How to Predict Projectile, Orbital, Electric, and Magnetic Motion

A ball leaving a ledge, a satellite circling Earth, and a charged particle bending in a field can all trace curved paths. But curvature alone does not tell you which model to use. Predict this before reading on: which field can change a particle’s direction without changing its speed? A magnetic field can, if magnetic force is the only force; gravity does this throughout a circular orbit, but not every orbit, while an electric field can change both direction and speed. The decision starts with the field, the initial velocity, and the direction of the force relative to that velocity.

01Choose the right model in 10 seconds

Use this sequence whenever a question asks you to predict an object’s motion.

1. Identify the field. Is it uniform gravity, central gravitational attraction, an electric field, or a magnetic field?
2. Check the initial velocity. Is it parallel, perpendicular, or at an angle to the field or the direction towards the gravitational centre?
3. Find the force and acceleration. Remember that acceleration follows the net force, not necessarily the velocity.
4. Check whether the acceleration stays constant. Constant acceleration and acceleration that continually changes direction lead to different models.
5. Predict the path and speed separately. A changing direction does not automatically mean a changing speed.

Use the table as your first reference when solving a problem.

02The rule that connects all four models

Newton’s second law gives the acceleration produced by the net force:

\[ \vec a=\frac{\vec F_{\text{net}}}{m} \]

Here, \(\vec a\) is acceleration in \(\mathrm{m\,s^{-2}}\), \(\vec F_{\text{net}}\) is the vector sum of all forces in newtons, and \(m\) is mass in kilograms.

The important word is net. If several forces act, combine them before predicting the motion.

There is also a useful way to decide what happens to the speed.

• A force component parallel to the velocity changes the speed.
• A force component perpendicular to the velocity changes its direction.
• If the force stays perpendicular to the velocity throughout the motion, it changes direction without changing speed.

A force can therefore produce acceleration even when the speed remains constant. Acceleration means a change in velocity, and velocity includes both speed and direction.

Keep this distinction in mind as you work through the four models.

01Projectile motion: uniform gravity near Earth

Imagine throwing a ball horizontally from a ledge. It moves forwards while falling downwards.

You might predict that its horizontal motion gradually runs out as gravity pulls it down. That would make sense if gravity acted against the horizontal motion. But it doesn’t.

In the ideal projectile model, gravity acts vertically downwards. There is no horizontal force, so the horizontal velocity remains constant. Meanwhile, the downward velocity increases as the ball falls.

This assumes air resistance is negligible and the gravitational field is approximately uniform, as it is over ordinary distances near Earth’s surface.

The equations

Choose the horizontal direction as \(x\) and upwards as \(y\). Let \(v_{0x}\) and \(v_{0y}\) be the initial horizontal and vertical velocity components, respectively. Let \(t\) be the elapsed time and \(g=9.8\,\mathrm{m\,s^{-2}}\) the magnitude of gravitational acceleration.

The acceleration components are

\[ a_x=0,\qquad a_y=-g \]

The velocity components are

\[ v_x=v_{0x} \]

\[ v_y=v_{0y}-gt \]

The displacement equations are

\[ \Delta x=v_{0x}t \]

\[ \Delta y=v_{0y}t-\frac12 gt^2 \]

These equations describe two simultaneous motions:

• Horizontally, displacement changes at a constant rate.
• Vertically, displacement depends on time squared because the velocity is changing.

To see why the path is parabolic, eliminate time using \(t=\Delta x/v_{0x}\), assuming \(v_{0x}\ne0\).

\[ \begin{aligned} y-y_0={}&\frac{v_{0y}}{v_{0x}}(x-x_0)\\ &-\frac{g}{2v_{0x}^2}(x-x_0)^2 \end{aligned} \]

This is a quadratic relationship between \(y\) and \(x\), so the path is a parabola.

There is one important qualification: if the object is launched straight upwards or downwards, its horizontal velocity is zero. Its path is vertical, not a visible parabola.

Notice what happens at the highest point of a projectile’s path. The vertical velocity becomes zero, but gravity still acts. The object has not stopped accelerating; its velocity is simply horizontal at that instant.

Its speed is

\[ v=\sqrt{v_x^2+v_y^2} \]

Because \(v_y\) changes with time, the speed generally changes too, even though \(v_x\) remains constant.

02Orbital motion: gravity towards a central body

A satellite in orbit is also accelerating because of gravity. So why doesn’t it follow a projectile’s parabola?

The difference is the direction of gravity.

For a projectile travelling a short distance near Earth’s surface, gravity is approximately constant in both magnitude and direction. For a satellite, gravity points towards the centre of the planet at every point. As the satellite moves, the direction of the gravitational force changes.

The resulting acceleration continually turns the satellite’s velocity.

The gravitational force

For two objects of masses \(M\) and \(m\), separated by a centre-to-centre distance \(r\), Newton’s law of universal gravitation gives the magnitude of the force:

\[ F_g=\frac{GMm}{r^2} \]

Here:

• \(G=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\) is the universal gravitational constant.
• \(M\) is the mass of the central body, in kilograms.
• \(m\) is the satellite’s mass, in kilograms.
• \(r\) is the distance between their centres, in metres.

Dividing the force by the satellite’s mass gives the gravitational acceleration:

\[ a_g=\frac{GM}{r^2} \]

The acceleration points towards the central body. Its magnitude depends on the distance \(r\).

When does the orbit become circular?

For a circular orbit, the satellite’s velocity is tangential to its path, while gravity points towards the centre. These directions are perpendicular.

The gravitational force supplies the centripetal force required for circular motion:

\[ \frac{GMm}{r^2}=\frac{mv^2}{r} \]

Cancel the satellite mass \(m\) and rearrange:

\[ v_c=\sqrt{\frac{GM}{r}} \]

The symbol \(v_c\) is the speed required for a circular orbit at radius \(r\), assuming gravity is the only significant force.

This equation gives you a useful test. At a particular radius, a circular orbit requires the correct tangential speed. Simply moving sideways around a planet does not guarantee a circular orbit.

• At the correct tangential speed, the satellite follows a circular orbit.
• With different initial conditions, it may follow an elliptical path, collide with the planet, or escape, depending on its velocity and energy.

For a circular orbit, gravity is always perpendicular to the velocity. It changes the velocity’s direction but does no work, so the speed stays constant.

An elliptical orbit is different. The satellite moves faster near the planet and slower when farther away. Gravity is not perpendicular to the velocity throughout the orbit, so it can change the speed.

Think of a satellite as continually falling towards the planet while moving sideways fast enough to keep going around it. For a circular orbit, its distance from the planet’s centre stays constant even though gravity is pulling it inwards. The analogy describes the acceleration, not a gradual loss of altitude.

03Electric-field motion: constant acceleration

An electric field can also bend a particle’s path. The key difference from a magnetic field is that a uniform electric field produces a constant force on a particle with fixed charge.

The electric force is

\[ \vec F_E=q\vec E \]

Here, \(q\) is the particle’s signed electric charge in coulombs, and \(\vec E\) is the electric field in \(\mathrm{N\,C^{-1}}\), equivalent to \(\mathrm{V\,m^{-1}}\).

The electric field points in the direction of the force on a positive test charge. Therefore:

• A positive charge accelerates along the electric field.
• A negative charge accelerates opposite to the electric field.

Using Newton’s second law,

\[ \vec a=\frac{q\vec E}{m} \]

If the electric field is uniform and the particle’s charge and mass remain constant, its acceleration is constant.

How the initial velocity determines the path

Suppose a particle enters a uniform electric field with its initial velocity perpendicular to the field.

The electric force acts sideways relative to the initial motion. The particle continues moving in its original direction while gaining velocity in the direction of the force. This produces a parabolic path, just as uniform gravity produces a parabola for a projectile.

The difference is that electric acceleration depends on the charge-to-mass ratio:

\[ a_E=\frac{|q|E}{m} \]

A particle with a larger charge magnitude experiences greater acceleration at the same mass and field strength. A more massive particle experiences less acceleration at the same charge and field strength.

If the initial velocity is parallel or antiparallel to the electric field, the motion remains along a straight line. The particle may speed up or slow down, depending on whether the electric force acts with or against its velocity.

There is a subtle point when the initial velocity is perpendicular to the field. At the instant of entry, the force is perpendicular to the velocity, so the instantaneous rate of change of speed is zero. But the force soon gives the particle a velocity component along itself. The electric force then does work, and the particle’s speed increases.

A uniform electric field therefore produces constant acceleration, but it does not generally produce constant speed.

04Magnetic-field motion: changing direction without changing speed

A magnetic field behaves differently from the three cases above.

A magnetic field exerts a force on a moving charged particle. Its magnitude depends on the particle’s speed and on the angle between its velocity and the magnetic field:

\[ F_B=|q|vB\sin\theta \]

Here:

• \(q\) is the particle’s charge in coulombs.
• \(v\) is its speed in \(\mathrm{m\,s^{-1}}\).
• \(B\) is the magnetic field strength in tesla.
• \(\theta\) is the angle between velocity and the magnetic field.

The magnetic force is always perpendicular to the instantaneous velocity. That is the crucial property.

Three possible paths

Case 1: Velocity parallel to the field

When \(\theta=0^\circ\) or \(180^\circ\), \(\sin\theta=0\). There is no magnetic force, so the particle continues in a straight line at constant velocity, assuming no other forces act.

Case 2: Velocity perpendicular to the field

When \(\theta=90^\circ\), the magnetic force has its maximum magnitude for that speed and field strength:

\[ F_B=|q|vB \]

Because the force is perpendicular to velocity, it continually turns the particle without changing its speed. In a uniform magnetic field, this produces circular motion.

The magnetic force supplies the centripetal force:

\[ |q|vB=\frac{mv^2}{r} \]

Cancel one factor of \(v\) and rearrange:

\[ r=\frac{mv}{|q|B} \]

Here, \(r\) is the radius of the circular path in metres. The particle’s mass is \(m\) in kilograms, its speed is \(v\) in metres per second, and the magnitude of its charge is \(|q|\) in coulombs.

This equation gives several useful predictions:

• Increasing the particle’s mass increases the radius.
• Increasing its speed increases the radius.
• Increasing the magnetic field strength decreases the radius.
• Increasing the magnitude of the charge decreases the radius.

Case 3: Velocity at an angle to the field

If the initial velocity has both parallel and perpendicular components, split it into

\[ v_\parallel \quad\text{and}\quad v_\perp \]

The parallel component remains unchanged, while the perpendicular component produces circular motion. Together, they produce a helical path.

The radius of the circular part is

\[ r=\frac{mv_\perp}{|q|B} \]

Use \(v_\perp\), not the total speed, when the velocity is oblique to the field.

Why magnetic force does not change speed

A magnetic force is always perpendicular to the particle’s instantaneous velocity. It therefore does no work on the particle.

Work changes kinetic energy. Since magnetic force alone does no work, the particle’s kinetic energy and speed remain constant.

A useful analogy is to think of magnetic force as steering rather than accelerating like a motor. The analogy has a limit: the magnetic force depends on the direction of motion and disappears when the velocity is parallel to the field.

This conclusion applies when magnetic force is the only force doing work. If an electric field also acts, the electric force can change the particle’s speed.

05When more than one field acts

Real problems may involve more than one field. You cannot always choose a single model just by looking at the path.

For example, a charged particle in both electric and magnetic fields experiences

\[ \vec F_{\text{net}} =q\vec E+q(\vec v\times\vec B) \]

If gravity also matters, add its force as well. Then use

\[ \vec a=\frac{\vec F_{\text{net}}}{m} \]

The magnetic force depends on the particle’s velocity, so even constant electric and magnetic fields do not necessarily produce constant net acceleration.

One particularly useful case occurs when electric and magnetic forces cancel.

\[ q\vec E+q(\vec v\times\vec B)=0 \]

For a charged particle, this requires

\[ \vec E=-\vec v\times\vec B \]

When the velocity is perpendicular to both fields, the magnitude condition is

\[ E=vB \]

The net force is zero, even though both fields and both individual forces can be non-zero. The particle travels straight through without deflection.

This cancellation works for either sign of charge, provided the velocity and fields satisfy the same vector relationship. Both forces reverse when the charge changes sign, so they still cancel.

08Common mistakes

Mistake 1: Assuming every curved path is a projectile.

A projectile follows a parabola only when the acceleration is constant and the horizontal velocity is non-zero, under the usual ideal assumptions. Central gravity and magnetic force produce different acceleration patterns.

Mistake 2: Treating gravity as a constant downward force in an orbit.

Near Earth’s surface, the direction of gravity is approximately fixed. Around a planet, gravity points towards the planet’s centre, so its direction changes as the satellite moves.

Mistake 3: Assuming electric fields affect positive and negative charges in the same direction.

The signed charge matters in \(\vec a=q\vec E/m\). Reversing the sign of \(q\) reverses the acceleration direction.

Mistake 4: Assuming a magnetic force changes speed because it changes velocity.

Velocity includes direction as well as speed. A magnetic force changes direction without changing speed when it is the only force acting.

Mistake 5: Using the magnetic-radius formula for any angle.

The formula \(r=mv/(|q|B)\) assumes velocity is perpendicular to the magnetic field. For an oblique velocity, use the perpendicular component \(v_\perp\).

Mistake 6: Assuming a force perpendicular to velocity at one instant cannot change speed later.

The force must remain perpendicular to the velocity throughout the motion for speed to remain constant. Gravity is perpendicular to the velocity at the highest point of a projectile’s path, but the projectile’s speed changes before and after that instant.

Mistake 7: Treating centripetal force as an additional force.

Centripetal force is the name for the net inward force required for circular motion. It can be supplied by gravity, magnetic force, tension, or another real force. Do not add a separate centripetal force to forces already acting.

09Questions and solutions

Question 1

An ion with mass \(3.3\times10^{-27}\,\mathrm{kg}\) and charge \(+1.6\times10^{-19}\,\mathrm C\) enters a uniform magnetic field of \(0.50\,\mathrm T\) at \(2.0\times10^6\,\mathrm{m\,s^{-1}}\). Its velocity is perpendicular to the field.

Calculate the radius of its circular path and state whether its speed changes.

Solution 1

The radius is \(8.3\times10^{-2}\,\mathrm m\), and the ion’s speed remains constant if magnetic force is the only force acting.

Because the velocity is perpendicular to the magnetic field, the magnetic force supplies the centripetal force. Use

\[ r=\frac{mv}{|q|B} \]

Substitute the values:

\[ \begin{aligned} r&=\frac{(3.3\times10^{-27})(2.0\times10^6)} {(1.6\times10^{-19})(0.50)}\\ &=\frac{6.6\times10^{-21}}{8.0\times10^{-20}}\\ &=0.0825\,\mathrm m\\ &\approx0.083\,\mathrm m \end{aligned} \]

The units reduce to metres, as required for a radius.

Although the magnetic force continually changes the ion’s direction, it remains perpendicular to the velocity. It does no work, so the speed stays at \(2.0\times10^6\,\mathrm{m\,s^{-1}}\).

Question 2

An ideal satellite travels in a circular orbit of radius \(r\), speed \(v\), and period \(T\) around a planet of mass \(M\).

It is transferred into another circular orbit of radius \(4r\) around the same planet. Ignore all forces except gravity.

Determine the ratios:

(a) \(v_2/v_1\), the new orbital speed divided by the original speed.

(b) \(a_2/a_1\), the new centripetal acceleration divided by the original acceleration.

(c) \(T_2/T_1\), the new orbital period divided by the original period.

Solution 2

The satellite moves at half its original speed, experiences one-sixteenth of its original acceleration, and takes eight times as long to complete an orbit.

Part (a): Orbital speed

For a circular orbit,

\[ v=\sqrt{\frac{GM}{r}} \]

Therefore,

\[ \begin{aligned} \frac{v_2}{v_1} &=\sqrt{\frac{GM/(4r)}{GM/r}}\\ &=\sqrt{\frac14}\\ &=\frac12 \end{aligned} \]

The new orbital speed is half the original speed.

Part (b): Centripetal acceleration

Gravitational acceleration is

\[ a=\frac{GM}{r^2} \]

Hence,

\[ \frac{a_2}{a_1} =\frac{GM/(4r)^2}{GM/r^2} =\frac1{16} \]

The acceleration is one-sixteenth of its original value because gravitational acceleration decreases with the square of the orbital radius.

Part (c): Orbital period

The period is the circumference divided by the orbital speed:

\[ T=\frac{2\pi r}{v} \]

Using the new radius and the speed found in part (a),

\[ \begin{aligned} \frac{T_2}{T_1} &=\frac{2\pi(4r)/(v/2)}{2\pi r/v}\\ &=8 \end{aligned} \]

The new orbit has four times the circumference, but the satellite travels at half the speed. It therefore takes eight times as long to complete an orbit.

The tempting prediction is that a larger orbit must mean a faster satellite because it has farther to travel. The gravitational force requirement reverses that prediction: a circular orbit farther from the planet requires a lower speed.

Question 3

A particle with charge \(+q\), mass \(m\), and initial speed \(v\) enters two separate regions.

• In region A, it enters a uniform electric field perpendicular to its initial velocity.
• In region B, it enters a uniform magnetic field perpendicular to its initial velocity.

The field strengths are chosen so that the two forces have the same magnitude, \(F_0\), at the instant of entry.

(a) Compare the initial accelerations.

(b) Explain why the particle follows a parabolic path and changes speed in region A, but follows a circular path at constant speed in region B.

(c) The initial speed is now doubled, with both field strengths unchanged. Compare the electric path’s radius of curvature at entry with the magnetic circular radius. Use \(a_\perp=v^2/R\) for the local radius of curvature \(R\) when acceleration is perpendicular to velocity.

Solution 3

The initial accelerations are equal, but doubling the initial speed makes the electric path’s radius of curvature at entry four times larger and the magnetic circular radius twice as large.

Part (a): Initial acceleration

Newton’s second law gives

\[ a=\frac{F_{\text{net}}}{m} \]

Both forces initially have magnitude \(F_0\), and the particle has the same mass in both regions. Therefore,

\[ a_A=a_B=\frac{F_0}{m} \]

Both initial accelerations are perpendicular to the initial velocity.

Part (b): Different paths and speed behaviour

In region A, the electric field produces a constant force and constant acceleration:

\[ a_E=\frac{qE}{m} \]

The acceleration direction stays fixed. The particle therefore follows a parabolic path, and its velocity develops a component along the electric force. The force does work, so the particle’s speed changes.

In region B, the magnetic force stays perpendicular to the instantaneous velocity. In a uniform magnetic field, the force supplies centripetal acceleration, producing circular motion at constant speed.

The important distinction is that the electric acceleration has a constant direction, whereas the magnetic force continually changes direction as the particle moves.

Part (c): Doubling the initial speed

In the electric field, the acceleration remains unchanged because \(a_E=qE/m\) does not depend on speed. At entry,

\[ R_E=\frac{v^2}{a_E} \]

If the speed becomes \(2v\),

\[ R'_E=\frac{(2v)^2}{a_E}=4R_E \]

The electric path’s radius of curvature at entry becomes four times larger.

For the magnetic field,

\[ r_B=\frac{mv}{qB} \]

Doubling the speed gives

\[ r'_B=\frac{m(2v)}{qB}=2r_B \]

The magnetic circular radius doubles.

The trap is to assume that doubling speed has the same effect in both fields. Electric force is independent of speed, while magnetic force is proportional to speed when the velocity is perpendicular to the field. The electric and magnetic forces therefore no longer have equal magnitudes at entry after the speed is doubled.

Question 4

An ideal satellite travels in an elliptical orbit around a planet. Its farthest point, A, is at distance \(3r\) from the planet’s centre, and its closest point, P, is at distance \(r\).

(a) Compare the magnitudes of the gravitational force and acceleration at P and A.

(b) Which point has the greater satellite speed?

(c) At both A and P, the satellite’s velocity is tangential to the orbit, so gravity is perpendicular to the velocity at those exact points. A student concludes that the satellite’s speed must therefore be constant throughout the orbit. Explain the error.

Solution 4

At P, both the gravitational force and acceleration are nine times their values at A, and the satellite moves faster at P. The student’s conclusion is incorrect because gravity is perpendicular to the velocity at the two endpoints, not throughout the entire elliptical orbit.

Part (a): Gravitational force and acceleration

The gravitational force magnitude is

\[ F_g=\frac{GMm}{r^2} \]

At A, the orbital radius is \(3r\). At P, it is \(r\).

Therefore,

\[ \begin{aligned} \frac{F_P}{F_A} &=\frac{GMm/r^2}{GMm/(3r)^2}\\ &=9 \end{aligned} \]

Gravitational acceleration follows the same inverse-square relationship:

\[ a_g=\frac{GM}{r^2} \]

Thus,

\[ \frac{a_P}{a_A}=9 \]

Both the gravitational force and acceleration are nine times larger at P.

Part (b): Satellite speed

The satellite moves faster at P, the closest point to the planet.

As the satellite moves from A towards P, gravitational potential energy decreases. Gravity does positive work over this part of the orbit, increasing kinetic energy and speed.

Part (c): Why perpendicular force at two points is not enough

The student’s reasoning uses a true observation but applies it too broadly.

At A and P, the velocity is tangential while gravity points towards the planet’s centre. The force is perpendicular to the velocity at those instants, so the instantaneous rate of change of kinetic energy is zero.

However, between A and P, the satellite has a radial component of velocity directed towards the planet. Gravity has a component along that motion and does positive work.

The speed therefore changes over the journey even though the force is perpendicular to the velocity at both endpoints. To conclude that speed is constant, the force must remain perpendicular to the velocity throughout the motion, as it does in a circular orbit or in magnetic-only motion.

Question 5

An electron and a proton are launched with the same velocity vector and speed \(v\) through region 1. The region contains uniform electric and magnetic fields, \(\vec E\) and \(\vec B\), which are perpendicular to each other. The initial velocity is perpendicular to both fields.

Both particles travel straight through region 1 without deflection. An independent measurement confirms that the magnetic field is non-zero.

The particles then enter region 2, where the electric field is absent but the same magnetic field remains. Ignore gravity and other forces.

(a) Explain why the straight paths through region 1 do not imply that the electric and magnetic forces are absent. If the non-zero fields cancel, derive the relationship between \(E\), \(B\), and \(v\).

(b) In region 2, compare the electron’s and proton’s speed, direction of curvature, and radii of curvature.

(c) A neutral particle with the same initial velocity also travels straight through region 1. Does this provide evidence that the electric and magnetic fields are zero? Explain.

Solution 5

The straight paths in region 1 are caused by cancellation of two non-zero forces. In region 2, both charged particles maintain their speeds but curve in opposite directions, and the proton’s radius is approximately 1,836 times the electron’s. The neutral particle’s straight path does not indicate whether the fields are present.

Part (a): Force cancellation

The net electromagnetic force on a charged particle is

\[ \vec F_{\text{net}} =q\vec E+q(\vec v\times\vec B) \]

Factor out the charge:

\[ \vec F_{\text{net}} =q(\vec E+\vec v\times\vec B) \]

Because the magnetic field is non-zero and the velocity is perpendicular to it, the magnetic force is non-zero for either charged particle.

Yet neither particle deflects. The electric force must therefore cancel the magnetic force:

\[ \vec E+\vec v\times\vec B=0 \]

Rearranging gives

\[ \vec E=-\vec v\times\vec B \]

Since the three directions are mutually perpendicular, the magnitudes satisfy

\[ E=vB \]

The same condition cancels the forces on both particles. Changing the sign of the charge reverses both forces simultaneously because \(q\) multiplies the entire expression. Their cancellation is therefore independent of whether the particle is positive or negative.

The tempting conclusion is that a straight path proves there is no force. In fact, a straight path establishes that the net force is zero, not that every individual force is zero.

Part (b): Motion in the magnetic-only region

Both particles maintain their speeds because the magnetic force is perpendicular to their instantaneous velocities and does no work.

Their paths curve in opposite directions. The magnetic force direction depends on the sign of the charge, so the electron and proton experience opposite force directions for the same initial velocity and magnetic field.

The radius of circular motion is

\[ r=\frac{mv}{|q|B} \]

The electron and proton have the same speed, the same magnetic field, and equal charge magnitudes. Therefore,

\[ \frac{r_p}{r_e}=\frac{m_p}{m_e} \]

Using the approximate proton-to-electron mass ratio,

\[ \frac{r_p}{r_e}\approx1836 \]

The proton’s radius is about 1,836 times the electron’s radius. The proton follows a much broader curve, while the electron bends much more sharply.

The key constraint is that the magnetic force changes the direction of motion, not the speed. The radius depends on mass, speed, charge magnitude, and magnetic field strength, so equal speeds do not imply equal radii.

Part (c): The neutral particle

The neutral particle’s straight path provides no evidence that the fields are zero.

In this model, a neutral particle has \(q=0\), so its electromagnetic force is

\[ \vec F=q(\vec E+\vec v\times\vec B)=0 \]

It would travel straight through region 1 whether the electric and magnetic fields were present or absent, assuming no other forces act.

The charged particles provide more information: because the magnetic field is independently known to be non-zero, their undeflected paths show that the electric and magnetic forces cancel at their particular velocity.

A straight path is therefore not enough to identify the individual forces. You must combine the field information with the initial velocity and the net-force condition.

The next useful decision is what happens when a particle enters a new region or encounters several fields at once. Write down each force, add the forces to find the net force, and then use \(\vec a=\vec F_{\text{net}}/m\). This method works even when the path looks unfamiliar, and it prevents you from choosing a motion equation based on the shape alone.