2022 A/L Physics Past Paper · Question 45 (English Medium)

Answer: (1) Figure (1)

What the question asks

What we know:

A neutral particle sits at rest at $O$. It decays into two charged particles $P$ and $Q$ with the same mass $m$.

The figure shows $P$ leaving $O$ upwards and $Q$ leaving downwards.

A uniform magnetic field $B$ points into the page. We neglect the push of $P$ and $Q$ on each other.

There are no numbers. So we keep the letters, and later we check with toy sample values: $m = 2 \times 10^{-3}\ \text{kg}$, $|q| = 1 \times 10^{-3}\ \text{C}$, $v = 5\ \text{m s}^{-1}$ and $B = 0.5\ \text{T}$.

We want: which picture shows the paths of $P$ and $Q$.

What each letter means:

$m$ ($\text{kg}$): the mass of each particle.

$v$ ($\text{m s}^{-1}$): the speed of each particle.

$q$ ($\text{C}$): the size of the charge of each particle.

$B$ ($\text{T}$): the flux density of the magnetic field.

$F$ ($\text{N}$): the magnetic force on a particle.

$r$ ($\text{m}$): the radius of the path of a particle.

Relevant theory

What is happening:

Before the decay the particle is at rest, so its momentum is zero. After the decay the momentum of the two particles must still add up to zero. So their momenta are equal in size and opposite in direction. Their masses are the same, so their speeds are the same (at speeds far below the speed of light).

The particle was neutral, so after the decay the two charges still add up to zero. They are equal in size and opposite in sign.

A magnetic force acts on a moving charge, at right angles to its velocity. It turns the path and does not speed the particle up. With a constant speed and a force of constant size, the particle goes round a circle.

We use this to find the radius and the side of the circle for $P$ and for $Q$.

For a system of interacting objects, the momentum remains constant, provided no external force acts on the system.

No force acts on an electric charge which is at rest in a magnetic field. However if the charge is moving it is equivalent to an electric current and hence a force would act on it.

$F = BqV \sin\theta$

Thus the acceleration of a body performing circular motion is directed towards the centre of the circle and is hence known as "centripetal acceleration".

The magnetic force on a charge $q$ moving at right angles to $B$ has the size $F = Bqv$. It points at right angles to the velocity, towards the centre of the circle. So it plays the part of the centripetal force $\dfrac{mv^2}{r}$.

For example, two skaters on smooth ice who push each other apart from rest. They glide away in opposite directions. If they have the same mass they have the same speed.

The momenta of $P$ and $Q$ are like that.

Applying it to this question

$mv_P = mv_Q$

$F = |q|vB$

$|q|vB = \dfrac{mv^2}{r}$

$r = \dfrac{mv}{|q|B}$

The unit of $\dfrac{mv}{|q|B}$ is $\dfrac{\text{kg m s}^{-1}}{\text{C T}}$, and $\text{C T} = \text{N s m}^{-1}$. So the unit is $\dfrac{\text{kg m s}^{-1}}{\text{kg m s}^{-2} \times \text{s m}^{-1}} = \text{m}$, a length.

$20$

$r = \dfrac{mv}{|q|B} = \dfrac{2 \times 10^{-3} \times 5}{1 \times 10^{-3} \times 0.5}$

Now the side.

A negative charge moving in one direction is pushed the opposite way to a positive charge moving in the same direction.

$Q$ moves in the direction opposite to $P$. That turns the force round. $Q$ has the opposite sign of charge to $P$. That turns the force round again.

So the force on $Q$ points the same way as the force on $P$. Both are pushed to the same side of $O$.

Both circles have radius $r$, both pass through $O$, and both have their centre on the same side of $O$ at the distance $r$. So they are the same circle.

Reasoning and calculation

$P$ and $Q$ go round one circle through $O$. $P$ goes up and round, $Q$ goes down and round, and they meet again on the far side of the circle.

That is picture (1).

Reading the options

Check the five pictures.

Pictures (2) and (3) have circles of different sizes. Picture (4) has circles on opposite sides of $O$. In picture (5) both particles leave $O$ in the same direction.

Picture (1) has one circle, with $P$ above $O$ and $Q$ below $O$. It matches.

Why the other options fail

Pictures (2) and (3) have circles of different sizes.

Picture (4) has circles on opposite sides of $O$.

In picture (5) both particles leave $O$ in the same direction.

Checks

$m * v / (q * b)$

Does it make sense?

The radius is bigger for a bigger speed or a bigger mass. It is smaller for a bigger charge or a stronger field. That is what we expect: a heavier or faster particle is harder to turn, and a stronger force turns it more.

The paths of $P$ and $Q$ are the same circle, so they must be at opposite ends of a diameter after half a turn. They have the same speed, so they get there together.

Conclusion

$\text{figure }(1)$

$P$ and $Q$ go round one circle through $O$ in opposite senses.

The answer is (1).

Use momentum to compare the speeds, and the neutral particle to compare the charges.

The radius is $\dfrac{mv}{|q|B}$. For the side, count the reversals: a reversed velocity turns the force round, and a reversed sign of charge turns it round again.

Exam tip