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# A-level Physics (Advancing Physics)/Force

Magnetic fields exert a force on a charge when the charge is moving. If the charge is stationary, no force is exerted. This force is given by:
${\overrightarrow {F}}=q({\overrightarrow {v}}\times {\overrightarrow {B}})$,
where q is the charge on the point charge, v is its velocity and B is the magnetic field strength. This involves a vector cross product, which you don't need to know about for A-level. However, you do need to know a simplified version of this. The magnitude of this force F is given by:
$F=Bqv\sin {\theta }$,
where θ is the angle between the direction of motion of the point charge and the direction of the magnetic field. If the velocity and the magnetic field are in the same direction, the θ = 0, so sin θ = 0 and F = 0. If the velocity and the magnetic field are perpendicular to each other, θ = ½ π, so sin θ = 1. This means that, in the special case where velocity is perpendicular to the magnetic field:
$F=Bqv$
If q is negative (for example, for an electron), the force is in the opposite direction.

## Current

A current is just a flow of moving electrons, and so a magnetic field will exert a force on a wire with a current flowing through it. The case you need to know about is when the magnetic field is perpendicular to the wire. In this case, the magnitude of the force on the wire is given by:
$F=BIL$,
where I is current, and L is the length of the wire.

## Direction

The direction of the force on either a point charge or on a wire can be worked out using Fleming's left-hand rule, as shown in the diagram on the right. The direction of the thumb is that of the force (or thrust), the direction of the first finger is that of the magnetic field, and the direction of the second finger is that of the current (or the motion of the point charge).
On a 2D diagram, the direction of a magnetic field is represented by one of two symbols, which resemble the point and fletchings of an arrow pointing in the direction of the magnetic field. The symbol $\bigodot$ means that the field is pointing towards you (just as the arrow would be, if you were looking at the point). The symbol $\bigotimes$ means that the field is pointing away from you (just as the arrow would be, if you were looking at the fletching).

## Questions

1. What force is exerted by a 1T magnetic field on an electron (of charge -1.6 x 10$^{−19}$C) moving at 5% of the speed of light (3 x 10$^8$ ms$^{−1}$)?
2. What force is exerted by a 5mT magnetic field on a 20 cm wire with resistance 1μΩ attached to a 9V battery?
3. The following diagram shows a positive charge moving through a magnetic field. Draw an arrow representing the direction of the force on the charge.
4. The following diagram shows a wire in a magnetic field. Draw an arrow representing the direction of the force on the wire.
Worked Solutions
# A-level Physics (Advancing Physics)/Forces and Impulse in Collisions/Worked Solutions

1. Escape velocity from the Earth is 11.2 km$^{−1}$. How much impulse must be exerted on a 47000 kg payload to get it to travel away from the Earth?
$I=m(v-u)=47000(11200-0)=526.4{\mbox{ MNs}}$
2. Two billiard balls, of mass 10g, collide. One is moving at 5ms$^{−1}$, and the other at 2ms$^{−1}$. After the collision, the first billiard ball is moving backwards at 4ms$^{−1}$. The collision takes 1 ms. What force was exerted on this ball?
$I=m(v-u)=0.01(5-(-4))=0.09{\mbox{ Ns}}$
$F={\frac {I}{\Delta t}}={\frac {0.09}{0.001}}=90{\mbox{ N}}$
3. What impulse and force were exerted on the second ball?
The impulse was -0.09Ns and the force was -90N.
4. A 60 kg spacewalker uses a jet of gas to exert an impulse of 10Ns. How many times would he have to do this to reach a speed of 1 ms$^{−1}$ from stationary?
$\Delta v={\frac {I}{m}}={\frac {10}{60}}={\frac {1}{6}}{\mbox{ ms}}^{-1}$
So, the spacewalker would have to do this 6 times to reach a speed of 1ms$^{−1}$.
5. A 5 kg bowling ball collides with a stationary tennis ball of mass 0.1 kg at 3ms$^{−1}$, slowing to 2.5ms$^{−1}$. It exerts a force of 100N on the ball. How long did the collision take?
$F={\frac {m(v-u)}{\Delta t}}$
$\Delta t={\frac {m(v-u)}{F}}={\frac {5(3-2.5)}{100}}=0.025{\mbox{ s}}$
# A-level Physics (Advancing Physics)/Gravitational Forces/Worked Solutions

1. Jupiter orbits the Sun at a radius of around 7.8 x 10$^{11}$m. The mass of Jupiter is 1.9 x 10$^{27}$kg, and the mass of the Sun is 2.0 x 10$^{30}$kg. What is the gravitational force acting on Jupiter? What is the gravitational force acting on the Sun?
$F_{grav}={\frac {-GMm}{r^{2}}}={\frac {-6.67\times 10^{-11}\times 2\times 10^{30}\times 1.9\times 10^{27}}{(7.8\times 10^{11})^{2}}}=-4.17\times 10^{23}{\mbox{ N}}$
2. The force exerted by the Sun on an object at a certain distance is 10$^6$N. The object travels half the distance to the Sun. What is the force exerted by the Sun on the object now?
${\frac {1}{\left({\frac {1}{2}}\right)^{2}}}=4$
So, the new force is 4 MN.
3. How much gravitational force do two 1 kg weights 5 cm apart exert on each other?
$F_{grav}={\frac {-GMm}{r^{2}}}={\frac {-6.67\times 10^{-11}\times 1\times 1}{0.05^{2}}}=-2.67\times 10^{-8}{\mbox{ N}}$
In other words, ordinary objects exert negligible gravitational force.
4. The radius of the Earth is 6360 km, and its mass is 5.97 x 10$^{24}$kg. What is the difference between the gravitational force on 1 kg at the top of your body, and on 1 kg at your head, and 1 kg at your feet? (Assume that you are 2m tall.)
$\Delta F_{grav}=GMm\left({\frac {1}{6360000^{2}}}-{\frac {1}{6360002^{2}}}\right)=(1.55\times 10^{-20})GMm=1.55\times 10^{-20}\times 6.67\times 10^{-11}\times 5.97\times 10^{24}\times 1=6.19{\mbox{ }}\mu {\mbox{N}}$
This is why it is acceptable to approximate the acceleration due to gravity as constant over small distances.
# A-level Physics (Advancing Physics)/Heat and Energy/Worked Solutions

1. Carbon dioxide sublimes at 195°K. Roughly what energy per. particle does this correspond to?
$E\approx kT=1.38\times 10^{-23}\times 195\approx 10^{-21}{\mbox{ J particle}}^{-1}$
2. A certain chemical reaction requires particles with mass of the order 10$^{−26}$ to move, on average, at 10ms$^{−1}$. Roughly what temperature does this correspond to?
$E_{k}={\frac {1}{2}}mv^{2}=0.5\times 10^{-26}\times 10^{2}=5\times 10^{-25}$
$5\times 10^{-25}=kT$
$T={\frac {5\times 10^{-25}}{1.38\times 10^{-23}}}\approx 10^{-2}$ °K
3. The boiling point of water is 100 °C. Roughly what energy per. particle does this correspond to?
100 °C = 373°K
$E\approx kT=1.38\times 10^{-23}\times 373\approx 5\times 10^{-21}{\mbox{ J particle}}^{-1}$
4. Thermionic emission from copper requires around 5eV of energy per. particle. How hot will the wire be at this energy level?
5eV = 5 x 1.6 x 10$^{−19}$ = 8 x 10$^{−19}$ J particle$^{−1}$
$T={\frac {8\times 10^{-19}}{1.38\times 10^{-23}}}\approx 6\times 10^{4}$ °K
# A-level Physics (Advancing Physics)/Gravitational Potential/Worked Solutions

G = 6.67 x 10$^{−11}$ m$^3$kg$^{−1}$s$^{−2}$
g = 9.81 ms$^{−2}$
1. What is the gravitational potential at the Earth's surface? (mass of Earth = 5.97 x 10$^{24}$ kg,radius of Earth = 6371 km)
$V_{grav}={\frac {-GM}{r}}={\frac {-6.67\times 10^{-11}\times 5.97\times 10^{24}{MJkg}^{-1}}{6371000}}=62.5{\mbox{ MJkg}}^{-1}$
2. Taking the Earth's surface as V$_{grav}$ = 0, what is the gravitational potential 2m above the ground?
$V_{grav}\approx g\Delta h=9.81\times 2=19.62{\mbox{ Jkg}}^{-1}$
3. A 0.2 kg firework reaches a gravitational potential relative to the ground of 500Jkg$^{−1}$. If the firework is 30% efficient, how much energy was expended to get there?
$V_{grav}={\frac {E_{grav}}{m}}$
$E_{grav}=mV_{grav}=0.2\times 500=100{\mbox{ J}}$
However, this is only 30% of the energy expended, so:
$E_{expended}={\frac {100}{0.3}}\approx 333{\mbox{ J}}$
4. Express gravitational potential in terms of gravitational force.
$V_{grav}={\frac {\int {F_{grav}}\;dr}{m}}$
5. Draw the equipotentials and field lines surrounding the Earth.
# A-level Physics (Advancing Physics)/Half-lives

The half life of something that is decaying exponentially is the time taken for the value of a decaying variable to halve.

## Half Life of a Radioisotope

The most common use of half-lives is in radioactive decay. The activity is given by the equation:
$A_{t}=A_{0}e^{-\lambda t}$
At t=t$_½$, A$_t$ = ½A$_0$, so:
${\frac {A_{0}}{2}}=A_{0}e^{-\lambda t_{\frac {1}{2}}}={\frac {A_{0}}{e^{\lambda t_{\frac {1}{2}}}}}$
$2=e^{\lambda t_{\frac {1}{2}}}$
$\ln {2}=\lambda t_{\frac {1}{2}}$
Therefore:
$t_{\frac {1}{2}}={\frac {\ln {2}}{\lambda }}$
It is important to note that the half-life is completely unrelated to the variable which is decaying. At the end of the half-life, all decaying variables will have halved. This also means that you can start at any point in the decay, with any value of any decaying variable, and the time taken for the value of that variable to halve from that time will be the half-life.

## Half-Life of a Capacitor

You can also use this formula for other forms of decay simply by replacing the decay constant λ with the constant that was in front of the t in the exponential relationship. So, for the charge on a capacitor, given by the relationship:
$Q_{t}=Q_{0}e^{\frac {-t}{RC}}$
So, substitute:
$\lambda ={\frac {1}{RC}}$
Therefore, the half-life of a capacitor is given by:
$t_{\frac {1}{2}}=RC\ln {2}$

## Time Constant of a Capacitor

However, when dealing with capacitors, it is more common to use the time constant, commonly denoted τ, where:
$\tau =RC={\frac {t_{\frac {1}{2}}}{\ln {2}}}$
At t = τ:
$Q_{t}=Q_{0}e^{\frac {-RC}{RC}}={\frac {Q_{0}}{e}}$
So, the time constant of a capacitor can be defined as the time taken for the charge, current or voltage from the capacitor to decay to the reciprocal of e (36.8%) of the original charge, current or voltage.

## Questions

1. Radon-222 has a decay constant of 2.1μs$^{−1}$. What is its half-life?
2. Uranium-238 has a half-life of 4.5 billion years. How long will it take for a 5 gram sample of U-238 to decay to contain 1.25 grams of U-238?
3. How long will it be until it contains 0.5 grams of U-238?
4. Tritium, a radioisotope of Hydrogen, decays into Helium-3. After 1 year, 94.5% is left. What is the half-life of tritium (H-3)?
5. A large capacitor has capacitance 0.5F. It is placed in series with a 5Ω resistor and contains 5C of charge. What is its time constant?
6. How long will it take for the charge in the capacitor to reach 0.677C? ($0.677={\frac {5}{e^{2}}}$)
Worked Solutions
# A-level Physics (Advancing Physics)/Half-lives/Worked Solutions

1. Radon-222 has a decay constant of 2.1μs$^{−1}$. What is its half-life?
$t_{\frac {1}{2}}={\frac {\ln {2}}{2.1\times 10^{-6}}}=330070{\mbox{ s }}=3.82{\mbox{ days}}$
2. Uranium-238 has a half-life of 4.5 billion years. How long will it take for a 5 gram sample of U-238 to decay to contain 1.25 grams of U-238?
2 half-lives, since 1.25 is a quarter of 5. 2 x 4.5 = 9 billion years.
3. How long will it be until it contains 0.5 grams of U-238?
First calculate the decay constant:
$\lambda ={\frac {\ln {2}}{t_{\frac {1}{2}}}}={\frac {\ln {2}}{4.5\times 10^{9}}}=1.54\times 10^{-10}{\mbox{ yr}}^{-1}$
$0.5=5e^{-1.54\times 10^{-10}t}$
$0.1=e^{-1.54\times 10^{-10}t}$
$\ln {0.1}=-1.54\times 10^{-10}t$
$t={\frac {\ln {0.1}}{-1.54\times 10^{-10}}}=14.9{\mbox{ Gyr}}$
4. Tritium, a radioisotope of Hydrogen, decays into Helium-3. After 1 year, 94.5% is left. What is the half-life of tritium (H-3)?
$0.945=e^{-\lambda \times 1}$ (if λ is measured in yr$^{−1}$)
$\lambda =-\ln {0.945}=0.0566{\mbox{ yr}}^{-1}={\frac {\ln {2}}{t_{\frac {1}{2}}}}$
$t_{\frac {1}{2}}={\frac {\ln {2}}{0.0566}}=12.3{\mbox{ yr}}$
5. A large capacitor has capacitance 0.5F. It is placed in series with a 5Ω resistor and contains 5C of charge. What is its time constant?
$\tau =RC=5\times 0.5=2.5{\mbox{ s}}$
6. How long will it take for the charge in the capacitor to reach 0.677C? ($0.677={\frac {5}{e^{2}}}$)
2 x τ = 5s
# A-level Physics (Advancing Physics)/Gravitational Potential

Gravitational potential is the gravitational potential energy given to objects per unit mass:
$V_{grav}={\frac {E_{grav}}{m}}={\frac {-{\frac {GMm}{r}}}{m}}={\frac {-GM}{r}}$
For short changes in distance Δh (Δh << r), gravitational potential energy is given by:
$E_{grav}\approx mg\Delta h$,
where
$g={\frac {-GM}{r^{2}}}$
So, for short changes of distance, gravitational potential is equal to gΔh. Gravitational potential is measured in Jkg$^{−1}$.

## Equipotentials

On a field diagram, lines can be drawn which, like contours on a map, show all the points which have the same gravitational potential. These lines are known as equipotentials. Equipotentials are always perpendicular to field lines, and get closer together as the field strength increases, and the density of field lines increases. Over short distances, equipotentials are evenly spaced.

## Summary of Gravity

You should now know (if you did the gravity section in the right order) about four attributes of gravitational fields: force, field strength, potential energy and potential. These can be summarised by the following table:
| Force $F_{grav}={\frac {-GMm}{r^{2}}}$ | → integrate → with respect to r and × by −1 | Potential Energy $E_{grav}={\frac {-GMm}{r}}$ |
| ↓ per unit mass ↓ |||
| Field Strength $g={\frac {-GM}{r^{2}}}$ | → integrate → with respect to r and × by −1 | Potential $V_{grav}={\frac {-GM}{r}}$ |

## Questions

G = 6.67 x 10$^{−11}$ m$^3$kg$^{−1}$s$^{−2}$
g = 9.81 ms$^{−2}$
1. What is the gravitational potential at the Earth's surface? (mass of Earth = 5.97 x 10$^{24}$ kg,radius of Earth = 6371 km)
2. Taking the Earth's surface as V$_{grav}$ = 0, what is the gravitational potential 2m above the ground?
3. A 0.2 kg firework reaches a gravitational potential relative to the ground of 500Jkg$^{−1}$. If the firework is 30% efficient, how much energy was expended to get there?
4. Express gravitational potential in terms of gravitational force.
5. Draw the equipotentials and field lines surrounding the Earth.
Worked Solutions
# A-level Physics (Advancing Physics)/Gravitational Potential Energy/Worked Solutions

1. A ball rolls down a 3m-high smooth ramp. What speed does it have at the bottom?
$mgh={\frac {1}{2}}mv^{2}$
$gh={\frac {1}{2}}v^{2}$
$v={\sqrt {2gh}}={\sqrt {2\times 9.81\times 3}}=7.67{\mbox{ ms}}^{-1}$
2. In an otherwise empty universe, two planets of mass 10$^{25}$ kg are 10$^{12}$ m apart. Both the planets have a radius of 10$^6$ m. What are their speeds when they collide?
Let $M_{1}$ be the mass of planet 1 and $M_{2}$ be the mass of planet 2. Both are $1\times 10^{25}$ kg
Assume planet 1 to be stationary and planet 2 to be accelerating towards it (relative).
Let D = $1\times 10^{12}$ meters = distance between the center of the two planets. Let d = $1\times 10^{6}$ meters = radius of planets.
$\int _{2d}^{D}{\frac {GM_{1}M_{2}}{r^{2}}}dr=$ Gravitational Potential Energy
$\left[{\frac {-GM_{1}M_{2}}{r}}\right]_{2d}^{D}=-GM_{1}M_{2}\left[{\frac {1}{D}}-{\frac {1}{2d}}\right]={\frac {1}{2}}M_{2}v^{2}$
${\sqrt {-2GM_{1}\left[{\frac {1}{D}}-{\frac {1}{2d}}\right]}}=v=25,800ms^{-1}$
3. What is the least work a 2000 kg car must do to drive up a 100m hill?
$mgh=2000\times 9.81\times 100=1.962{\mbox{ MJ}}$
4. How does the speed of a planet in an elliptical orbit change as it nears its star?
As it nears the star, it loses gravitational potential energy, and so gains kinetic energy, so its speed increases.
# Adventist Youth Honors Answer Book/Skill Level 1 Nature

* Adventist Youth Honors Answer Book/Nature/Bats
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# Adventist Youth Honors Answer Book/Skill Level 2 ADRA

* Adventist Youth Honors Answer Book/ADRA/Literacy
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* Adventist Youth Honors Answer Book/Health and Science/First Aid, Basic
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# Adventist Youth Honors Answer Book/Skill Level 1 Household Arts

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