Solving a Simple Problem: Frustration with Mechanics

[wuffle]

Once again I can't seem to do this simple problem, not sure why.

hate mechanics :(

Homework Statement

Charged particle enters a region containing constant magnetic field and leaves it after 708 micro seconds, what is Fx, the x component of the force after being 236 micro seconds long in the constant magnetic field
Also it travels a quarter circle(not sure if the right word, basically a half of a semicircle)

we're given:

B=1.2 T
R=0.95 m
m=5.7*10^-8 kg

Homework Equations

ma=qvb

The Attempt at a Solution

I calculated the constant velocity when the charge is in the magnetic field, the only problem i have is calculating Vx and Vy after 236μsec, particularly i can't find the angle, one way i tried solving it is realizing that 236/708 is 1/3, so distance that particle has traveled is

(2piR/4)/3 which i found to be 0.49742...then idk, i thought maybe the angle would be 60 deg or 30 deg since it has traveled 1/3 of the way so i tried finding Vx=V*cos(30) or Vx=V*cos(60) but it didn't work out, i think i know that on halfway the angle is 45(Vx=Vy), but that's it, basically the equation i got is

Fx=m*(V*cos(theta))^2/R and I can't find theta :(

 

[Simon Bridge]

When the particle enters the B field, it has circular motion.
So you can use the kinematics of circular motion to find the angle that rotates through by the time it leaves the field.

 

[Basic_Physics]

The force stays perpendicular to the motion so the speed stays constant. A quarter circle is 90 degrees so a third of it is 30 degrees.

 

[Simon Bridge]

Though you are encouraged to think of angles in radians.

The size of an angle is the distance around the circumference of a unit circle that is inside the angle.

The total distance around the whole circle is $2\pi$.
A half circle is therefore $\pi$ and a quarter circle is $\pi/2$.
The units here are "radius units" or "radians" for short.

Use these and a lot of physics equations will make more sense - like the arc-length inside angle $\theta$ at distance $r$ is $s=r\theta$. If you used degrees you have to say that $s=\pi r/180$ .

A lot of trig starts to make sense as well ... like the tangent of the angle is the distance along the tangent to the unit circle that is inside the angle, and the sine is the length of the cord that is inside the angle.

 

[Nickie]

I understand your frustration with mechanics, it can be a challenging subject to grasp. It seems like you have made some progress in solving this problem and have identified the key equations to use. However, it appears that you are struggling with finding the angle at which the particle exits the magnetic field.

One approach you could try is to use the equation for centripetal acceleration, ac = v^2/R, where v is the velocity of the particle and R is the radius of its circular motion. You have already calculated the velocity in the magnetic field, so you can use this equation to solve for the radius of the particle's circular motion. Then, using the given radius of 0.95 m, you can solve for the angle at which the particle exits the magnetic field.

Another approach is to use the fact that the particle travels a quarter circle (or half of a semicircle) in the magnetic field. This means that the angle at which it exits the field is equal to half of the angle at which it enters the field. You can use this information to solve for the angle and then use the equation you have for Fx.

I hope these suggestions help you solve the problem. Remember to take your time and carefully think through each step. Mechanics can be frustrating, but with patience and persistence, you can overcome any challenge. Keep up the good work!