Conversation started Apr 17, 2020 at 7:23.
Apr 17, 2020 07:23
@JohnRennie: Hi sir. Good morning :-)
Could you please reply to the following once you're free? It seems you're already busy in two rooms.
> In projectile motion, the modulus of rate of change of speed:
> (a) is constant
> (b) first increases then decreases
> (c) first decreases then increases
> (d) None of the above
My approach:
I decomposed the two dimensional motion to two orthogonal one dimensional motions - one along the horizontal and the other vertical. I'm calling the horizontal axis as the $x$ axis and the vertical as the $y$ axis.
The acceleration along the $x$ axis is zero whereas it's $g$ along the $y$ axis. So the velocity remains constant along the $x$ but it changes along the $y$ axis.
The speed of the projectile during the ascending phase decreases with time and it increases when it starts its descending phase. So rate of change of speed is $-g$ for half the flight time and $+g$ for the remaining half.
As the question asks for the modulus of rate of change of speed, the answer is always $g$. Or as per the options, the modulus of rate of change of speed remains constant (option a).
But the book states the answer to be (c). Could you please tell whether the answer in the book is incorrect or not?
(End of message)
Apr 17, 2020 07:56
Please ignore this message (but not the ones above :-) ): DCP-03-047-04
Apr 17, 2020 08:14
@GuruVishnu hi
Apr 17, 2020 08:48
@JohnRennie Hi sir. Could you please ping me after Aladdin's doubt?
@GuruVishnu hi, I think he has finished now.
Ok sir. If possible, can you please look at the previous block of messages? That's a simple question from kinematics however I'm facing issues with the answer provided by the book.
What you did was work out (correctly):
v_x is constant
v_y first decreases then increases
Yes?
Yes sir.
And it asks about the speed, $u$, which is $u = |\sqrt{v_x^2 + v_y^2}|$
i.e. the speed is the modulus of the velocity
Apr 17, 2020 08:55
Ok sir. I think I went wrong when I decomposed speed along two perpendicular axes.
Speed is a scalar not a vector. It doesn't have a direction so you cannot decompose it into components.
Fine. I understand that. I'm trying to do this by taking this into account which I didn't earlier.
Could you give any hints, sir?
@JohnRennie sir?
Are you free now, or shall we continue afterwards?
@GuruVishnu ah OK I thought it was obvious now. Are we agreed that $u = |\sqrt{v_x^2 + v_y^2}|$ ?
Yes sir.
And we know $v_x$ is constant
Apr 17, 2020 09:08
Yes sir.
And we know $|v_y|$ decreases to zero then increases again.
And $v_y^2$ decreases initially and then increases.
Yes
So that has to mean $u$ decreases at first then increases again
Yes sir. That's obvious.
But it was about the rate at which the speed varies. Let me differentiate the expression given by you.
Oh wait, sorry, I've just spotted that the question asks about the rate of change of speed.
Apr 17, 2020 09:10
No problem sir. But yes.
Let's draw the speed as a function of time (diagram incoming):
Ok sir.
I don't know exactly what shape the curve is because I'd need to differentiate the equation for $u$ to get that, but we know it starts off positive, goes to zero then goes positive again. So it's going to look something like this. Yes?
And the question asks about "modulus of rate of change of speed" i.e. $|du/dt|$
@JohnRennie Yes sir. Shouldn't that be something like the shifted graph of $|x|$?
@JohnRennie Yes sir. This is what I got: $$\frac{du}{dt}=\frac{v_yg}{\sqrt{v_x^2+v_y^2}}$$
So, $$\left|\frac{du}{dt}\right|=\frac{|v_y|g}{\sqrt{v_x^2+v_y^2}}$$
I would probably write that as:
$$ \left| \frac{du}{dt} \right| \propto \frac{1}{\sqrt{1 + v_x^2/v_y^2}} $$
Apr 17, 2020 09:18
Ok sir.
Yes, sorry, that was a typo.
No problem sir.
That is certainly not going to look like a graph of $|x|$, shifted or otherwise.
Fine sir. The reason I thought was, a velocity time graph is a straight line.
It's going to be some kind of smooth curve, possibly with a discontinuity at $u=0$ but it won't be two straight line segments.
Apr 17, 2020 09:21
Ok sir. I understand.
You can calculate the curve by substituing $v_y$ by the expression for it as a function of tim.
9 mins ago, by John Rennie
user image
So it's going to look something like this.
Ok sir.
I think we can also express the same thing as: $$\frac{du}{dt}=g\sin\theta$$ where $\theta$ is the angle between the velocity vector and the horizontal.
And the "modulus of rate of change of speed" that the question asks about is the modulus of the gradient.
As $\theta$ is easy to visualise in a projectile motion. First decreases then increases in the negative direction.
@JohnRennie Yes sir.
And the modulus of the gradient first decreases then increases again.
Apr 17, 2020 09:25
Ok sir. Thank you for the clarification. Now we got the correct answer.
I don't know why I got trouble with this simple question.
Apr 17, 2020 09:37
16 mins ago, by John Rennie
That is certainly not going to look like a graph of $|x|$, shifted or otherwise.
@JohnRennie: I agree with this point. But is that because of the constant horizontal component of the velocity which causes the variation from a linear graph?
Yes. If the motion were vertical i.e. $v_x = 0$ then we'd get $v_y(t) = v_0 - gt$
Then $u(t) = |v_y(t)| = |v_0 - gt|$
Fine. It's a modified |x| graph.
Ok sir. So even though one of the components of velocity has a linear relation and the other orthogonal component remains constant, the resultant velocity may not have a linear relation. I didn't realise this so far. Could you please tell whether this conclusion is correct or not, sir?
Yes that's correct
It's because $|v| = \sqrt{v_x^2 + v_y^2}$
Apr 17, 2020 09:47
Ok sir. Understood. I can take the mathematics from here.
i.e. the speed is not a linear function of $v_x$ and $v_y$.
Got it sir :-)
Thank you for your help :-)
 
Conversation ended Apr 17, 2020 at 9:48.