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