Will the pebble meet with the block according to the given condition?

[Mahfuz_Saim]

Homework Statement

In the figure below, AB= 2.5 m and AC= 8 m. The spring is stretched by 0.2 m and then released from rest at t=0 sec so that the spring executes SHM. At the time a pebble is thrown with a velocity V= 11.2 m/s from point C so that the pebble meets the block B at t= 1 sec. The angular frequency of the spring is π/3.
Question 1: Find the velocity of the spring at 1 sec.
Question 2: Will the pebble meet with block B according to the given condition?

Relevant Equations

v= Aω*cos(ωt+δ)
v= ω√(A^2-x^2)
t=x/ v cosθ

Question 1:
I have used v= Aω*cos(ωt+δ) where A= 0.2 m, ω= π/3, t=1 and δ=0. Are the values right in this case? I am confused.

Question 2:
From question 1 I have got the value of V which is 9 m/s. By using v= ω√(A^2-x^2), I have got the value of x. Now, do I need to add it with 2.5(distance between AB) or subtract from 2.5? I am confused. Please explain with the cause.

By using t=x/ v cosθ, I have got the value of x after t second of the projectile. By doing some simple calculation, I can get its distance from A. But there's no information about the height of the block. So can I say they will meet if the distance from A of them are the same?

I hope you have understood my question properly. Please help me. Thank you.

 

[haruspex]

I have used v= Aω*cos(ωt+δ) where A= 0.2 m, ω= π/3, t=1 and δ=0.

What does that give for the velocity at t=0?

 

[Delta2]

Your value of $\delta$ is not correct. It must be such , that at time t=0, the velocity v equals 0.

I think you are supposed to assume that the height of the block is zero. So block and projectile will meet(if they will do) when the projectile hits the ground. So find the time t that the projectile hits the ground, and then find at this time t where exactly the block is. you ll need to write down the equation for x of the block as $x=A\cos(\omega t+\delta_1)$. (now it will be $\delta_1=0$ i think).

 

[Mahfuz_Saim]

@haruspex, do I need to use δ= π/2? Maybe, yes. Why?

@Delta2, there is nothing about the height in the question. That's more confusing. The range of the projectile is more than the distance of AC. So, maybe if they stay at the same point on t= 1 sec, maybe the answer will be yes. Please let me know what you are thinking.

 

[haruspex]

@haruspex, do I need to use δ= π/2? Maybe, yes. Why?

To get the zero initial velocity it will have to be +/-π/2. To decide which, consider the displacement at t=0.

By using v= ω√(A^2-x^2)

Or you could use x=A*sin(ωt+δ) as displacement from B.

there is nothing about the height in the question

We are not told the block's width either, so take it as a point mass at ground level.

 

[Mahfuz_Saim]

Okay, but I am not talking about the width. The main problem is with height. If I consider the block as a point, the question becomes meaningless as the projectile's range is more than AC.

 

[haruspex]

the projectile's range is more than AC.

And substantially more than the furthest the block ever gets from C. So the answer to that part is...?

 

[hutchphd]

Okay, but I am not talking about the width. The main problem is with height. If I consider the block as a point, the question becomes meaningless as the projectile's range is more than AC.

Not be my reckoning. Please justify

 

[Mahfuz_Saim]

If I am not wrong, the horizontal range= (v^2*sin2θ)/g. Then it will be 11.08. So, the pebble will land 11.085 m away from C. Is it correct?
Please tell me if I am going in the right way or not. If not, please tell me the right way to solve this question.

 

[hutchphd]

sin2$\theta$= 0.5
v=11m/s
$g=10m/s^2$

 

[Mahfuz_Saim]

The angle is 60°. So, sin120°= √3/2. Am I correct?

 

[hutchphd]

Oh I guess so! It works out so well with that error.
My apologies this is more interesting (or not!) . Let me regroup.

 

[hutchphd]

So where is the pebble at t=1s? How high up is it?
Where is the block center at t=1s
I guess you could calculate the minimum size of the block to make a collision at 1s?Otherwise I don't understand the question.

 

[Mahfuz_Saim]

By using v= ω√(A^2-x^2), I will get a value of x. Do I need to add with the distance of AB? As the block is stretched, I think I need to add the value of x as I am keeping the value of A positive. Am I right? Please reply. Thank you.

 

[Mahfuz_Saim]

I guess you could calculate the minimum size of the block to make a collision at 1s

Yes, I want to add that with my answer. Please make an answer to my last question. Thank you.

 

[hutchphd]

I would point out that the easier way to find x is to use the explicit harmonic oscillator solution $x=A\sin(\omega t +\pi /2)$. This sets your origin at the original point B. But your way will give the same result with a little more work.

 

[Mahfuz_Saim]

Okay, thank you very much, @hutchphd . This was my second question here and have learned a lot from it.

 

[hutchphd]

Did you figure a height for the block?...I would like to know (I didn't work it all the way)

 

[Mahfuz_Saim]

One more thing, do I need to add the value of x with AB or subtract in order to get the distance from A and why? Please answer.

 

[hutchphd]

x is measured from the resting center of spring. So draw the system (for yourself) approximately at the time=1s with arrows showing the speed. If the phase and constants are correct ( I think they are ) then that should make it clear. In particular the block should be moving left but not yet to the spring center I believe.

 

[Mahfuz_Saim]

As it is on the left side, they will meet at a point which is 2.4 m away from A. And the minimum height of the block should be 4.7995 m. (If I have not done any mathematical error). Thanks.

 

[hutchphd]

I was also wondering where is the center of the block when that happens. it will be to the right of point B so the block will need to be at least wide enough to accommodate...
Good work numbers seem right to me ( but I did not calculate )