Bullet Enters/Exits Mass, Mass Slides

In summary, a 25 g bullet traveling horizontally at 1200 m/s goes through a 35 cm thick 350 kg stationary target and emerges with a speed of 850 m/s. The bullet is in the target for approximately 0.0003 seconds. The average force exerted on the target can be determined using the equation Δp/Δt, where Δp is the change in momentum and Δt is the change in time. To find the deceleration of the bullet, use the equation Vf2 = Vi2 - 2*a*x, where Vf is the final velocity, Vi is the initial velocity, a is the deceleration, and x is the distance traveled.
  • #1
myxomatosii
80
0

Homework Statement



In a ballistics test, a 25 g bullet traveling horizontally at 1200 m/s goes through a 35 cm thick 350 kg stationary target and emerges with a speed of 850 m/s. The target is free to slide on a smooth horizontal surface.

(a) How long is the bullet in the target?

(b)What average force does it exert on the target?

(c) What is the target's speed just after the bullet emerges?

Homework Equations



It would be nice to have some more equations, my professor gave the class four pages of equations scribbled in his horrible hand-writing, to help us... I guess. Instead of giving us the basic equations that we can use to plug-in to create our own equations he already solved them all out into the forms he understands, which means we have to un-solve them to try to make sense of them, so really I have no idea what other equations to put here since he has no clear division between which equations go with which section of the book.

momentum of system before = momentum of system after ||| Pi=Pf

The Attempt at a Solution



I already solved part C using the equation provided.

Part A I attempted something that I knew wouldn't work but gave it a try anyway and I know I need part A to solve B so I haven't gotten around to trying to solve B yet.

I seem not to be able to resist applying simple mechanics to complex problems just to see if I can knock it out of the way. Anyway I did this to solve for B, it was obviously going to be wrong but I submitted it anyway.

B: I tried to say that the

DT=0.35m

Vavg/DT=? Then that gave me seconds so I took s-1. Yea, it obviously was just bs, but that is how lost I am.
 
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  • #3
LowlyPion said:
Here are some kinematic equations anyway.
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

As for determining the force, recognize that Δ in momentum divided by the Δt should give you your average force.

Thanks, getting to work on things now. Hopefully I can make some progress without having to ask too much here.
 
  • #4
LowlyPion said:
Here are some kinematic equations anyway.
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

As for determining the force, recognize that Δ in momentum divided by the Δt should give you your average force.

My problem is finding the time, the Δp (momentum) is a joke to find.

0.001seconds doesn't seem like a bad impulse time, considering the average of the problems I've been working with is 0.005 seconds, but it won't accept it, I'm not sure what else to do even since I've posted this because no one mentioned time.

I'm given it decelerates from 1200m/s to 850m/s in the distance of 0.35m.

Finding the time should be cut and dry like I found it, but its not, so I'm confused.
 
  • #5
Tried.

a) Use the equation Δx=1/2(Vo+V)t, so t=(2Δx)/(Vo+V)

From.

http://qaboard.cramster.com/physics-topic-5-218055-cpi0.aspx

Got 0.0003seconds, didn't accept it.
Also read all of

https://www.physicsforums.com/showthread.php?t=200285

He seems to think the part I think is hard is elementary =(

So I learned nothing from that as well.

I recalculated the time he declared to be right using the formula I tried using on mine(the same one I tried right above this in this very post), it gave his same answer. Meaning I am calculating time right, assuming he was, and he did get the right answer. Anyway, I am getting really annoyed with my "online" system when I am almost positive my answer is correct.. am I wrong?
 
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  • #6
Start with determining the deceleration through the material.

That you can get from Vf2 = Vi2 - 2*a*x

In your case that would be 12002 - 8502 = 2*.35*a

Now armed with a you can determine t, from x = 1/2*a*t2

That arrives at a number a bit larger than .0003 sec.
 

1. How does the mass of an object affect the trajectory of a bullet entering or exiting it?

The mass of an object does not affect the trajectory of a bullet entering or exiting it. The trajectory is determined by the initial velocity and angle of the bullet, as well as external factors such as air resistance. The mass of the object may affect the amount of damage caused by the bullet, but not its trajectory.

2. Can a bullet exit a mass at a different angle than it entered?

Yes, it is possible for a bullet to exit a mass at a different angle than it entered. This can occur if the bullet encounters resistance or obstacles within the mass, causing it to change direction.

3. How does the density of an object affect the penetration of a bullet?

The density of an object can affect the penetration of a bullet. Generally, denser objects will offer more resistance to the bullet and may cause it to penetrate less, while less dense objects may allow the bullet to pass through more easily.

4. Is there a difference in the force of a bullet when it enters versus exits a mass?

The force of a bullet when it enters a mass will be greater than when it exits. This is due to the energy that is transferred from the bullet to the mass upon impact. As the bullet exits, it will have lost some of its energy, resulting in a decrease in force.

5. Can the shape of an object affect the trajectory of a bullet entering or exiting it?

Yes, the shape of an object can affect the trajectory of a bullet entering or exiting it. Objects with irregular or angled surfaces may cause the bullet to deflect or change direction. Additionally, the shape of the object may also affect the amount of resistance the bullet encounters, potentially altering its trajectory.

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