Kinetic Energy, Momentum, Projectile Motion ( A real stumper )

[xaer04]

Homework Statement

From the text:

"While standing on frictionless ice, you (mass 65.0 kg) toss a 4.50 kg rock with initial speed 12.0 m/s. If the rock is 15.2 m from you when it lands, (a) at what angle did you toss it? (b) How fast are you moving?"

mass of me = 65.0 kg
mass of rock = 4.50 kg
initial velocity = 12.0 m/s
initial height = ?
initial angle = ?
distance = 15.2 m
speed of my body (recoil) = ?
acceleration of gravity = 9.8 m/s^2
time of flight = ?
xo = 0
vo = 0

There is no energy lost to outside forces.

Homework Equations

These are simply equations I've tried toying with

Newton's Second Law
F= Ma

Projectile Motion Disregarding time
v^2 = vo^2 +2a(x-xo)

Definition of work
Wnet = Integral(F*dx)


And these are ones i know i'll need later

Conservation of Momentum
m1(v1o) + m2(2o) = m1(v1f) + m2(v2f)

Conservation of kinetic energy
(1/2)m1*(v1o)^2 + (1/2)m2*(v2o)^2 = (1/2)m1*(v1f)^2 + (1/2)m2*(v2f)^2

The Attempt at a Solution

I have ideas, but none of them seem to work because I'm not given an angle or an initial height to work with (i'm actually supposed to find the initial angle). I just don't know where to begin on this problem. I know that in order to make that specific distance that initial velocity will need to be at a specific angle, and I have the notion that this is either going to be simple and I'm going to kick myself for overlooking it or it's going to be overly complicated... although the latter is less likely. Please help by giving me a concept to use, steer me in the right direction.

 

[arildno]

Remember that the ratio between your horizontal velocity and the rock's horizontal velocity equals the negative reciprocal mass ratio, due to conservation of (horizontal) momentum.

There being no horizontal forces acting upon either of you, the ratio between your traveled horizontal distance and the rock's traveled horizontal distance equals the same negative reciprocal mass ratio.

Use this to determine the RANGE of the rock first.
The range of the rock is the horizontal distance from the point it was thrown, to the point where it landed, relative to the ground.

 

[xaer04]

alright, i worked out the reciprocal mass ratio, but the range thing... is that not the distance i already gave? (it was given that it landed at 15.2 m) or are you speaking in terms of it being at the same height it was thrown at after a little fly time?

i've thought of something - i think this problem may be unsolvable. I'm asked to give an initial angle, but without an initial height or a flight time that's impossible because it can fly to 15.2 m from different heights, which all would require different angles (within a range).

 

[Doc Al]

The problem is quite solvable. Start by figuring out how long the rock was in the air (in terms of its initial speed and angle). Then use that to calculate:
(1) the horizontal distance the rock traveled from the start point
(2) the horizontal distance you traveled from the starting point

Then you can apply the given data to solve for the angle.

 

[xaer04]

*sigh*

my point is, the rock can travel 15.2meters being thrown at 12.0 m/s at different heights with a different angle at each height. flight time, launch height, and launch angle are all dependant on each other, and none are given, so it's impossible to state that for this scenario there is one solveable answer for either part(since part b relies on part a - different launch angles will yield different horizontal momentums, obviously). i think the best i can do is find an equation that works with what i have and what i need, and input different variables to show that without the necessary information there are many answers for this problem.

 

[arildno]

Okay, let's continue!

Due to mass conservation, your velocity wrt. to the ground is:
v_{you}=-\frac{4.5}{65.0}*v_{rock}
Your position at the time of impact relative to the origin is therefore:
p_{you}=-\frac{4.5}{65.0}p_{rock}
The distance between you and the rock was given to be 15.2m, that is:
p_{rock}-p_{you}=15.2\to{p}_{rock}=\frac{65.0}{4.5+65.0}*15.2m

p_{rock} is the sought RANGE of the rock.
Did you follow this?

 

[xaer04]

oh my, i feel like such a ditz. I'm sorry if got on your nerves... I'm always overlooking how things are worded. i imagined the distance being the distance the rock travelled, not including the displacement of my body:(

*yikes*

i'll rework it and let you guys know how it goes, lol...

 

[arildno]

To give you a further hint:
Remember that the SPEED of the rock contains both the horizontal&vertical velocity components of it!
Thus, when you are given that the initial speed of the rock is 12m/s, you have to make one of the following interpretations:
1. This is the speed the rock has relative to YOU. This results in nasty implicit expressions

2. 12m/s is the initial speed the rock has relative to the GROUND. This makes your problem relatively easy to solve.

So, choose interpretation 2.
In this case, when solving the rock problem, you have:
v_{0}=12.0,v_{x,0}=v_{0}\cos\theta_{0},v_{y,0}=v_{0}\sin\theta_{0}
where the initial velocity components are determined up to a sine or cosine of the sought launch angle \theta_{0}

 

[Luke1294]

Alrighty, I'm working on this problem as well. However, I'm running into a good bit of problem.

I solved time to be t=\frac{12sin\Theta}{g}, and V_{0x} to be 12cos\theta. But now, I'm running into a bit of a problem.

When i try to plus these values into the equation x= x_0 + V_{0x} t, I wind up with 11.6=cos\theta * sin \theta, and haven't the slightest clue where I went wrong or what to do to fix it. Any pointers would be greatly appriciated.

 

[xXmarkXx]

Alrighty, I'm working on this problem as well. However, I'm running into a good bit of problem.

I solved time to be t=\frac{12sin\Theta}{g}, and V_{0x} to be 12cos\theta. But now, I'm running into a bit of a problem.

When i try to plus these values into the equation x= x_0 + V_{0x} t, I wind up with 11.6=cos\theta * sin \theta, and haven't the slightest clue where I went wrong or what to do to fix it. Any pointers would be greatly appriciated.

Luke, I'm having the same problem too. I'm a little lost. I read through this whole post plenty of times and i still can't figure it out.

 

[Doc Al]

I solved time to be t=\frac{12sin\Theta}{g}, and V_{0x} to be 12cos\theta.

Is that the total time that it's in the air?

But now, I'm running into a bit of a problem.

When i try to plus these values into the equation x= x_0 + V_{0x} t, I wind up with 11.6=cos\theta * sin \theta, and haven't the slightest clue where I went wrong or what to do to fix it.

Hint: Review your trig identities; you can replace cos\theta * sin \theta with a single trig function. (You are almost there! )

 

[Luke1294]

Is that the total time that it's in the air?Hint: Review your trig identities; you can replace cos\theta * sin \theta with a single trig function. (You are almost there! )

I do believe it should be the total time in the air. I solved that from v_y = v_y0 - gt , so...Yes, yes it should be the total time in the air.

As for sin*cos, that can be rewritten as sin(2\theta)= 2sin\theta*cos\theta, yes? Maybe my algebra/trig is a bit rusty, but I keep getting domain errors and it is making my TI-89 cry...as well as myself.

 

[Luke1294]

I guess I should write out what I'm doing so you can steer me in the right direction, haha. Because of the aforementioned identity, i multiplied both sides by 2- giving me 23.6=2sin\theta*cos\theta, then rewrote it as 23.6=sin(2\theta). From here, my first instinct would be to take the inverse sin of 23.6, but that is a domain error in a big way. SO...help? PS, thanks for the help so far.

 

[xXmarkXx]

I guess I should write out what I'm doing so you can steer me in the right direction, haha. Because of the aforementioned identity, i multiplied both sides by 2- giving me 23.6=2sin\theta*cos\theta, then rewrote it as 23.6=sin(2\theta). From here, my first instinct would be to take the inverse sin of 23.6, but that is a domain error in a big way. SO...help? PS, thanks for the help so far.

Luke, i am completely lost. Could you help me out from what you have so far?

 

[Doc Al]

I do believe it should be the total time in the air. I solved that from v_y = v_y0 - gt , so...Yes, yes it should be the total time in the air.

What's the final velocity?

As for sin*cos, that can be rewritten as sin(2\theta)= 2sin\theta*cos\theta, yes?

That's what you'll need.

 

[Luke1294]

What's the final velocity?


That's what you'll need.

Final velocity...oh! because it's on frictionless ice, it should keep it's V_{0x} value for its final velocity, shouldn't it?

And I'm still hitting a brick wall with the trig.

 

[Doc Al]

Final velocity...oh! because it's on frictionless ice, it should keep it's V_{0x} value for its final velocity, shouldn't it?

I'm still talking about V_y that you used in calculating the time. (You made an error that I want you to find.)

And I'm still hitting a brick wall with the trig.

One step at a time.

 

[xXmarkXx]

can anybody help me with this problem. I can't find the angle. I understand how to solve for the distance the rock was thrown from the origin. I then solve for time, and i am stuck. any suggestions?

 

[Luke1294]

Doc, thank you so much for your help. Unfortunatly, I keep staring at this and can't find my error. I sset it up so that V_y, the final Y component of the velocity, is zero. As such, 0 = V_{0x} - gt, or 0=12sin\theta - gt. I then add the gt to the opposite side, divide by g to get my t value. I honestly haven't the slightest clue where I'm going wrong. A gentle nudge would be much appriciated.

 

[Doc Al]

Doc, thank you so much for your help. Unfortunatly, I keep staring at this and can't find my error. I sset it up so that V_y, the final Y component of the velocity, is zero.

So, when you toss a ball in the air its speed is zero when it lands back into your hand?

 

[xXmarkXx]

so nobody would like to help me solve for the angle?

 

[Doc Al]

so nobody would like to help me solve for the angle?

If you want specific help, show what you did step by step. (Or at least describe the specific step you are stuck on.)

 

[Luke1294]

No, you're right. It's speed is most certainly not zero. It's the same it started out with.

But that just seems to be leading me in a weird direction, probably because I'm so frazzled. I rewrote the expression as -12sin\theta=12sin\theta-gt. From there, i added 12sin, and arrove at 0=24sin\theta-gt, which brought me to t=\frac{24sin\theta}{g}.

I kept going, and plugged that into x= x_0 + V_{0x} t, giving me 14.31=\frac{12cos\theta*24sin\theta}{g}.

I multiplied through by g, then divided by 12 to give me 11.68=12sin\theta*cos\theta.

I then divided through by 6... giving me 1.944=2sin\theta*cos\theta, or 1.944=sin(2\theta).Now, my double angle trig is a bit flakey...so this might be very illegal. I divided 1.944 by 2, and took the inverse sin...giving me 76.4 degrees. Does my math seem correct? Oh man, you have helped me a TREMENDOUS amount.

 

[xXmarkXx]

If you want specific help, show what you did step by step. (Or at least describe the specific step you are stuck on.)

for projectile problems, i assume we would use t=x/v₀cos(theta)₀ so i pluged in the x distance from the origin to the landing (14.2) and i also pluged in 12.0 for v₀. Now i am left with two unknowns.

 

[Doc Al]

No, you're right. It's speed is most certainly not zero. It's the same it started out with.

But that just seems to be leading me in a weird direction, probably because I'm so frazzled. I rewrote the expression as -12sin\theta=12sin\theta-gt. From there, i added 12sin, and arrove at 24sin\theta-gt, which brought me to t=\frac{24sin\theta}{g}.

Good!

I kept going, and plugged that into x= x_0 + V_{0x} t, giving me 14.31=\frac{12cos\theta*24sin\theta}{g}.

Where did you get the 14.31?

I multiplied through by g, then divided by 12 to give me 11.68=12sin\theta*cos\theta.

Check your arithmetic.

 

[Doc Al]

for projectile problems, i assume we would use t=x/v₀cos(theta)₀ so i pluged in the x distance from the origin to the landing (14.2) and i also pluged in 12.0 for v₀. Now i am left with two unknowns.

Find the time by considering the vertical motion. Then use that time in calculating the horizontal motion.

 

[Luke1294]

The 14.31 came from an earlier post, by arildno. He helped me find the range of the rock by playing with the conservation of momentum equation to give p_{rock}-p_{you}=15.2\to{p}_{rock}=\frac{65.0}{4.5+65.0}*15 .2m.

Haha, I see that arithmetic error. A big one ideed. Unfortunatly, now I'm a bit lost...as I now have 11.68=2sin\theta*cos\theta...but I can't take the inverse sin of 11.68, or 5.84...ARGGGGHHHHHH.

 

[Doc Al]

The 14.31 came from an earlier post, by arildno. He helped me find the range of the rock by playing with the conservation of momentum equation to give p_{rock}-p_{you}=15.2\to{p}_{rock}=\frac{65.0}{4.5+65.0}*15 .2m.

That's cool.

Haha, I see that arithmetic error. A big one ideed. Unfortunatly, now I'm a bit lost...as I now have 11.68=2sin\theta*cos\theta...but I can't take the inverse sin of 11.68, or 5.84...ARGGGGHHHHHH.

You still have an arithmetic error.

 

[xXmarkXx]

Find the time by considering the vertical motion. Then use that time in calculating the horizontal motion.

so v_y=v_y0-gt?

 

[Doc Al]

so v_y=v_y0-gt?

Sure. (Read the other posts in this thread!)

 

[xXmarkXx]

Sure. (Read the other posts in this thread!)

i have been..i'm really confused.

now i have...
(12sin(theta))/9.8=(14.2)/12cos(theta) does that look ok??

 

[Luke1294]

Doc...i'm not going to lie, I cannot find the arithmertic error for the life of me.

When I multiply through by g, i wind up with 140.238=12cos24sin. So I factor out a 12, and divide by that... which brings me to 11.68=2 sin x cos x.

Could you please give me a pointer as to where I'm going wrong? I don't just want the answer.

 

[Doc Al]

Doc...i'm not going to lie, I cannot find the arithmertic error for the life of me.

...giving me 14.31=\frac{12cos\theta*24sin\theta}{g}.

Do it step by step:
(1) multiply by g
(2) divide by whatever you need to to get 2 on the right

 

[xXmarkXx]

i got 23.2=sin2(theta), but i used 14.2 instead of 14.3

do you agree doc?

 

[Luke1294]

Hm. So, I was looking at that the wrong way for sure. That can definatly be written as 140.238= 288cos\theta sin\theta, can't it? So, I divide both sides by 144...giving me .973875=2sin\theta cos\theta, or .973875=sin(2\theta). I then divide the left by 2, take the inverse sin, and arrive at 29 degrees?

Am i error free?! Finally?!

 

[xXmarkXx]

Hm. So, I was looking at that the wrong way for sure. That can definatly be written as 140.238= 288cos\theta sin\theta, can't it? So, I divide both sides by 144...giving me .973875=2sin\theta cos\theta, or .973875=sin(2\theta). I then divide the left by 2, take the inverse sin, and arrive at 29 degrees?

Am i error free?! Finally?!

luke, how did you get 14.3 for x? doesn't 14.3 = the momentum of the rock?

 

[Doc Al]

Hm. So, I was looking at that the wrong way for sure. That can definatly be written as 140.238= 288cos\theta sin\theta, can't it? So, I divide both sides by 144...giving me .973875=2sin\theta cos\theta, or .973875=sin(2\theta).

Good.

I then divide the left by 2, take the inverse sin, and arrive at 29 degrees?

Take the inverse sine first, then divide by 2 to get the angle theta.

 

[Luke1294]

Doc,

Thank you so much for your help. This problem even stumped our professor- apparently, he was instructing students to use some "likely" values to make the problem much simpler. Once again, thank you for helping me work through it.To the OP-
We're in this class together, pal. Jus' sayin.

 

[Luke1294]

As I'm looking back through this problem, I'm not sure i understand one little part- Why is it that the denomenator is 4.5 +65? I thought I had a handle on it earlier, but it's escaping me.

The distance between you and the rock was given to be 15.2m, that is:
p_{rock}-p_{you}=15.2\to{p}_{rock}=\frac{65.0}{4.5+65.0}*15.2m

p_{rock} is the sought RANGE of the rock.
Did you follow this?

 

[xaer04]

you can find
p_{you}

by using conservation of momentum. set initial velocities to zero, and solve for
v_{you}

since momentum is constant after the throw (lack of friction, etc), that implies time is constant, so the M/S of the velocity becomes simply meters, and it can be used as a ratio of displacements, or
p

(arildno meant positions, btw) values. then, you know
p_{rock}-p_{you} = 15.2 m

and you can substitute in for
p_{you}

and the rest is just solving for p_rock.

i think it's really interesting that you guys spent more time solving this than me... :) i assume you're solving it for y_o = y_final (the starting height was the same as the landing height)? i still think there's not enough information though (i guess I'm just stubborn), but i am going to set up equations... heh, I'm just a computer science major anyway... physics isn't my strong subject (if that isnt' obvious already)

 

[Luke1294]

Thanks for the help. Like I said, you and I are in there with good old Mellendorf together...

If you read a ways up, you can see we actually found an angle, and then found the velocity of the man. Problem solved :)

 

[denverdoc]

hey we need something to do.

Actually I was surprised this one went on so long, as it was pretty straightforward by recognizing that the absolute velocity of separation was 3 percent larger than that of the rock alone. I agree tho, that the Yo issue was never resolved. I assumed it was thrown two handed as one might a bowling ball off the ground--after all it was close to 10#. Could have been shot put I suppose if the rock of that mass was small enuf to palm..

 

[xaer04]

denverdoc: heh... i just don't like holes in my problems, i suppose :)

luke: you wouldn't happen to be one of the guys that sits in the back left corner?

 

[Luke1294]

No sir, I happen to be upfront next to the deciptively cute blonde girl/ex girlfriend.

 

[xaer04]

ouch:(

so... would you happen to have been in cook's wednesday afternoon chem 130 lab last semester?

 

[Luke1294]

Yessir, that would also be me.

 

[denverdoc]

denverdoc: heh... i just don't like holes in my problems, i suppose :)

luke: you wouldn't happen to be one of the guys that sits in the back left corner?

Nor do I--so can someone tell me how this 140# guy threw from ice, and at rest, a 3.5 liter object some 50'. Thats 2/3'rds the size of a bowling ball, a distance of 2/3'rds the world shotput record!
J

 

[xaer04]

lol... got to love physics homework problems:)