references: THe physics of every day phenomena - A conceptual introduction to Physics - McGraw-Hill
Conceptual Physics, Paul Hewitt
http://www.staff.amu.edu.pl/~romangoc/M2-1-projectile-motion.html
in construction, please be patient.
PART I: THE BIG FIVE, REFER TO 11th grade lessons.
MOTION IN THE UNIFORM GRAVITATIONAL FIELD
Let's review the motion of a projectile in a new perspective. We say the projectile moves in an uniform field: the
gravitational field. Because the object we throw has mass, it experiences a constant force: its weight = m g @ down.
Its motion can be linear or parabolic.
1) LINEAR MOTION in an uniform gravitational field
The force of gravity W = m g@ down and since F = m a then a = g @ down or a = - g
we neglect any friction force like air resistance. The weight is the only force acting on a falling object. (in free-fall)
 We consider the gravitational field g to be uniform at the surface of the Earth. |  The motion can be linear. The acceleration a is constant and is a = (Vfinal - Vinitial )/ t equ1 here a = (-20 -20) / 4 = -10 m/s/s This is the acceleration due to gravity g acceleration is a vector. You give the projectile an initial velocity Vo = 20m/s The velocity drecreases at a constant rate : V = Vo + a t equ2 here: V = 20 - 10t so t = 1s, V = 10m/s t=2s , V = 0 t = 3s, V = -10m/s t=4s , V = -20m/s remember that the velocity V is a vector If the acceletation is constant you have: Vaverage = (Vinitial + V final)/2 and Vaverage = distance/t equ3 so between t =0 and t = 1s Vaverage = 10 + 20 / 2 = 15m/s distance = Vaverage x t = 15m displacement AB = Vot + 0.5a t2 equ4 AB is the vector connecting an inital point A to a final point B here : AB = 20 t - 0.5 10 t2 The displacement increases with the time squared So t = 4s, AB = 20(4) - 5 (16) = 0 . The displacement is 0 because the change in position is 0. t = 1 AB = 20(1) - 5 (1) = 15m@up Vfinal2 = Vinitial2 + 2 a |AB| equ5 This is the last equation for uniform motion |
2) PROJECTILE MOTION in an uniform gravitational field
The force of gravity W = m g@ down and since F = m a then a = g @ down or a = - g (only along vertical)
along the horizontal, there is no force (we neglect any friction force) so a = 0 along the horizontal
This is very simple. Galileo was the first one to understand that any 2D motion has 2 independent components.
the X-component and the Y-component. You just have to rewrite the first 5 equations along the X-axis and along the Y-axis.
You need to know the trig relations too. (SOH- CAH-TOA)
 | ALONG X-axis: acceleration = ax = 0 Vx = Vox + ax t so : Vx = Vox =Vo cos(α) x = Vox t + 0.5 ax t2 so x = Vox t = Vo cos(α) t (1) parametric equation ALONG Y-axis acceleration ay = - g = -9.8 m/s/s Vy = Voy +ayt so Vy = Voy - (g) t Vy= Vosin (α ) - g t y = Voy t + 0.5 ay t2 y = Vosin( α) t - 0.5 g t2 (2) parametric equation ------------------------------------ (1) gives t = x / Vo cos(α ) (2) y = Vosin( α) ( x /Vo cos(α )) - 0.5 g ( x / Vo cos(α ))2 so y = - 1/2 g / ( Vo2 cos(α ) 2 ) x2 + tan(α ) x |
y = - 1/2 g / ( Vo2 cos(α ) 2 ) x2 + tan(α ) x This is the equation of a parabola downward.
The range R is ( Vo2 /g ) sin (2α ) (x for y = 0 )
R is maximum when 2α = 180 degrees so α = 45 degrees (we saw that in 11th grade)
3) Your turn. A marble of mass m is thrown from a height of 2m in the plan Oyz with an inital velocity of 10m/s @ 30 degrees
A) Find the acceleration vector (Find its components ax, ay, az )
B) Find the velocity vectors (find its components Vx, Vy, Vz)
C) Find the displacement vectors x(t), y(t) and z (t)
D) Find the equation of motion of the marble z = z(y)
PART2: MOTION IN AN UNIFORM ELECTRIC FIELD
1) we can apply the same ideas to the linear motion of a charged particle in an uniform electric field, Fe = q E with q negative or positive
Because F = m a , a = qE/m = constant a can be negative or positive.
In a gravitational field a = g@down. The difference is that the acceleration now depends on the mass and that the
acceleration can be negative or negative. Otherwise, the equations of motions are very similar. The trajectory
can be a straight line if the initial velocity is along the field or a parabola if there is a component of the velocity
perpendicular to the field.
 Let's say you drop an electron (charge - e with e = 1.6 10-19C ) in an uniform field E . The charge q = -e experiences a Force Fe = q E = -e E The force is @ dowm and a = - F/m or a = - e E /m | Along the y-axis: a = - e E /m V = Vo - eE/m t here Vo = 0 then V = - eE/m t y = Vot - 0.5 eE/m t2 here Vo = 0 so y = - eE/2m t2 The speed increases at a constant rate, the displacement depends on the square on the time. the distance between the plates is d. so if y = - d, - d= -eE/2m t2 t = sqrt ( 2md / eE) time it takes for the electron to cross the capacitor. Let's find the speed of the electron when it get to the bottom. V = - eE/m t with t = sqrt ( 2md / eE ) or V = - eE/m sqrt (2md / eE) V = - sqrt (2m e2 E2 d / m2 e E) = - sqrt ( 2e E d / m ) But E = U/d with U the voltage between the plates So V = - sqrt ( 2 e U / m ) Find the kinetic energy KE = 0.5 m V2 with V = sqrt ( 2 e U / m ) V2 = 2eU / m so KE = eU This is what we found previously !! eU is also the electric potential energy of the electron before falling. |
4) You can also get a parabola when the electron enters an uniform field with an initial velocity Vo perperndicular to the field.
You sill have a = e E / m
 | along the X-axis: ax = 0 Vx = Vo x = Vo t (1) along the Y-axis ay = eE / m Vy = eE / m t y = 0.5 eE / m t2 (2) (1) in (2) y = eE / 2mVo2 x2 The motion is therefore a parabola. Once the electron leaves the electric field, the motion is linear and the velocty is constant. of an oscilloscope. To learn more about how an oscilloscope works. |
oscilloscope. The beam can be deviated sideway or up and down. The voltage between the plates control the deviation.
5)
From the point O (0,0,0) a particle is let go with no inital speed. The tension between the plates is UAB = 1000V .
The plates are 10cm apart. Find the acceleration of the particle and its final velocity when it reaches the plate B.
d = 10cm, m = 9.1 10-31kg, the charge is -1.610-19C.
(answer V = 1..875 107 m/s)
6)
A particle of mass m and of charge q moves between 2 plates A and B. The particle is ejected with an initial velocity Vo = 80km/s
The distance between the plate is 8cm and the plates are 15cm long.
A) Show that we can neglect the weight compared to the electric force.
B) Find the coordinates of particle when the particle leaves the set up.
(answer 15cm, - 2.1cm)
7) part of an ocilloscope
An electron with an inital speed Vo = 107 m/s is accelerated between 2 horizontal plates because of the electric field E = 600 V/m.
The plates are 15 cm long. See picture. Once the electron leaves the plates at S (l, YS), it keeps going at a constant speed, in a straight line,
and hits the screen at M (L, Y). L = 40cm. m = 1.67 10 -27kg , q = 1.6 10-19 C .
A) Show again that Ys = qE l2 / 2 mVo2 and thet Vs = qE l /mVo
B) Using the fact that OC = l/2 and using the similar trianle find Y.
(hint: Y/YS =( L - l/2 )/ l/2 )
8) Use the work energy theorem. (change in kinetic energy = work done by the electric force = change in potential energy)
The electrons are accelerated by a tension U = 5000V. The inital speed is 100kim/s. q = -1.6 10-19C and m=9.1 10-31kg.
Find the final speed of the electrons.
PART3 CIRCULAR MOTION IN AN UNIFORM MAGNETIC FIELD
COMPARE TO CIRCULAR MOTIONS OF PLANETS (moon, satellite)
1)REmember ? An electric charge (stationary or not) in an electric field will experience a force Fe. The same way a mass
in a gravitational field will experience a force Fg. The trajectory can be a straight line or a parabola.
In a gravitaitona field : a = g@ down (the force depends on the mass and the field, acceleration only on the mass )
In an electric field : a = q E / m with a >0 or a < 0 (the force depends on the charge and the electric field, the acceleration on the field, the mass and the charge)
What about a magnetic field ? A charge in a magnetic fielf will experience a force only if it is moving.
The magnetic force Fm depends on the magnetic field, the charge and the velocity of the charge (both direction and magntitude).
More precisely, the force depends on the component of the velocity perpendicular to the magnetic field.
If the component is zero, there is no magnetic force!
The charges (the current I) are moving in a magnetic field. The velocity is perpendicular
to the field and therefore a magnetic force is produced.
The magnetic field F has to be perpendicular to both V and B. (V, B , F) have the same
directions than an (X, Y, Z) system/.
 charges moving in a magnetic field will experience a magnetic force. The direction of the force is given by the right hand rule. The force is perpendicular to both the velocity of the charges and to the magnetic field. |  right hand rule. current I (positive chrages) along the thumb, B (magnetic field) along the fingersand F our of the hand. |
 A moving charge in a magnetic field will experience a force perpendicular to both V (initial velocity) and B. B exerts a torque on the charge and the charge rotates./ |  One consequence : if you move a wire in a magnetic field, the electrons will experience a force. A curent will be induced and a voltate will build up at the ends of the wires. This is EM induction. |
2) If the moving charges are negative instead of positive (like electrons), then you have to reverse the force.
or you use your left hand !!!
2) Experiments also show that the force is zero when the velocity is parallel to the field.
With varying angle of orentation theta and a given charge q, we find that the magnitude of the magnetic force
is given by: (magnitude) Fm = BqV sin (Φ)
Or simply: Fm = Bq (component of V perpendicular to B) or Fm = Bq Vperpendicular
|  | sin(Φ) = qV perpendicular / qV so qVperpendicular = sin(Φ) qV (in the right anlge triangle) |
This formula shows that the units for B are Newtons seconds / coulomb meter
WE can think of the magnetic field as a measure of the impulse given to 1 coulomb of charge over
1 meter. In the SI system , this combination is called a telsa. 1 telsa = 1 newton per ampere. meter.
3) lets study the motion of a charge in a magnetic field. Suppose the velocity is perpendicular to the magnetic field.
The magnetic force is therefore perpendicular to both the velocity and to the magnetic field. (see first image below)
use right hand rule.
This is very much like the moon orbiting the Earth. REmember ?
The gravitational force Fg is perpendicular to the velocity of the moon.
(see 2nd image). The force changes the direction of the moon (so there is a centripetal acceleration) but not its speed V (so no tangential acceleration)
Remember ? according to the law of inertia, the moon wants to keep going at the same speed (no force along
the tangent) but can' t keep the same direction because of Fg. It keeps falling toward the EArth.
The resulting motion is a circle. The circular path has a radius R such as F = ma with a = ac = mV2/R
ac is the centripetal acceleration. (see lecture 11rh grade). For the moon or a sattelite F = mg = mV2/R so R = V2/g.
The radius depends on the speed (magnitude of velocity) and on the gravitational field. not on the mass.
See third image.
Same idea for a charge moving in a magnetif field. Fm = qVB = mV2/R
R = m V / qB R depends on the speed, the field and the mass.
 |  |  |
try this applet
4) A charge moves in a circular orbit of radius R due to a uniform magnetic field.
If the velocity of the charge is doubled , the orbital radius become
2R ? R ? R/2 ? R ? R/4 ?
5) F =qV x B = qVB sin(α ) α is the angle between V and B
An electron has a velocity V along the x-axis. A magnetic field B in the plan (o,x,y) makes an angle α = 30 degrees with the x axis.
Find the magnetic force B (direction and magnitude).
V = 30m/s B = 7mT q = -1.6 10-19C.
6) Sketch the missing vector.
7) A particle of mass m and of charge q nter a magnetic field B = 0.1T and is curved by it.
The sign of the particle is unknown. the magnitude is 3.2 10-19C. the mass is unknown.
R = 2.08m et Vc = 107 m/s
A) according to the picture find the sign of the charge
B) find the mass of the particle.
( acceleration = force/m = qVB/m and acceleration = V2/R )
8) in a TV tube.
An electron is defected by a magnetic field an an small angle alpha α.
out of the field , the particle keeps in a straight line at a constant speed.
The angle is very small so tan(α) = sin (α) and l << L
L=40cm ; l=.5cm ; B= 510-3 T ; q = - 1.6 10-19C ; m = 9.1 10-31kg ; Vo = 2 107 m/s
A) Find the radius R
B) Find sin(α) in term of l and T . find its numerical value.
C) Find the expression of tan(α) in term of D and L.
if α is small tan(α) = sin (α), derive D the magnetic deviation
9) spectrograph.
In a spectrograph ions are ejected and deflected by a magentic field. Find the masses of the ions of OI = 20.9cm and OJ = 24.3 cm
B=0.3 T ; Vo=106 m/s q = 1.610 19C.
8) In both cases, motion of a charge in a magnetic field or motion of a planet/sattelite in circular orbit,
the force is perpendicular to the trajectory. No work is done by Fm or Fg.
So the kinetic energy stays constant (Same magnitude for V) .