@JohnRennie ... Yes...

@NehalSamee Hi

Hello ...

So , have you read my problem ..?

spring is hanging at 20m height from the ground . A ball of mass 0.2kg is shot towards the spring with velocity 49m/s such that extension of spring is 3m . What will be the rebound velocity of ball on the ground ?

Yup...

I've tried it using forces and found a result...

Bit when I used energy .

.I couldn't ...

So the ball is launched upwards at 49 m/s, then it hits the spring and compresses it, then the spring shoots the ball back down again and it hits the ground with velocity $v$. And we're trying to calculate $v$. Is that correct?

Yes ...

It will hit the ground with the same velocity that it was launched ...

But how ...?

The spring gives it an extra push , which is the restoring force ...

If it hits the ground at a different velocity some energy has been lost

Oh, wait a minute. Is the end of the spring fixed, or is the whole spring free to move upwards?

Look , if I threw it in the air only , then it will stop and fall with same velocity ...

Ahhh....There you go...

Yes ...

The spring is fixed...

@JohnRennie Is the question clear 🤔🤔

?

@NehalSamee The ball hits the spring at some velocity $v'$. As it compresses the spring the KE of the ball turns to PE until the ball stops. Then the spring expands again and the PE turns back into KE. If no energy is lost, when the ball leaves the spring it willhave the same speed that it had when it ht the spring.

What I did : $v²=u_o ² -2g.20$... v is the velocity at the moment of touching spring .... Then I found kinetic energy and said that it will turn into elastic potential energy ...

Then I found k and then restoring force ... I divided by mass and found the acceleration for spring ...

How can there be extension in the spring?

I've clearly misunderstood the problem ...

When you throw something towards it?

That , I added with $g$ and used v²=2(a+g) . 20 ...

@MadhuchhandaMandal I think it must mean compression.

Yes ... The spring is compressed 3m ...

Or may be they are trying to say that due to some fault in the spring a permanent extension of 3m happens (permanent set)

Who knows

🤔

Mmm...The ball *compresses spring...

@MadhuchhandaMandal ...

The set up us in a vertical plane ...

After compressing...The ball rebounds ...

@JohnRennie...

I understand... Okay

Then what I said in my calculation ... Is there anything wrong ....?

@NehalSamee Inbound velocity not equal to outbound velocity

That's what I'm trying to say ... The spring gives a push ... What say @JohnRennie ...

@NehalSamee not due to that

Is it like this?

Yes....

@JohnRennie i don't think so

I drew a rectangle for the spring cos I can't draw springs :-)

@JohnRennie they must mean at the 3m compressed position, the ball rebounds

That's the picture ...I'm using Android ...So I can't post that picture...

I think

@NehalSamee if that picture is correct then the ball must return to the ground with the same speed it left the ground.

@JohnRennie ofcourse

But think of the situation I said

So , what about the other case ...?

I mean , at the compressed position the ball rebounds

Right ...

What would be the solution then ...?

(Although seems impractical)

You have to subtract the elastic potential energy of the spring

And calculate the new velocity

@NehalSamee the problem is we don't seem to be agreeing about what the question is.

Like how ...?

I think I've made mistake from translating into English ... But I got it right now ...

@JohnRennie ... We can proceed with the argument stated by @MadhuchhandaMandal...

The ball readily rebounds after compressing the ball up to 3m ...

I that case the final KE is the initial KE minus 0.5k(3)^2.

Why minus ?

@JohnRennie - 3mg

i.e. the energy required to compress the spring a distance x is 0.5kx^2 and tat energy has to come from the KE of the ball

@MadhuchhandaMandal no, the ball loses 3mg on the way up and gets back the 3mg on the way down.

Then what about the gravitational energy at 23m ...?

@JohnRennie okay in that case initial refers to at 20m height.. not initially after rebound

@NehalSamee the ball starts with a KE of 0.5 m (49)^2. Yes?

Yes ...

That's 240.1 J

Now what's 0.5kx²...?

And it ends with a KE of 0.5 m v^2, where we are trying to calculate v. Yes?

Yes ...

What about the elastic potential energy ...?

Calculation ...

And if v < 49 m/s then the ball has lost some energy. Yes?

@NehalSamee need not worry about that

Yes ...@JohnRennie ...

@NehalSamee so where did that lost energy go?

But give the calculation of 0.5kx²...

I understood that elastic potential has used up lost energy ...

But I'm concerned regarding the calculation of it ...

Yes I guess that's what the question means. The lost energy is just 0.5 k (3)^2

@NehalSamee Epe has caused the lost energy

Now ...What is k ???

It's not given...

The spring constant

If it's not given you can't answer the question.

But I'm still not convinced we have understood what the question is asking.

What about :

I calculate the final velocity while throwing at 20m ...

It seems very impractical situation

Then I use it for calculating kinetic energy ...

@MadhuchhandaMandal there has to be something we have misunderstood about the question.

That kinetic energy is stored as epe...

@JohnRennie ... In the above mentioned process , we can calculate epe...

@NehalSamee But why do you think it should work? (In that case we are taken back to model 1 of the question)

That is , no change in velocity

But what I'm doing is before rebound...

Ah, I see what you mean. The KE when the ball touches the spring is 0.5 mV^2 (where we can work out V from the height). Then 3mg + 0.5 k (3)^2 = 0.5 mV^2

@MadhuchhandaMandal ... The confusion will arise if it is after rebound ...

@NehalSamee you don't know exactly when it should rebound, so total Kinetic energy won't be stored in it

Is that what you meant?

A fraction of it would be stored

@JohnRennie...Yes...

@JohnRennie that's model 1 of the question. No change in vel

Isn't there a much easier way to do this?

If we're saying the spring doesn't rebound then the ball simply falls from rest a distance of 23 metres

So 23 m g = 0.5 m v^2

:p 3rd interpretation

You see ... The ball compresses up to the limit ... So that's the question said...

If we see it like this way ...After compression when the ball starts to rebound ..It's not in free air ...The restoring force at the very beginning might give a push ...

My book calculated it that way ...They took the acceleration of restoring force into account ..And added with acceleration due to gravity ...

@NehalSamee if the spring returns to its original length then the energy of the spring hasn't changed.

Still ... The body might have got some extra force ..

@NehalSamee How can they take Acceleration due to spring ?? It will be good if you Post a picture of the solution

then the energy of the spring hasn't changed

@NehalSamee and thus violate conservation of energy

If it would not ... Then with spring and with air is the same...

That means the total energy (KE+PE) of the ball cannot have changed either

@NehalSamee of course that's same here

Until you take into account 2&3rd interpretation of the question

Then considering which condition will they differ ...

@NehalSamee 2nd and 3rd

Ya...I'm taking into the 2nd one , where ball rebounds instantly...

@NehalSamee k must be known

To solve that case

But I'm having difficulty to understand why not should be known ...

(or might be its a text book with Young's modulus value given)

I mean ... Why not consider the math I posted...?

@NehalSamee which one?

The one for finding k...

With v ..

0.5 k 9 + 3mg = Initial KE at 20 m ?

That one?

Yes..

@Abcd @Tanuj @GaurangTandon Do you know any good Online resource to quickly revise formulas and properties of Conic sections (Parabola, Hyperbola, Circle , Ellipse and Straight lines)

@MadhuchhandaMandal Cengage Coordiinate geometry has all formulas.

Is it available online?

@Abcd

@MadhuchhandaMandal No, but I have noted them for Straight lines to Ellipse...Do you want them?

Yeah that would be extremely helpful

@Abcd

@MadhuchhandaMandal okay, I'll try to send it before sleeping (i.e in the morning at around 7 or 8). Bye for now :) !

Okay!! Bye.. Good night-study

:-p