knowing the mass and the before and after velocities, how much KE does it lose?

 

[Abro baba jhon]

Knowing the mass and the before and after velocities, how much KE does it lose?

yes, I want to calculate the energy lose (absorbed) knowing bullet wight before n after velocities,
as I have earlier stated the values of bullet mass initial (before) velocity and final (after) velocity

 

[Abro baba jhon]

yes, I want to calculate the energy lose (absorbed) knowing bullet wight before n after velocities,
as I have earlier stated the values of bullet mass initial (before) velocity and final (after) velocity

initial velocity of bullet = Vi = 755 f/s
final velocity of bullet = Vf = 343 f/s
weight (mass) of bullet = m = 1.75 grams

 

[haruspex]

initial velocity of bullet = Vi = 755 f/s
final velocity of bullet = Vf = 343 f/s
weight (mass) of bullet = m = 1.75 grams

Ok, but you are not answering my question. Do you know the formula for kinetic energy of a mass?

 

[ehild]

Homework Statement

a bullet 15 g travels 300 m/s then hit the plastic sheet that is thick 2 cm and pass through it. After the crash, the bullet travels 90 m/s ( continue to lose the velocity ) How much the average force acting on the bullet?

Homework Equations

Momentum = Mass x Velocity

The Attempt at a Solution

View attachment 105462
From the diagram, While the bullet is travelling, it has momentum (15x10^-3)(300) = 4.5 kg(m/s) and after the crash, it has momentum (15x10^-3)(90) = 1.35 kg(m/s)
4.5 - 1.35 = 3.15 kg(m/s)
the change in momentum is equal to Impulse
Impulse = 3.15 kg(m/s)
I found the time with S=ut+1/2at²
2x10^-2 = 300(t) + 1/2[(90-300)/t](t)²
t is about 1x10^-4 s

You can calculate the exact time. Simplify the second term with t: 0.02 = 300(t) + 1/2[(90-300)](t)=((90+300)/2)t So what is the time the bullet travels inside the sheet?

 

[ehild]

These problems that ask for the "average" force are ambiguous: does that mean the average over time or the average over distance? If you want to calculate the average over time, then you do have to assume constant acceleration, as otherwise the time is variable.

Otherwise, you can calculate the average over distance: using energy, as you suggest.

Aren't they the same in this problem?

 

[jbriggs444]

Aren't they the same in this problem?

No. The elapsed time and, hence, the average velocity while penetrating the sheet depends on the acceleration profile. Front load the deceleration and the time is higher. Back load the deceleration and the time is lower.

 

[PeroK]

Aren't they the same in this problem?

They are only the same if the average velocity (wrt time, of course) is half way between the initial and final velocities.

 

[Abro baba jhon]

Ok, but you are not answering my question. Do you know the formula for kinetic energy of a mass?

I did not know the formula... I'm asking for thee formula too

 

[ehild]

No. The elapsed time and, hence, the average velocity while penetrating the sheet depends on the acceleration profile. Front load the deceleration and the time is higher. Back load the deceleration and the time is lower.

I meant that the energy method will result in the same average force as the method the OP used, assuming constant acceleration. Which might not be true, but there is no way to calculate the time without knowing how the velocity depends on time. The time averaged force is $F_{av}= \frac{\int _0^T{madt}}{T}=m\frac{v_f-v_i}{T}=\frac{p_f-p_i}{T}$. As the distance traveled is given, T can be calculated knowing the time dependence of v: $s=\int_0^T{v(t)dt}$
Using energy results in the force averaged for distance, and it is the same as the constant force which would produce the same final velocity over the same distance.

 

[haruspex]

I did not know the formula... I'm asking for thee formula too

Ah, ok. The kinetic energy of a mass m moving with velocity v is ½mv².
But this is extremely basic.. I don't understand how you could be studying this subject and not have encountered this formula.

 

[haruspex]

I meant that the energy method will result in the same average force as the method the OP used,

Unfortunately that is not the impression post #23 gives in its context.