Hi @JohnRennie :)

Hi :-)

When we want the point of impact in the projectile motion

I saw that the formula tan(theta) = |vy|/vx is used

But I've seen that sometimes using a different formula than tan(theta) = vy / vx , I get a different result

Yes, you could use vₓ = v cosθ

It should work out the same.

I had the data, vx = 5.56, v = 12.71, vy = 11.44

We can work through the problem if you want ...

I get the same result, I'm the one who typed the numbers wrong 🤦‍♂️

on the calculator

OK :-)

In fact, it seemed strange to me that it was different

If you are free i can send an exercise

OK, go ahead

A block of mass M = 1kg moves in a straight line on a rough horizontal plane with a dynamic friction coefficient μ = 0.5. At the instant in which the block has a speed Vo = 3 m/s, it is hit by a projectile, with mass m = 10 g, which travels horizontally in the same direction as the block with a speed v = 117 m/s.

The projectile, after the supposedly instantaneous impact, remains stuck in the block. Determine:

a) the speed of the block-projectile system immediately after the impact; b) the time elapsed between the moment of the impact and the moment the system stops; c) the distance traveled by the block-projectile system before stopping.

I did it if you want we can discuss it together

to see if the steps are correct

I hope the translation of the text is clear

OK, go ahead and explain how you did part (a)

M = mass of the block, m = mass of the projectile

M • v0 + m • v1 = (m+M)v

Yes, you use conservation of momentum.

And i found v = 4,12

you cannot use conservation of energy because the collision of the bullet and the block is inelastic, but momentum is still conserved.

Yes

Now point (b)

I used suvat equations

v = u + at

So I had to find a , I did it using the force diagram

y: N = (m+M)g , x: -Ff = (m+M)a

-μ (m+M)g = (m+M)a

a = -4,9

Yes :-)

Is it correct that the acceleration comes out negative?

The acceleration is in the opposite direction to the velocity (i.e. it is a deceleration)

So if we are taking the direction of the velocity to be the positive direction, which seems reasonable, then the acceleration is indeed a negative number.

Ah yes yes

It's negative because the velocity is decreasing.

If a was positive it would mean the velocity was increasing.

@Pizza Yes

So when the system stops, v = 0

Yes

0 = u + at , 0 = 4,12 - 4,9t , t = 0,84

I haven't checked the arithmetic, but yes that's the correct method.

point (c)

@JohnRennie It should be right at least that I didn't make any mistakes on the calculator, like before -_-

KE = Wf

1/2 (m+M)v^2 = -Fa • d

1/2 ( 1 + 0,010 ) • 4,12^2 = -4,949 • d

Yes.

The problem is that I get d = -1.72

Shouldn't it be positive?

You got mixed up about the sign of the work. This is very easy to do because in any situation like this we have the work done by the block and the work done on the block, and they are equal and opposite.

Shall I go into more detail about this?

Yes

Suppose you do work, e.g. pick up something heavy, then it's your energy that goes into doing the work. So your energy decreases by the amount of work done by you.

Yes?

Yes

In this case the block+bullet have some initial KE and that KE goes into the work done by the block.

Yes

And the work done by the block is the force exerted on the floor by the block times the distance moved by the block.

Now, we say the frictional force is to the left, because it is slowing the block. But this is the force exerted by the floor on the block.

Does this make sense so far?

Yes

And by Newton's third law the force exerted by the block on the floor must be equal and opposite, so the force exerted by the block is to the right i.e. positive.

Yes?

Aaaa

In fact, it all makes sense

OK :-)

I had considered the force exerted by the floor on the block.

This is super confusing and even I get confused about after all these years :-)

When in fact it was the opposite!

Yes :-)

So wait

1/2 (m+M)v^2 = Fa • d

d = 1,72!

@JohnRennie I tried this I get the same result

Again, I haven't checked the arithmetic but this is the correct method.

Yes

@Pizza Yes, but this time the sign is correct i.e. the displacement vector is to the right (in the direction of the velocity).

Friction is already "present" in the acceleration we calculated previously, right?

And therefore we can also use this other formula

Yes

In suvat equations , i always need 3 variables. If i know three, then i can plug those in and find the fourth

Anyway, thanks so much for the help!

You're welcome :-)

Hi :)

But couldn't we use that Wff=Final Mechanical energy - Initial Mechanical Energy?

@Pizza here

@BinkyMcSquigglebottom It works out the same.

@JohnRennie 👍

A box of mass m=20 kg is allowed to slide from the top of a guide (A) placed at a height h=3m and inclined at an angle θ compared to the horizontal. At the end of the guide (B), the chest continues along the plane until the end of it (C). The distance between point C and the projection onto the plane of A is L=5m. Be along the guide that along the plane the case is affected by sliding friction, with dynamic friction coefficient μD=0.3

Determine the work W of the forces acting on the ball in the motion from A to C (Solution : W=294.3 J)

I just found out that the name SUVAT, would be all the variables contained in the equations

I can't do it, it seems I'm missing some data 😔

@BinkyMcSquigglebottom How far have you got with it?

I don't have theta, I don't know what to do

Is the friction the same both along the ramp and along the floor?

Yes

In that case we may find that θ cancels out and the final result is independent of θ

Or of course it may be a mistake in the question!

If you want we can go through the calculation and see what happens ...

Okay

@BinkyMcSquigglebottom I've drawn in some values we know on the diagram.

Does this all make sense so far?

Where does h/tan(th) come from?

Take the distance from A to B i.e. the length of the ramp. Let's call this 𝑦.

Okay

Yes

So y = h/sinθ

Yes

Yes

Ah ...

Right

@JohnRennie yes

OK :-)

So let's start by calculating the work done from A to B.

Mmm

Friction work and weight-force work?

👍

@BinkyMcSquigglebottom Hmm, the question doesn't make clear exactly what work it means. Let's find the friction work since the work done by gravity is just mgh and we can add it later if we need to.

Okay

The work done by friction is where θ will cancel out if it does cancel out. So let's just do the friction work for now.

Wff=Ek-Pe(Initial pe)

@BinkyMcSquigglebottom Let's go with my method for now ...

@JohnRennie right

@JohnRennie okay

Now let's find the work done from B to C

Can you do this?

Okay

Wbc = μmg × [L - h/tanθ]

Yes, so what happens when we add Wab and Wbc to get the total work done by friction?

Theta is erased!!!

Yes :-)

Wtot_ff=μmgL

If you think a question does not give you all the information you need it's usually because the missing parameters will cancel out. Though you do sometimes encounter questions that are wrong.

@BinkyMcSquigglebottom Yes, and in fact there is a quick way to see this.

Right

If θ does not matter then we can choose θ = 90° i.e. the block falls straight down to D then slides from D to C.

Okay

Then there is no friction from A to D because the block is just falling straight down and there is no normal force.

Yes

And in that case the work is just μmgL

Correct

And that's the same result we got when we did the long calculation that did include θ

Yes, in fact doing the calculations, using g=9.81 I obtain precisely W=294,3 J

Which was the solution to the problem :⁠-⁠)

So it did just mean just the work done by friction i.e. not including the change in gravitational PE.

But what would have happened if I had also calculated the work of the weight force?

How did I know that I only had to calculate the friction one

The way the question is written is not clear.

Oh ...

Unfortunately not all questions are as well written as they could be.

But I would have guessed it mean only the friction since that's usually the hard bit to calculate.

Maybe theta wouldn't have canceled out?

The work done by gravity is just mgh. Yes?

Yes

And that does not depend on θ.

So whether you include the work done by gravity or not makes no difference to θ cancelling.

But since work is a dot product, what happens to cos(theta)?

The gravitational force is mg, and the PE change is indeed ΔU = mg s cosφ where φ is the angle between the force vector and the displacement vector.

OK so far?

Yes

I've drawn the two vectors and the angle φ in blue.

Does this make sense so far?

Yes

Now ADB is a right angle triangle so φ = 90 - θ

That means cos(φ) = cos(90 - θ)

Yes?

Yes

Yes

Yes

Can you see where this is going now?

Yes

OK :-)

s=h/sin(theta)

so

Making the substitution

mg (h/sin(theta)) • sin(theta) = mgh

Yes. When we are considering changes in gravitational PE we only need to consider the vertical displacement. We don't care what track the body followed between the initial and final heights. Only the change in the height matters.

Right

Thanks so much for the help!

You're welcome :-)

@JohnRennie Are you by any chance still available for another exercise? It's quick

If you're busy, don't worry though

I'm just having doubts, it's not that I can't solve the problem

I'm busy at the moment, but if you want to post the question I can look at it as soon as I'm free.

Yes, I will also write everything I have done

A solid cylinder of mass m=1kg and radius R=30cm, rolls without sliding along an inclined plane of height h=3.1 m. The plane is rough, and has a slope with respect to the horizontal equal to an angle $\theta$ = 30°. To calculate:

Point (a) : $\frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 = mgh$

Where $I = \frac{1}{2}mR^2, \quad \omega = \frac{v}{R}$

Solving this equation I find $v$ at the base (B)

Point (b) : $fR = I \alpha$

Where $\alpha = \frac{a}{R}$

$fR = \frac{mR^2}{2} \cdot \frac{a}{R}$

$f = \frac{ma}{2}$

Now I write the force diagram

On the x axis : $-f + mg\sin(\theta) = ma$

On the y axis: $N = mg cos(\theta)$

So : $-\frac{ma}{2} + mgsin(\theta) = ma$

$a = \frac{2}{3}g \sin(\theta)$

$f = f_s = \mu_s \cdot N$

But : $f = \frac{ma}{2}$

$\mu_s mg \cos(\theta) = \frac{ma}{2}$

$\mu_s = \frac{ma}{2(mg\cos(\theta))}$

Point (c): $\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2 = Wf$

$Wf = \mu_d mg d$

$d = \frac{3v^2}{4(\mu_d g)}$

It should be so

I noticed that for point (c) , if I use $-\mu_d mg = ma$

$a = -\mu_d g$

And then I use this acceleration, in the suvat equation:

$v^2 = u^2 + 2ad$

Where u in this case is the v that i found before

I get a different result

@Pizza yeah... these two do give different results. but im not sure why. i will reply if get something

i feel the second solution must be the correct one, but it doesnt take into account the rotational energy

@RyderRude 👍

@Pizza ok so the work done by friction on a rolling object is a complicated topic. it's not simply F*d

where $d$ is the overall displacement of the body

see this physics.stackexchange.com/questions/806487/… @Pizza

so the first method is incorrect

Now I'll see thank you very much!

note that friction directly acts on the point of contact of the cylinder and the ground... the work done depends on the relative motion of that point wrt the ground

@RyderRude But if the cylinder slipped, would it have been correct to use the first method?

@Pizza ok.. im not sure anymore. ill let someone else answer

the friction should be zero if it's rolling without slipping.. so it doesnt make sense for it to stop

@JohnRennie You can take a look above as soon as you're there pls :)