Impulse of force correct here?

[normal_force]

Im sure Imma mess this up, probably because I am really tired...but

I have a bullet that weights .015kg's and is moving at 285m/s, so 609.18J's as with KE.

I hit a 70kg block, penetrates .3048 meters thus exerting 1998.62N's of force, which is a lot of force.
.3048/285m/s so 0.00106 seconds, now we can that this with impulse of force for...

Favg•∆t = m • ∆v -> (1998.62N's)•(0.00106) = 2.1374N•s impulse of force, now we use this to calculate the change in velocity ∆v= 2.1374N•s/ .015kg = ∆v:142.5m/s Vi:285m/s Vf:427.5 m/s
.015kg • 142.5 = 2.1374 N•s

If I take my impulse of 2.1374N•s over 70kg's...the velocity is .03053m/s and we divide by 0.00106 seconds we get 28.55 m/s^2 acceleration...

We can rework this obviously, 1998.62N's over 70kg's = 28.55m/s

d=(0m/s)(.00106s)+1/2(28.55m/s^2)(0.00106s^2)
d=0.000016328 meters

Object of 70kg Vf=(0m/s)(s)+(28.55m/s)(0.00106s)
Object of 70kg Vf=0.03053m/s

momentum p=70kg • 0.03053m/s = 2.13kg-m/s
KE=0.032622 Joules...

The force applied by bullet is constant...

So, what type of collision is this? why did the Vf of the bullet become 427 when it is stopped...and I am doing this correct RIGHT?

This is a concept check, not homework.

 

[A.T.]

...moving at 285m/s...penetrates .3048 meters...3048/285m/s so 0.00106 seconds...

Is the velocity constant during penetration?

 

[normal_force]

Is the velocity constant during penetration?

I should have mentioned that, the force applied by the bullet is constant thus same with velocity...

 

[haruspex]

I should have mentioned that, the force applied by the bullet is constant thus same with velocity...

No, constant force does not lead to constant velocity. What does it lead to?

 

[normal_force]

No, constant force does not lead to constant velocity. What does it lead to?

well, meant to say we just have constant force..not constant velocity, my bad.

 

[haruspex]

well, meant to say we just have constant force..not constant velocity, my bad.

So the way you calculated the time is wrong, as A.T. points out.

 

[normal_force]

So the way you calculated the time is wrong, as A.T. points out.

How so? t=.3048m/285m/s that is .00106 Seconds...correct? If it is wrong, care to correct it?

 

[haruspex]

How so? t=.3048m/285m/s that is .00106 Seconds...correct? If it is wrong, care to correct it?

The relationship speed=distance/time is only true for finding average speed. The average speed over the .3048 seconds is not 285m/s, that's the max speed.

 

[normal_force]

The relationship speed=distance/time is only true for finding average speed. The average speed over the .3048 seconds is not 285m/s, that's the max speed.

Okay, Okay so how do I calculate the time applied speed for impulse?

 

[haruspex]

Okay, Okay so how do I calculate the time applied speed for impulse?

Usually, with these 'bullet enters block' questions, the first step is to find the speed of the combination just after impact.

 

[normal_force]

Usually, with these 'bullet enters block' questions, the first step is to find the speed of the combination just after impact.

you also said it was .3048 seconds, it was .3048 meters...

 

[normal_force]

Usually, with these 'bullet enters block' questions, the first step is to find the speed of the combination just after impact.

Second, we don't know the speed after...we are calculating that.

 

[haruspex]

it was .3048 meters

Sure, but the comment stands.

we don't know the speed after...we are calculating that.

It would have helped to have mentioned that in the problem statement.
The way to find it is by a conservation law. Which one?

 

[normal_force]

Sure, but the comment stands.

It would have helped to have mentioned that in the problem statement.
The way to find it is by a conservation law. Which one?

look, look...We are looking at a collision, the collision is inelastic...this is the conservation of energy and conservation of momentum.

requires two equations I am guessing?...

 

[haruspex]

look, look...We are looking at a collision, the collision is inelastic...this is the conservation of energy and conservation of momentum.

requires two equations I am guessing?...

No, to find the velocity of bullet+block at the point where the bullet reaches maximum penetration, given the two masses and the bullet's initial velocity, only requires one equation.

 

[normal_force]

No, to find the velocity of bullet+block at the point where the bullet reaches maximum penetration, given the two masses and the bullet's initial velocity, only requires one equation.

Okay, your asking...never mind, I was thinking elastic collision...okay...but we are digressing...We know the bullet penetrates the block by .3048 meters, thus giving us an impact force...

 

[haruspex]

Okay, your asking...never mind, I was thinking elastic collision...okay...but we are digressing...We know the bullet penetrates the block by .3048 meters, thus giving us an impact force...

No, don't get stuck into forces again. Think conservation laws.

 

[normal_force]

No, don't get stuck into forces again. Think conservation laws.

okay, okay...We need conservation of momentum...its an inelastic collision...all I need to know is the impact time...or time the force of impact lasts...we can find impulse...and find displacement after..

 

[haruspex]

okay, okay...We need conservation of momentum...its an inelastic collision...all I need to know is the impact time...or time the force of impact lasts...we can find impulse...and find displacement after..

You wrote in post #12 that you are trying to calculate the speed after impact. To do that you do not need to find the force or the time, just apply conservation of momentum.
If that is not what you are trying to find, please state clearly what the objective is.

 

[normal_force]

You wrote in post #12 that you are trying to calculate the speed after impact. To do that you do not need to find the force or the time, just apply conservation of momentum.
If that is not what you are trying to find, please state clearly what the objective is.

Im asking if my calculations are correct but since my time was incorrect...how do I calculate impulse time...to multiple my force too, would I just use Mass times ∆v to get impulse?

 

[haruspex]

Im asking if my calculations are correct but since my time was incorrect...how do I calculate impulse time...to multiple my force too, would I just use Mass times ∆v to get impulse?

First steps first: let's find the resulting velocity, that's easy. Then we can discuss force and time.

 

[normal_force]

First steps first: let's find the resulting velocity, that's easy. Then we can discuss force and time.

My point exactly, we know the bullet has 285m/s, it stops in .3048 meters within the block, shouldn't that be 0.00106 seconds? We can also find the time by acceleration, or deceleration in this case t=√2d/a we could just use that...

 

[haruspex]

we know the bullet has 285m/s, it stops in .3048 meters within the block, shouldn't that be 0.00106 seconds

No, we've been down that rabbit hole. Please stop mentioning force or time until we've found the velocity.
I shall desist from responding to this thread until you make an attempt to find the velocity by applying conservation of momentum. I cannot think why you are so resistant to it.

 

[normal_force]

No, we've been down that rabbit hole. Please stop mentioning force or time until we've found the velocity.
I shall desist from responding to this thread until you make an attempt to find the velocity by applying conservation of momentum. I cannot think why you are so resistant to it.

Well, okay...just show me...this can be solved by work-energy and kinematics...cant it?

however, the equation m1+v1=(m1+m2)v2

v2=(m1/m1+m2)v1

 

[haruspex]

Well, okay...just show me...this can be solved by work-energy and kinematics...cant it?

however, the equation m1+v1=(m1+m2)v2

v2=(m1/m1+m2)v1

Good.
Next, you have to make some assumption about how the force varies during penetration. It is reasonable to assume that it will be constant, so knowing the distance it penetrates we can find the force from the energy absorbed by it.
Now, what work was done in penetrating the block? Think carefully.

 

[normal_force]

Good.
Next, you have to make some assumption about how the force varies during penetration. It is reasonable to assume that it will be constant, so knowing the distance it penetrates we can find the force from the energy absorbed by it.
Now, what work was done in penetrating the block? Think carefully.

Okay, the thing is the bullet has the ability to do work and will do work by exerting a force, the reality a bullet exerts a force, that is a fact. KE -> mechanical energy, in between that is work, which is done by force over a distance, which requires the bullets energy to exert a force, transferring energy.

IF we understand the bullet hasn't over penetrated, we know that the energy expended is 609.18Joules of KE, the amount of KE lost on the collision.

We know our .015 kg bullet which has enough Kinetic energy to do 609.18 Joules of work(which most of that energy won't last very long, conservation of energy but we understand the force can be deadly)

WE can understand ratios of KE before and after...but I am too lazy to do the work on this, so back to the basics...

to get v2...> 0.015kg m1/.015kg m1 + 70kg m2 •285m/s v1; we get v2 of 0.061058m/s...thus 1/2(70kg)(0.061058m/s)^2 we get .130 J's and 4.27kg*s momentum...

but work energy theorem can be applied...but the only problem I am having is finding time, using that to calculate impulse...

Now, we both know what happens with KE as it is turned into force within a penetrated body...material contact with bullet can have accelerated and expanded material...a cavity and/or crater.