Cannon connected to spring fires projectile

[~christina~]

Homework Statement

A cannon is rigidly attached to a carriage which can move along horizontal rails but is connected to a post by a large spring, initially unstretched and with a force constant k= 2.00 x 10^4 N/m, as shown in figure below. The cannon fires a 200kg projectile at a velocity of 125m/s directed 45 deg above the horizontal.

a) If the mass of the cannon and it’s carriage is 5,000kg find the recoil speed of the cannon.

b)Determine the maximum extension of the spring

c)Find the maximum force the spring exerts on the carriage.

d)Considered the system consisting of the cannon, carriage, and the shell. Is the
momentum conserved during the firing? Explain

e)At the maximum height of it’s trajectory the projectile explodes into 2 fragments. The smaller fragment, one third the projectile’s total mass drops straight down after the explosion. What is the velocity of the larger fragment when it returns to the level of the cannon?

f)What is the horizontal distance from the canon to the larger fragment at this level?

g)Is the energy of the projectile conserved during the explosion? Use appropriate physics principles to explain your response.

Homework Equations

I'm not quite sure however...

$$K= .5mv^2$$

$$U_s= .5 kx^2$$

The Attempt at a Solution

I don't even know how to start this.

a) for this I have to find the recoil velocity of the cannon.

know:
k= 2.00x10^4 N/m ===> is this the spring constant? (I'm wondering b/c it says it's the
force constant)
projectile veloctity = 125m/s
theta= 45 deg

m cannon = 5000kg

PLEASE HELP ME ...I need help and would like to get past part a)

Thanks

 

[~christina~]

I forgot to attatch a picture of the pic given of the cannon.
http://img130.imageshack.us/img130/1396/28734583cj0.th.jpg

 

[aq1q]

h
a)you have to use conservation of momentum..
b) when you have that velocity.. find the x component of that velocity.. and do 1/2(Mc)(Vx)^2= 1/2 (K)x^2 (find x)
c)once you find this x.. F=-kx will give u the maximum force..(the negative is relative)
d)-e) (are different problems, figure out a-c first) :) gl

 

[~christina~]

Since you say that I have to use the conservation of momentum so would it be

$$m_1v_1= -m_2v_2$$
however since the object is fired at an angle wouldn't it be

$$m_1v_1 + m_2v_2cos theta= 0$$ ?

thus
m1(cannon)= 5000kg- 200kg= 4800kg
v1= ?

m2 (projectile)= 200kg
v2= 125m/s cos 45

thus...
$$m_1v_1= - m_2v_2 cos theta$$

4800kg (v1)= - (200kg)(125m/s cos 45)

v1= -3.68 m/s ====> recoil speed of cannon


Is this alright? I'll do the rest and post it as I do that.

 

[aq1q]

very nice.. but why did u subtract 200 from it.. ? it says "cannon and it’s carriage is 5,000kg" so I am guessing that doesn't include the 200.. but I am not sure.. english isn't my strongest subject. haha but u get the idea

 

[aq1q]

and remember there is a momentum in the y direction.. but the ground is in the way

 

[~christina~]

very nice.. but why did u subtract 200 from it.. ? it says "cannon and it’s carriage is 5,000kg" so I am guessing that doesn't include the 200.. but I am not sure.. english isn't my strongest subject. haha but u get the idea

$$m_1v_1= -m_2v_2$$
however since the object is fired at an angle wouldn't it be

$$m_1v_1 + m_2v_2cos theta= 0$$ ?

thus
m1(cannon)= 5000kg
v1= ?

m2 (projectile)= 200kg
v2= 125m/s cos 45

thus...
$$m_1v_1= - m_2v_2 cos theta$$

5000kg (v1)= - (200kg)(125m/s cos 45)

v1= -3.54 m/s ====> recoil speed of cannon
_____________________________________________
Hm If there is momentum in the y direction but the ground is in the way how would the equation look like if I included the y value?

m1 v1 = m2 v2 cos theta

 

[aq1q]

thing is.. u don't need the Y value. But, since the ground blocks it.. the momentum isn't conserved, right? we are assuming that it doesn't bounce back up

P.S. I'm pretty sure that that the momentum in the Y direction isn't conserved... but I a student in high school. I am good in physics but doesn't mean i can't be wrong..

 

[catkin]

Momentum is always conserved! Except when you draw a system boundary cutting part of the system off, as in this question. The reaction to the Y component of the impulse given to the projectile is on the ground. In effect the Earth is accelerated downwards. This is hard to grasp intuitively because the Earth is so massive (relative to projectiles and such like) that changes to its momentum are imperceptible.

 

[aq1q]

ya u are right.. duh .. momentum is always conserved.. i doubted myself haha

 

[~christina~]

$$m_1v_1= -m_2v_2$$
however since the object is fired at an angle wouldn't it be

$$m_1v_1 + m_2v_2cos theta= 0$$ ?

thus
m1(cannon)= 5000kg
v1= ?

m2 (projectile)= 200kg
v2= 125m/s cos 45

thus...
$$m_1v_1= - m_2v_2 cos theta$$

5000kg (v1)= - (200kg)(125m/s cos 45)

v1= -3.54 m/s ====> recoil speed of cannon
_____________________________________________
Hm If there is momentum in the y direction but the ground is in the way how would the equation look like if I included the y value?

m1 v1 = m2 v2 cos theta

I'm still confused as to what was said about the momentum before and after the cannon fires the projectile.

Is the momentum conserved?
I didn't include the y component of the velocity though so does that make it incorrect?


HELP!

 

[aq1q]

no momentum is conserved..i'm sorry about the previous statement..momentum is always conserved.. read catkin's post... but anyway, you did the problem right... its -3.54m/s go on from there.

 

[~christina~]

h
a)you have to use conservation of momentum..
b) when you have that velocity.. find the x component of that velocity.. and do 1/2(Mc)(Vx)^2= 1/2 (K)x^2 (find x)
c)once you find this x.. F=-kx will give u the maximum force..(the negative is relative)
d)-e) (are different problems, figure out a-c first) :) gl

well
for b) V1= -3.54m/s ===> v of the cannon found in part a)

$$Ki + Ui_s = Kf + Uf_s$$

0 + 1/2 mv^2 = 1/2 kx^2 + 0

m1= 5000kg
vx= -3.54m/s
k= 2.00x10^4 N/m

[.5(5000kg)(-3.54m/s)^2]/ .5(2.00x10^4N/m) = x^2

$$x_{max}= \sqrt{} 3.1329 = 1.77m$$


d) Is the momentum conserved during the firing?

~I don't quite know however...I think it would be no from what was said before about

how the system is cut off from the Earth but the Earth actually accelerates down. The

earth would exert a normal force on the cannon and carriage as it sits on the Earth and

wouldn't the normal force that the Earth exerts on the cannon be increased when the

cannon fires the projectile? Thus since there is a x and y component of the fired

projectile wouldn't that mean that the system's momentum is NOT conserved b/c of the

forces that act on the earth?

However I also think YES it is conserved since from my calculations of finding the recoil velocity is that momentum of the system before is equal to the momentum after...

so I'm really really confused as to this situation.


e)At the maximum height of the trajectory the projectile explodes into 2 fragments. The smaller fragment, 1/3 the projectile's mass drops straight down after the explosion. What is the velocity of the larger fragment when it returns to the level of the cannon?

I guess I assume that the cannon is Sy= 0 ?
I'm not given the height of the cannon so I guess I'll have to assume it is 0

I do know that it would have the same trajectory though as if no explosion had happened

Would I use the conservation of momentum law for this since the mass before is one mass then becomes 2 and it is described that the one that falls is 1/3 the mass of of the original projectile.

I guess I could find the max height first and go from there..

well according to my thinking...

V= 125m/s
theta= 45 deg

Voy= 125 sin 45= 88.39m/s

$$Vy^2= Voy^2 + 2a_y (Sy- Soy)$$

Vy= 0 at max height

[tex]0= (88.39 m/s)^2 + 2(-9.8m/s) (Sy)

Sy= -7812.79/ -19.6 = 398.61m

[tex]Sy_{max}=398.61m[/tex]

Then again thinking about it...I think since it explodes as it is in max height there would only be a x velocity ( the same throughout the projectile's path)
Thus the momentum assuming that right before the explosion it's momentum is..
initially

m = 200kg ==> projectile's mass
vx= vxi = 125 cos 45= 88.39

$$p_i= mvi= 200kg (88.39i m/s)$$

$$p_i = 176780 kg*m/s$$

this would equal to the momentum of the system after the

explosion...however I'm confused as to what the velocity of that particle that drops

down... I know I have to find the velocity of the other part of the particle in the y

direction but before the projectile explodes.. Vy= 0 of the particle that drops down and

the Vx would be only due to the acceleration due to gravity but what would that be? I

know the angle would be 90 but other than that I'm sort of confused..

[tex] p_f= = 200kg /3 ( ? j m/s) + 200kg- (200kg/3) (Vf)

I would equate that..to the momentum before the explosion

pi= pf
176780kg*m/s = 200kg /3 ( ? j m/s) + 200kg- (200kg/3) (Vf)

I would solve for Vf but I'm not sure as to what the velocity of a falling object would be...


f) what is the horizontal distanc from the cannon to the larger fragment at this level
I'd need to find the velocity of that particle first then plug in for the y velocity ...then find the time it takes to reach 0 then..go and plug in the time that I found into the x distance equation...

g) Is the energy of the projectile conserved during the explosion?

I'm not sure what they mean by energy...though I know that some kinetic energy is changed to heat by the explosion so would that be no?


Oh my goodness I almost get the problem...

Please please help me out... where I need it

 

[Doc Al]

well
for b) V1= -3.54m/s ===> v of the cannon found in part a)

$$Ki + Ui_s = Kf + Uf_s$$

0 + 1/2 mv^2 = 1/2 kx^2 + 0

m1= 5000kg
vx= -3.54m/s
k= 2.00x10^4 N/m

[.5(5000kg)(-3.54m/s)^2]/ .5(2.00x10^4N/m) = x^2

$$x_{max}= \sqrt{} 3.1329 = 1.77m$$

Good.

Seems that you left out part c, the maximum force of the spring. Use the answer from b to solve this.

d) Is the momentum conserved during the firing?

~I don't quite know however...I think it would be no from what was said before about

how the system is cut off from the Earth but the Earth actually accelerates down. The

earth would exert a normal force on the cannon and carriage as it sits on the Earth and

wouldn't the normal force that the Earth exerts on the cannon be increased when the

cannon fires the projectile? Thus since there is a x and y component of the fired

projectile wouldn't that mean that the system's momentum is NOT conserved b/c of the

forces that act on the earth?

However I also think YES it is conserved since from my calculations of finding the recoil velocity is that momentum of the system before is equal to the momentum after...

so I'm really really confused as to this situation.

The system you are analyzing is the cannon, carriage, and shell. Since the rails exert a vertical force on the system, momentum is not conserved. (It's not an isolated system.) However, it's true that the horizontal component of momentum is conserved.

e)At the maximum height of the trajectory the projectile explodes into 2 fragments. The smaller fragment, 1/3 the projectile's mass drops straight down after the explosion. What is the velocity of the larger fragment when it returns to the level of the cannon?

I guess I assume that the cannon is Sy= 0 ?
I'm not given the height of the cannon so I guess I'll have to assume it is 0

I do know that it would have the same trajectory though as if no explosion had happened

Would I use the conservation of momentum law for this since the mass before is one mass then becomes 2 and it is described that the one that falls is 1/3 the mass of of the original projectile.

I guess I could find the max height first and go from there..

well according to my thinking...

V= 125m/s
theta= 45 deg

Voy= 125 sin 45= 88.39m/s

$$Vy^2= Voy^2 + 2a_y (Sy- Soy)$$

Vy= 0 at max height

[tex]0= (88.39 m/s)^2 + 2(-9.8m/s) (Sy)

Sy= -7812.79/ -19.6 = 398.61m

[tex]Sy_{max}=398.61m[/tex]

Then again thinking about it...I think since it explodes as it is in max height there would only be a x velocity ( the same throughout the projectile's path)
Thus the momentum assuming that right before the explosion it's momentum is..
initially

m = 200kg ==> projectile's mass
vx= vxi = 125 cos 45= 88.39

$$p_i= mvi= 200kg (88.39i m/s)$$

$$p_i = 176780 kg*m/s$$

this would equal to the momentum of the system after the

explosion...however I'm confused as to what the velocity of that particle that drops

down... I know I have to find the velocity of the other part of the particle in the y

direction but before the projectile explodes.. Vy= 0 of the particle that drops down and

the Vx would be only due to the acceleration due to gravity but what would that be? I

know the angle would be 90 but other than that I'm sort of confused..

[tex] p_f= = 200kg /3 ( ? j m/s) + 200kg- (200kg/3) (Vf)

I would equate that..to the momentum before the explosion

pi= pf
176780kg*m/s = 200kg /3 ( ? j m/s) + 200kg- (200kg/3) (Vf)

I would solve for Vf but I'm not sure as to what the velocity of a falling object would be...

The first thing to do is find the speed of the larger piece immediately after the explosion. During the explosion, momentum is conserved. Since the smaller piece is seen to just "drop straight down", we can assume that its speed is zero after the explosion.

f) what is the horizontal distanc from the cannon to the larger fragment at this level
I'd need to find the velocity of that particle first then plug in for the y velocity ...then find the time it takes to reach 0 then..go and plug in the time that I found into the x distance equation...

g) Is the energy of the projectile conserved during the explosion?

I'm not sure what they mean by energy...though I know that some kinetic energy is changed to heat by the explosion so would that be no?

I suspect that they are asking about the kinetic energy of the projectile (and its pieces). Once you find the post-explosion speeds you can calculate the total KE before and after the explosion and compare. But since the shell explodes, that should give you a clue that some potential energy (chemical energy in the explosive) has been transformed into KE of the pieces.

 

[~christina~]

Seems that you left out part c, the maximum force of the spring. Use the answer from b to solve this.

Oh..I didn't notice that I didn't post it...

x= 1.77m
$$k= 2.00 x10^4 N/m$$

F= -kx since the force from the spring is opposite but equal to the force that the
carriage applies to it will be negative thus..

$$F= -(2.00x10^4 N/m)(1.77m)$$

F= -35,400N

The system you are analyzing is the cannon, carriage, and shell. Since the rails exert a vertical force on the system, momentum is not conserved. (It's not an isolated system.) However, it's true that the horizontal component of momentum is conserved.

I forgot that the cannon's carriage wasn't on the ground but I guess the normal force is from the rails then.
I'm still confused about a what a "isolated system" would be since if there is a car that crashes into a wall since the driver drives into it.

The momentum in the horizontal direction would be conserved I think but the car is also affected by the normal force that the ground exerts on it and also by the gravitational pull of the Earth on it. These factors combined would tell me that the momentum isn't conserved over all?

This is an example in my book but they state:
" the gravitational force nad the normal force exerted by the road on the car are perpendicular to the motion and therefore do not affect the horizontal momentum"

But in this problem where the projectile is fired from the cannon is the gravitational force and the normal force on the cannon still perpendicular to the motion?
I don't think it is.

Does it change anything since it was fired not straight with the angle at 0 deg but with a angle of 45 deg?

(this is what I'm confused about: if firing something at an angle will change compared with the object being fired with no angle)

e)

You didn't say whether it was correct or not..but assuming it was..

pi= pf

pi= mvi

at max height vy= 0 so only vx= 125 cos 45= 88.39m/s

pi= 200kg (88.39i m/s)= 17,678 kg*m/s

pf= 200kg/3 (0m/s) + 200kg- (200kg/3) (vf)

pi= pf

17,678kg*m/s = 200kg- (200kg/3) vf

vf= 132.58m/s ===> this looks reasonable but I'm not quite sure if I went about
getting the answer the correct way

Velocity of the fragment when it returns to level of cannon...
how do I find that??
I'm confused.. do I use the velocity right after the explosion as the initial velocity??


f) what is the horizontal distance from the cannon to the larger fragment at this level

hm..the horizontal distance from cannon to larger fragment ...

time to reach max height= ?
Vy= 0 ==> at max height

Voy= 125 sin 45= 88.39m/s

Vy= Voy + ayt

$$0= 88.39m/s + -9.81m/s^2 (t)$$

$$t_{max}= 9.01s$$

I think that I can't assume that the time it takes to get to the ground from the max height is double the time it takes to reach t max since the velocity of the particle is greater than what was started with.

Do I take that velocity right after the explosion as the initial velocity and see how long it takes for the projectile to reach 0 height then go and find the horizontal distance and add it to the distance from the initial point to the tmax horizontal distance?


THANK YOU

 

[Doc Al]

Oh..I didn't notice that I didn't post it...

x= 1.77m
$$k= 2.00 x10^4 N/m$$

F= -kx since the force from the spring is opposite but equal to the force that the
carriage applies to it will be negative thus..

$$F= -(2.00x10^4 N/m)(1.77m)$$

F= -35,400N

That sounds good. Drop the minus sign, which is only used to give direction of the force with respect to the displacement from the unstretched position.

I forgot that the cannon's carriage wasn't on the ground but I guess the normal force is from the rails then.

You can consider the rails to be part of the "ground" if you like.

I'm still confused about a what a "isolated system" would be since if there is a car that crashes into a wall since the driver drives into it.

I'm not following your example. If a car crashed into a wall, its momentum is certainly changed.

The momentum in the horizontal direction would be conserved I think but the car is also affected by the normal force that the ground exerts on it and also by the gravitational pull of the Earth on it. These factors combined would tell me that the momentum isn't conserved over all?

Again, I'm unsure of your example. The vertical forces on the car cancel out.

This is an example in my book but they state:
" the gravitational force nad the normal force exerted by the road on the car are perpendicular to the motion and therefore do not affect the horizontal momentum"

This is true.

But in this problem where the projectile is fired from the cannon is the gravitational force and the normal force on the cannon still perpendicular to the motion?
I don't think it is.

You are to assume that the cannon+carriage is free to move horizontally. The only force the rails can exert on it is a vertical force. So, yes, the gravitational and normal forces are perpendicular to the motion.

Does it change anything since it was fired not straight with the angle at 0 deg but with a angle of 45 deg?

Sure it makes a difference in the final speed of the cannon, but in both cases horizontal momentum is conserved.

(this is what I'm confused about: if firing something at an angle will change compared with the object being fired with no angle)

Sure things will change. What if the cannon fired straight up? After the explosion, the cannon wouldn't move at all. Regardless, the horizontal component of momentum is conserved because there's no external horizontal force acting on the system.

e)

You didn't say whether it was correct or not..but assuming it was..

pi= pf

pi= mvi

at max height vy= 0 so only vx= 125 cos 45= 88.39m/s

pi= 200kg (88.39i m/s)= 17,678 kg*m/s

pf= 200kg/3 (0m/s) + 200kg- (200kg/3) (vf)

pi= pf

17,678kg*m/s = 200kg- (200kg/3) vf

vf= 132.58m/s ===> this looks reasonable but I'm not quite sure if I went about
getting the answer the correct way

Looks good to me.

It's always a good policy to solve the problem symbolically as much as possible before plugging in numbers. Things may cancel, reducing the errors associated with arithmetic. I'd write it this way:
pi = m vi
pf = (m/3)*0 + (2m/3)vf

vf = (3/2)vi

(The mass cancels out.)

Velocity of the fragment when it returns to level of cannon...
how do I find that??
I'm confused.. do I use the velocity right after the explosion as the initial velocity??

Yes. What direction is the fragment moving in?

f) what is the horizontal distance from the cannon to the larger fragment at this level

hm..the horizontal distance from cannon to larger fragment ...

time to reach max height= ?
Vy= 0 ==> at max height

Voy= 125 sin 45= 88.39m/s

This thing starts out at its maximum height. What's the initial velocity immediately after the explosion? What's the vertical component of that velocity?

Vy= Voy + ayt

$$0= 88.39m/s + -9.81m/s^2 (t)$$

$$t_{max}= 9.01s$$

No, 88.39 m/s is not the initial speed in the vertical direction. You need to recompute the time it takes this thing to hit the ground. What's the height of the shell at the time of the explosion?

I think that I can't assume that the time it takes to get to the ground from the max height is double the time it takes to reach t max since the velocity of the particle is greater than what was started with.

You definitely cannot assume that, since the original shell exploded at the top.

Do I take that velocity right after the explosion as the initial velocity and see how long it takes for the projectile to reach 0 height then go and find the horizontal distance and add it to the distance from the initial point to the tmax horizontal distance?

Yes.

 

[~christina~]

Yes. What direction is the fragment moving in?

I would think that it was going down b/c I do know that the fragment would follow a path as if no explosion had occurred so it wouldn't just be in 1 direction..right?

To find the velocity when it just hits the ground I would need
1. max y distance
so...

Vy^2 = Voy^2 + 2ay(Sy- Soy)

at max Vy= 0

0= (125 sin 45)^2 + 2(-9.8m/s^2)(Sy)

Sy= -7,812.8/-19.6

Sy= 398.61m

Then I know the velocity at the max height right after the explosion but I would think that would be the x component of the velocity right after the explosion since the velocity originally was the x component BUT if the projectile exploded wouldn't that change the x velocity?? I'm confused once again...

V= 132.58m/s

I think I use the y velocity but I don't see how I can get the y component of velocity if I don't know the angle ...

but if I had that I would use the distance to the ground in the equation

Vx= Vix

Vy= Voy + ay(t)

This thing starts out at its maximum height. What's the initial velocity immediately after the explosion? What's the vertical component of that velocity?

Like I said above here in this post..I'm not sure about that vertical component since I would take the velocity right after the explosion which I found was 132.58m/s
and:

Voy= 132.58m/s sin theta ====> but what is theta ??

Is it 0 ? since at max height technically there wouldn't be any vertical component of velocity only horizontal unless...the explosion altered that but I don't see how that would happen.


I'll take care of this part first...


Thanks

 

[Doc Al]

I would think that it was going down b/c I do know that the fragment would follow a path as if no explosion had occurred so it wouldn't just be in 1 direction..right?

Not right. While the center of mass of the fragments will follow a path as if no explosion had occured, each fragment will follow a path based on its initial velocity after the explosion.

To find the velocity when it just hits the ground I would need
1. max y distance
so...

Vy^2 = Voy^2 + 2ay(Sy- Soy)

at max Vy= 0

0= (125 sin 45)^2 + 2(-9.8m/s^2)(Sy)

Sy= -7,812.8/-19.6

Sy= 398.61m

Good. Looks like you correctly calculated the height at which the explosion takes place. (What's the horizontal distance of the explosion from the original starting point?)

Then I know the velocity at the max height right after the explosion but I would think that would be the x component of the velocity right after the explosion since the velocity originally was the x component BUT if the projectile exploded wouldn't that change the x velocity?? I'm confused once again...

You have to figure out the new speed and direction of the fragment after the explosion using conservation of momentum. You did this, but don't fully understand what you did--otherwise you'd know what direction the larger fragment was moving.

Let's redo that momentum calculation more carefully:

What's the initial (total) momentum (just before the explosion)?
(x-direction): (200 kg)*(125 cos45)
(y-direction): 0

What's the final momentum of the small piece (just after the explosion)?
(x-direction): 0
(y-direction): 0

What's the final momentum of the larger piece?
(x-direction): (2/3)m*vfx
(y-direction): (2/3)m*vfy

Applying conservation of momentum:
(y-direction): 0 = 0 + (2/3)m*vfy ==> so vfy = 0! (the large piece moves horizontally!)
(x-direction): (200 kg)*(125 cos45) = (2/3)m*vfx

V= 132.58m/s

That's the horizontal speed of the larger piece just after the explosion. (The vertical component is zero.)

 

[~christina~]

Not right. While the center of mass of the fragments will follow a path as if no explosion had occured, each fragment will follow a path based on its initial velocity after the explosion.

oh..so the center of mass will follow that path only...okay

Good. Looks like you correctly calculated the height at which the explosion takes place. (What's the horizontal distance of the explosion from the original starting point?)

well the time it takes to get to the max height in the y direction

Sy= 398.61m

but I decided to go and get the time it takes to get to the max height with this eqzn instead of having to use the quadradic formula:

Vy= Voy + at
0= 88.39m/s + (-9.8m/s)(t)

t= 9.02s

then to get the horizontal distance to the max height...using that time

Sx= Sox + Vxt

Sx= 0 + 88.39m/s (9.02s)

Sx= 797.28m

You have to figure out the new speed and direction of the fragment after the explosion using conservation of momentum. You did this, but don't fully understand what you did--otherwise you'd know what direction the larger fragment was moving.

Let's redo that momentum calculation more carefully:

What's the initial (total) momentum (just before the explosion)?
(x-direction): (200 kg)*(125 cos45)
(y-direction): 0

What's the final momentum of the small piece (just after the explosion)?
(x-direction): 0
(y-direction): 0

What's the final momentum of the larger piece?
(x-direction): (2/3)m*vfx
(y-direction): (2/3)m*vfy

Applying conservation of momentum:
(y-direction): 0 = 0 + (2/3)m*vfy ==> so vfy = 0! (the large piece moves horizontally!)
(x-direction): (200 kg)*(125 cos45) = (2/3)m*vfx


That's the horizontal speed of the larger piece just after the explosion. (The vertical component is zero.)

[/QUOTE]
Oh..I didn't think about it in terms of momentum...but I did suspect that it was just moving horizontally.

So after the explosion :

Sy= 398.61m
Sx=0 => start from max heigh to ground so at max height I'm considering it to be 0
Voy= 0
Soy= 0
Vox= 132.58m/s

t to reach ground from max height

Sy= Soy + Vyt + .5 at^2

plugging in..

398.61m = 0 + 0t + 4.9 t^2

t= rad (398.61m / 4.9) = 9.02s

t = 9.02s ==> to reach ground

If I'm not incorrect...

The Velocity when the projectile reaches the ground...

Vx= Vox = 132.58m/s
Voy= 0

Vy = Voy + at

Vy= 0 + 9.8(9.02s)

Vy= 88.396m/s

so..

v= $$\sqrt{} (88.396)^2 + (132.58m/s)^2$$

v= 159.35m/s

Is this fine for part e) ?

Thanks

 

[Doc Al]

well the time it takes to get to the max height in the y direction

Sy= 398.61m

but I decided to go and get the time it takes to get to the max height with this eqzn instead of having to use the quadradic formula:

Vy= Voy + at
0= 88.39m/s + (-9.8m/s)(t)

t= 9.02s

Good. But realize that the other way of doing it (using the distance equation) turns out to be just as easy. (No quadratic formula needed.) In fact you use that method later in the problem.

then to get the horizontal distance to the max height...using that time

Sx= Sox + Vxt

Sx= 0 + 88.39m/s (9.02s)

Sx= 797.28m

Good.

So after the explosion :

Sy= 398.61m
Sx=0 => start from max heigh to ground so at max height I'm considering it to be 0
Voy= 0
Soy= 0
Vox= 132.58m/s

t to reach ground from max height

Sy= Soy + Vyt + .5 at^2

plugging in..

398.61m = 0 + 0t + 4.9 t^2

t= rad (398.61m / 4.9) = 9.02s

t = 9.02s ==> to reach ground

Good. (Note that it's no coincidence that the time to go up equals the time to go down, since vertical speed wasn't affected by the explosion.)

If I'm not incorrect...

The Velocity when the projectile reaches the ground...

Vx= Vox = 132.58m/s
Voy= 0

Vy = Voy + at

Vy= 0 + 9.8(9.02s)

Vy= 88.396m/s

so..

v= $$\sqrt{} (88.396)^2 + (132.58m/s)^2$$

v= 159.35m/s

Is this fine for part e) ?

Looks good to me.

 

[~christina~]

Good. But realize that the other way of doing it (using the distance equation) turns out to be just as easy. (No quadratic formula needed.) In fact you use that method later in the problem.

Yes that's only b/c the Voy was 0.

Good. (Note that it's no coincidence that the time to go up equals the time to go down, since vertical speed wasn't affected by the explosion.)

I was wondering about that. So I guess that if the vertical speed isn't affected, the time isn't going to change.

________________________________________

well for part f) find the horizontal distance from the cannon to the projectile when it oes back to the height of the cannon

distance from cannon to max height:

Sx= Sox + Vox(t)

Sx= 0 + 125cos 45( 9.02s)

Six= 797.28m

distance from max height to ground:

Sx= Sox + Vox (t)

t= 9.02s
Vox= 132.585m/s

Sx= 0 + 132.585m/s(9.02s)

Sfx= 1,195.92m

total distance from cannon to projectile in x direction:

Sx= Six + Sfx

Sx= 797.28m + 1,195.92m= 1,993.2m
________________________________________________
g) Is the E of the projectile conserved?

I'm comparing the kinetic E before with the kinetic E after to see...

$$0.5 mv^2= 0.5(m_1v_1^2 + m_2v_2 ^2)$$

Vy= 0 initially and finally
Vix= 88.39m/s
Vfx= 132.585m/s
m= 200kg

falling projectile
Vy= 0
Vx= 0
$$0.5 (200kg)(88.39m/s)^2= 0.5((200kg/3)(0m/s)^2 + (200kg-[200kg/3])(132.585m/s) ^2)$$

781,279,21 J = 1,171,889.5 J

I'd have to say that the potential E or chemical E in the shell was changed into $$E_k$$ thus the kinetic E after was more than before the explosion.


Is this fine?

THANKS Very Much !

 

[Doc Al]

well for part f) find the horizontal distance from the cannon to the projectile when it oes back to the height of the cannon

distance from cannon to max height:

Sx= Sox + Vox(t)

Sx= 0 + 125cos 45( 9.02s)

Six= 797.28m

distance from max height to ground:

Sx= Sox + Vox (t)

t= 9.02s
Vox= 132.585m/s

Sx= 0 + 132.585m/s(9.02s)

Sfx= 1,195.92m

total distance from cannon to projectile in x direction:

Sx= Six + Sfx

Sx= 797.28m + 1,195.92m= 1,993.2m

Careful. What you've calculated here is the distance of the larger fragment from its starting point (where it was shot out of the cannon). But the cannon has moved during the flight of the projectile. (How far does the cannon move back in that time?) [Edit: Ignore this comment--the cannon is attached to a spring.]

g) Is the E of the projectile conserved?

I'm comparing the kinetic E before with the kinetic E after to see...

$$0.5 mv^2= 0.5(m_1v_1^2 + m_2v_2 ^2)$$

Vy= 0 initially and finally
Vix= 88.39m/s
Vfx= 132.585m/s
m= 200kg

falling projectile
Vy= 0
Vx= 0
$$0.5 (200kg)(88.39m/s)^2= 0.5((200kg/3)(0m/s)^2 + (200kg-[200kg/3])(132.585m/s) ^2)$$

781,279,21 J = 1,171,889.5 J

I'd have to say that the potential E or chemical E in the shell was changed into $$E_k$$ thus the kinetic E after was more than before the explosion.

Looks good. But don't write KEi = KEf, even though you are just testing whether it's true. Simply compare KEi to KEf, then point out that KEf > KEi and why that makes sense.

 

[~christina~]

well for part f) find the horizontal distance from the cannon to the projectile when it oes back to the height of the cannon

distance from cannon to max height:

Sx= Sox + Vox(t)

Sx= 0 + 125cos 45( 9.02s)

Six= 797.28m

distance from max height to ground:

Sx= Sox + Vox (t)

t= 9.02s
Vox= 132.585m/s

Sx= 0 + 132.585m/s(9.02s)

Sfx= 1,195.92m

total distance from cannon to projectile in x direction:

Sx= Six + Sfx

Sx= 797.28m + 1,195.92m= 1,993.2m

Careful. What you've calculated here is the distance of the larger fragment from its starting point (where it was shot out of the cannon). But the cannon has moved during the flight of the projectile. (How far does the cannon move back in that time?)

After thinking about it I think that the distance the cannon moves is equal to the distance the spring is stretched isn't it?

That's all I could think of since the cannon is connected to a spring and the spring is connected to a post. The spring didn't break so wouldn't it go back to the initial position it was in at the beginning by the time the projectile reaches the ground?

If it was the distance the spring is stretched the current distance I measured would have 1.77m added to it.

Looks good. But don't write KEi = KEf, even though you are just testing whether it's true. Simply compare KEi to KEf, then point out that KEf > KEi and why that makes sense.

Alright, I actually went and had a question mark over the equivalent sign but I couldn't figure out how to do that here or if it was even possible but next time I'll just find the values for the kinetic energy for the initial and the final values and then put the greater than/ less than or equivalent sign.



Thanks Doc Al

 

[Doc Al]

After thinking about it I think that the distance the cannon moves is equal to the distance the spring is stretched isn't it?

That's all I could think of since the cannon is connected to a spring and the spring is connected to a post. The spring didn't break so wouldn't it go back to the initial position it was in at the beginning by the time the projectile reaches the ground?

If it was the distance the spring is stretched the current distance I measured would have 1.77m added to it.

Forget about what I said--I forgot that the cannon was attached to a spring. Just assume the cannon ends up where it started. Your original thinking (and answer) was correct.

Alright, I actually went and had a question mark over the equivalent sign but I couldn't figure out how to do that here or if it was even possible but next time I'll just find the values for the kinetic energy for the initial and the final values and then put the greater than/ less than or equivalent sign.

Good.

Is this what you were trying for?
$${KE}_i \stackrel{?}{=} {KE}_f$$

 

[~christina~]

Forget about what I said--I forgot that the cannon was attached to a spring. Just assume the cannon ends up where it started. Your original thinking (and answer) was correct.

If the cannon wasn't connected to the spring how would I calculate the distance it travels backwards?

Would I use the recoil velocity
for example in this situation vx= -3.54m/s ?

Good.

Is this what you were trying for?
$${KE}_i \stackrel{?}{=} {KE}_f$$

Actually, that's exactly what I had in mind. Unfortunately I'm just learning to use Latex.
I think that's what it's called from what I've read in the other places on the forum.

How'd one learn to use it though? It's not like there's a manual or anything.

 

[Doc Al]

If the cannon wasn't connected to the spring how would I calculate the distance it travels backwards?

Would I use the recoil velocity
for example in this situation vx= -3.54m/s ?

Exactly.

Actually, that's exactly what I had in mind. Unfortunately I'm just learning to use Latex.
I think that's what it's called from what I've read in the other places on the forum.

How'd one learn to use it though? It's not like there's a manual or anything.

Two ways to go:

(1) Read this: https://www.physicsforums.com/showthread.php?t=8997".

(2) Click on the $$\Sigma$$ symbol in the post edit window. (Look for it among all the other formatting symbols.)

 

[~christina~]

Well I kind of don't know how to go about getting that distance it travels though.

I have the Vx= -3.54m/s

I'm not sure but I don't think I have

acceleration ax

time

I do know that vx will be 0 when it stops though

distance => I would need to find...

Hm...it seems like I have a bunch of unknowns but I would think that I'm wrong of course. which is most likely the case.

Exactly.

Two ways to go:

(1) Read this: https://www.physicsforums.com/showthread.php?t=8997".

(2) Click on the $$\Sigma$$ symbol in the post edit window. (Look for it among all the other formatting symbols.)

When I click the $$\Sigma$$ which is what I usually do and I try to put something together but if I make it too complicated I usually don't get what I want as a result and thus that's why I usually make the divison symbol with / sign.

But I shall read that post about using LaTex when I have time.
It should help me.


Thanks