Projectile motion, potential and kinetic energy

[Suyash Singh]

Homework Statement

https://www.physicsforums.com/attachments/upload_2018-4-29_12-40-30-png.224871/

Homework Equations

The Attempt at a Solution

1/2mv^2=10
1/2m(ucos60)^2=10
1/2mu^2/4=10
mu^2=80

what to do now?

 

[PeroK]

This is not an easy problem. You'll need to think about the relationship between kinetic and potential energy.

Hint: the 10J is irrelevant. You can forget about that.

 

[haruspex]

Homework Statement

https://www.physicsforums.com/attachments/224871

Homework Equations

The Attempt at a Solution

1/2mv^2=10
1/2m(ucos60)^2=10
1/2mu^2/4=10
mu^2=80

what to do now?

So what was the initial KE? What will it be at height y?

I see this is a duplicate of https://www.physicsforums.com/threads/projectile-motion.946149/

 

[Suyash Singh]

someone removed my attachment

 

[Suyash Singh]

Ke + pe = total energy

 

[Suyash Singh]

Ke remains same?
because horizontal velocity is same
I am not sure:(

But max height=u^2sin^2(theta)/2g

 

[haruspex]

In your post 1 you found

mu^2=80

So what was the initial KE?

Ke + pe = total energy

Yes, so how much KE will be lost in rising height y? And what will the KE be then?

 

[Suyash Singh]

KE(at y)=KE-mgy
KE lost is mgy

 

[Suyash Singh]

KE-mgy=mgy
KE=2mgy
1/2mv^2=2mgy
1/2v^2=2gy
y=(1/4) (v^2/g)

 

[haruspex]

KE-mgy=mgy
KE=2mgy
1/2mv^2=2mgy

Your lack of detail has confused you. What exactly is that KE in the first two lines? The KE when?

 

[Suyash Singh]

Your lack of detail has confused you. What exactly is that KE in the first two lines? The KE when?

the loss in ke is the gain in pe

 

[haruspex]

the loss in ke is the gain in pe

That is not what I asked.
You obtained an equation KE=2mgy, where y is the height you are trying to find. But what KE is that? The KE at what point in the process? Check back through your working to make sure.

 

[Suyash Singh]

That is not what I asked.
You obtained an equation KE=2mgy, where y is the height you are trying to find. But what KE is that? The KE at what point in the process? Check back through your working to make sure.

it is the KE at height y

 

[haruspex]

it is the KE at height y

No.

In posts 8 and 9 you wrote:

KE(at y)=KE-mgy
KE lost is mgy

KE-mgy=mgy
KE=2mgy

Rewrite those statements in a way that makes it clear what KE each of those KE terms refers to. E.g. KE(x) for the KE at height x.

 

[Suyash Singh]

No.

In posts 8 and 9 you wrote:
Rewrite those statements in a way that makes it clear what KE each of those KE terms refers to. E.g. KE(x) for the KE at height x.

oh sorry i made silly mistake
KE(y)=KE-mgy
when,
KE(y)=PE(y)
=mgy
KE=2mgy

so initial kinetic enregy is 2 times of potential energy at height y (y is the height where KE(y)=PE(y))
but how come kinetic energy decreases if horizontal velocity is same

 

[haruspex]

how come kinetic energy decreases if horizontal velocity is same

Because this is total KE, made up of horizontal and vertical contributions.

 

[Suyash Singh]

KE at height y
KE(y)=mgy
PE(y)=mgy

 

[haruspex]

KE at height y
KE(y)=mgy
PE(y)=mgy

Yes, but you already have

mu^2=80

(So the initial KE is?)
And

KE=2mgy

 

[Suyash Singh]

Yes, but you already have

(So the initial KE is?)
And

1/2mu^2=40
2mgy=40
y=20/mg
y=2/m

 

[haruspex]

1/2mu^2=40
2mgy=40
y=20/mg
y=2/m

It is almost always a bad idea to plug in numbers before the final step. Keep everything symbolic as long as you can. I wish now that I had got you to make that switch first.

The useful part of what you have so far is that y=E/4g, where E is the initial KE.
Your original calculation of that KE was wrong anyway. You had ½m(u cos 60)²=E_final. What is cos²60?

 

[Suyash Singh]

It is almost always a bad idea to plug in numbers before the final step. Keep everything symbolic as long as you can. I wish now that I had got you to make that switch first.

The useful part of what you have so far is that y=E/4g, where E is the initial KE.
Your original calculation of that KE was wrong anyway. You had ½m(u cos 60)²=E_final. What is cos²60?

1/4

 

[Suyash Singh]

But he said that 10 joules is useless
1/2m(u^2 1/4)=10
mu^2=80

 

[haruspex]

But he said that 10 joules is useless

Quite so, which is a good reason for not plugging that value into equations. You can keep it as a variable name, but it will cancel out later.

Let's start over...

If the launch speed is u and the mass m, the initial KE, KE_i=½mu².
As you say, the KE at the top, KE_f, is 1/4 of that, so mu²/8.
So, what is the PE at the top, PE_f?
So what is the height at the top in terms of u and g?

At the point of interest, PE_y=KE_y=½KE_i=¼mu².
So what is the height of that in terms of u and g?

 

[Suyash Singh]

Quite so, which is a good reason for not plugging that value into equations. You can keep it as a variable name, but it will cancel out later.

Let's start over...

If the launch speed is u and the mass m, the initial KE, KE_i=½mu².
As you say, the KE at the top, KE_f, is 1/4 of that, so mu²/8.
So, what is the PE at the top, PE_f?
So what is the height at the top in terms of u and g?

At the point of interest, PE_y=KE_y=½KE_i=¼mu².
So what is the height of that in terms of u and g?

Ok
KEi=1/2 mu^2
KEf=1/8 mu^2
PEi=0
PEf=mg(h)
h is the max height
mg(h)=1/2 mu^2 - 1/8mu^2
mgh=3mu^2/8
gh=3/8 u^2
h=(3u^2)/(8g)
mgy=1/2 mv^2 = 1/4mu^2
mgy=1/4mu^2
gy=1/4 u^2
y=u^2/4g

y/h=u^2/4g x 8g/3u^2
y=2h/3

 

[Suyash Singh]

Omg this question was so hard!

 

[haruspex]

Omg this question was so hard!

It was not easy, but perhaps you have learned the benefits of:

• Keeping everything symbolic
• Using a distinct symbol for each variable
• Clearly defining the variables