Why does the speed remain the same?

[lioric]

Homework Statement

Homework Equations

F= qvB sin theta
F=ma
[/B]

The Attempt at a Solution

A) 6.2x10^18N
B)9.5x10^8m/s^2
C) this is the problem when I searched lot of solutions they say that speed remains the same
Could someone explain this to me[/B]

 

[Orodruin]

How is the force direction related to the particle velocity?

 

[lioric]

How is the force direction related to the particle velocity?

Using the right hand rule, v and b are at 52 degrees and the force should be 90 degrees to the velocity. Am I right?

 

[Orodruin]

Using the right hand rule, v and b are at 52 degrees and the force should be 90 degrees to the velocity. Am I right?

Yes, but what conclusion can you draw from this? (The magnetic force is always at a right angle to both the field and the velocity!)

 

[lioric]

I don't understand that yet

 

[Orodruin]

So let me ask you this: how do you compute the speed?

 

[lioric]

So let me ask you this: how do you compute the speed?

It's already given in the question 550m/s

 

[physics user1]

Think about a circular motion: The velocity can change in direction but the speed remains the same..

 

[Orodruin]

It's already given in the question 550m/s

I mean in general, not necessarily in this problem.

 

[ehild]

It's already given in the question 550m/s

Orudruin asks, how do you get the speed (scalar) if you know the velocity ( a vector) .

 

[lioric]

I mean in general, not necessarily in this problem.

Speed = distance/ time
If you know the velocity the magnitude is the speed right?

 

[Orodruin]

If you know the velocity the magnitude is the speed right?

Yes, so how do you find out the magnitude of a vector? Or, more useful, how do you find the squared magnitude of a vector? (The magnitude is constant if the magnitude squared is constant!)

 

[lioric]

Yes, so how do you find out the magnitude of a vector? Or, more useful, how do you find the squared magnitude of a vector? (The magnitude is constant if the magnitude squared is constant!)

I didn't quite get that but I ll give it a shot
Magnitude squared in this case 550^2

 

[lioric]

I didn't understand the magnitude square thing

But I think this is what happened
An alpha particle with a velocity of 550m/s goes in a magnetic field in an angle of 52 degrees
So like in the right hand rule the alpha particle is deflected 90 degrees to the field so the acceleration is to change the velocity by changing the direction but the speed remains the same.

If that is right how would I know if the speed did change?

 

[Orodruin]

I didn't understand the magnitude square thing

This is something you will have to understand to understand the solution of the problem.

Magnitude squared in this case 550^2

We are not interested in this particular case at the moment, we are interested in the general case when you are given a velocity vector $\vec v$. If I give you a general velocity vector $\vec v = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z$, what is the speed of the object?

If that is right how would I know if the speed did change?

This is what we are trying to guide you to! This is why you should follow the steps I am giving you.

 

[lioric]

This is something you will have to understand to understand the solution of the problem.

We are not interested in this particular case at the moment, we are interested in the general case when you are given a velocity vector $\vec v$. If I give you a general velocity vector $\vec v = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z$, what is the speed of the object?

This is what we are trying to guide you to! This is why you should follow the steps I am giving you.

I'm really trying to get this in my head please continue. I'll follow the steps
As for the question with the 3D vector, I don't know but if it was just x and y only v would be found using Pythagoras theorem. I know that much. I don't know about the 3 D one

 

[Orodruin]

As for the question with the 3D vector, I don't know but if it was just x and y only v would be found using Pythagoras theorem. I know that much. I don't know about the 3 D one

Pythagoras' theorem holds in an arbitrary number of dimensions.

 

[ehild]

As for the question with the 3D vector, I don't know but if it was just x and y only v would be found using Pythagoras theorem. I know that much. I don't know about the 3 D one

Have you learned about the scalar product (dot product) of two vectors?

 

[lioric]

This is something you will have to understand to understand the solution of the problem.

We are not interested in this particular case at the moment, we are interested in the general case when you are given a velocity vector $\vec v$. If I give you a general velocity vector $\vec v = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z$, what is the speed of the object?

This is what we are trying to guide you to! This is why you should follow the steps I am giving you.

I think I got it tell me if this is right
V= square root Vx^2+Vy^2+Vz^2

Is this the speed?

 

[Orodruin]

I think I got it tell me if this is right
V= square root Vx^2+Vy^2+Vz^2

Is this the speed?

Yes, so what is the derivative of the speed with respect to time? You may use that dVx/dt = ax (x-component of acceleration) and so on.

 

[lioric]

Yes, so what is the derivative of the speed with respect to time? You may use that dVx/dt = ax (x-component of acceleration) and so on.

Acceleration? But it's derivative of velocity

 

[Orodruin]

Acceleration? But it's derivative of velocity

Yes, that is exactly what I said!

 

[Orodruin]

To clarify, the derivative of speed is not acceleration! You need to compute it instead of making wild guesses.

 

[lioric]

To clarify, the derivative of speed is not acceleration! You need to compute it instead of making wild guesses.

I m sorry
I know rate of change of velocity is acceleration but I don't know if there is a rate of change of speed

 

[Orodruin]

I m sorry
I know rate of change of velocity is acceleration but I don't know if there is a rate of change of speed

Of course there is, and it is what the problem wants you to compute! I am not asking you for a name (names are conventions), I am asking you to do the actual math.

 

[lioric]

Of course there is, and it is what the problem wants you to compute! I am not asking you for a name (names are conventions), I am asking you to do the actual math.

Ok let's see
If v = square root Vx^2+Vy^2+Vz^2
You want me to differentiate this with dv/dx?

 

[Orodruin]

You want me to differentiate this with dv/dx?

With respect to t, not x.

 

[lioric]

With respect to t, not x.

But there is no t in there

 

[Orodruin]

But there is no t in there

Yes there is. The velocity generally depends on time. The derivative of velocity wrt time is called acceleration.

 

[lioric]

Yes there is. The velocity generally depends on time. The derivative of velocity wrt time is called acceleration.

Ya but we are finding speed not velocity

 

[Orodruin]

Ya but we are finding speed not velocity

Seriously? You have gotten the speed as a function of velocity, now you need to differentiate this function with respect to time, the speed is generally going to depend on time because the velocity on which the speed depends does. Just apply the chain rule!

 

[lioric]

Seriously? You have gotten the speed as a function of velocity, now you need to differentiate this function with respect to time, the speed is generally going to depend on time because the velocity on which the speed depends does. Just apply the chain rule!

I think I understand
The speed doesn't change because the particle goes in a circular motion and no work is done on it

 

[Orodruin]

I think I understand
The speed doesn't change because the particle goes in a circular motion and no work is done on it

The motion is a spiral, not a circle, but the spirit of it is the same. What I want you to do is to work this out in mathematics and to see why the magnetic force does not do any work (in general, it does not matter if it is homogeneous or not - the motion may not be a spiral or circle in that case).

 

[lioric]

The motion is a spiral, not a circle, but the spirit of it is the same. What I want you to do is to work this out in mathematics and to see why the magnetic force does not do any work (in general, it does not matter if it is homogeneous or not - the motion may not be a spiral or circle in that case).

That's my problem I don't know how to do that in math

 

[ehild]

That's my problem I don't know how to do that in math

You learned the dot product (scalar product) of two vectors, and you should know that the dot product of a vector with itself is the square of the magnitude of the vector. So $\vec b \cdot \vec b = |\vec b|^2$, or you can write it in the form $a=\sqrt {{\vec b} ^2}$, $b$ meaning the magnitude of $\vec b$.

You have the time dependent velocity vector $\vec v(t)$ . The speed is its magnitude, $v(t)= \sqrt {{\vec v(t)} ^2}$
v(t) is just a function of $\vec v$, and $\vec v$ is a function of t, Determine dv/dt, using the chain rule. What do you get?
It might be confusing that $\vec v$ is vector. How would you differentiate a function F(t)=f(g^2(t)) with respect to time?

 

[lioric]

You learned the dot product (scalar product) of two vectors, and you should know that the dot product of a vector with itself is the square of the magnitude of the vector. So $\vec b \cdot \vec b = |\vec b|^2$, or you can write it in the form $a=\sqrt {{\vec b} ^2}$, $b$ meaning the magnitude of $\vec b$.

You have the time dependent velocity vector $\vec v(t)$ . The speed is its magnitude, $v(t)= \sqrt {{\vec v(t)} ^2}$
v(t) is just a function of $\vec v$, and $\vec v$ is a function of t, Determine dv/dt, using the chain rule. What do you get?
It might be confusing that $\vec v$ is vector. How would you differentiate a function F(t)=f(g^2(t)) with respect to time?

Df/dt right?
It would be g^2 since t has no power and the differentiation is with respect to t I don't need to do anything with g

 

[ehild]

Do you know the chain rule? What is the derivative of $f(t) = \sqrt{sin(t)}$ with respect to t, for example?

 

[lioric]

Do you know the chain rule? What is the derivative of $f(t) = \sqrt{sin(t)}$ with respect to t, for example?

Sorry my bad

Is this ok?

 

[Mark44]

Sorry my bad

View attachment 97676
Is this ok?

No. Somehow you ended up with the correct derivative, but the work you show is wrong.
In the second line of your attached image, you have
$$\frac{df(t)}{dt} = (\sin t)^{1/2} \cdot \sin t$$
This is wrong. On the right side, you haven't taken the derivative yet -- $(\sin t)^{1/2}$ is the same as $\sqrt{\sin t}$. Also, where did the extra factor of $\sin(t)$ come from?
The third line is correct, and so is the last line.

 

[ehild]

View attachment 97676
Is this ok?

The end is OK. Now you have the function $f(t)= {\vec v(t)} ^2$. What is df/dt? Do not worry about that the inner function is a vector, just apply the chain rule.

 

[lioric]

The end is OK. Now you have the function $f(t)= {\vec v(t)} ^2$. What is df/dt? Do not worry about that the inner function is a vector, just apply the chain rule.

Do/dt=2v(t)

 

[ehild]

Do/dt=2v(t)

No, you have to differentiate the inner function, too. If v(t)= sin(t), for example, what is the derivative of f(t)=(sin(t))² ?

 

[lioric]

No, you have to differentiate the inner function, too. If v(t)= sin(t), for example, what is the derivative of f(t)=(sin(t))² ?

2(sin (t))cos t

 

[ehild]

2(sin (t))cos t

Good. cos(t) is the derivative of the inner function. So how do you write the derivative of f(t)=(V(t))² in general?

 

[lioric]

Good. cos(t) is the derivative of the inner function. So how do you write the derivative of f(t)=(V(t))² in general?

2(V(t))V

 

[Orodruin]

2(V(t))V

No, you did not take the derivative for the inner derivative.

I am sorry to say so, but your attempts here indicate to me that you are not mathematically prepared to take the physics class you are in. A word of advice would therefore be to go back to your single and multi variable calculus before trying to go further. Continuing without the prerequisite math is only going to come back to bite you in the end.

 

[lioric]

No, you did not take the derivative for the inner derivative.

I am sorry to say so, but your attempts here indicate to me that you are not mathematically prepared to take the physics class you are in. A word of advice would therefore be to go back to your single and multi variable calculus before trying to go further. Continuing without the prerequisite math is only going to come back to bite you in the end.

I know you guys have tried very hard to help me. I'm sorry for being such a pain
And I know there is no escape from maths. I just have hard time understanding the question.
Thank you for all you have done I'll try to get by some other way.

 

[Orodruin]

I just have hard time understanding the question.

This here is exactly why I recommend you to revisit your calculus. If you do not even understand the question, how can you hope to understand the answer? There is no shame in going back to repeat the basics, you will need the basics to build on that knowledge.