Weight and motion under gravityEdexcel A-Level Maths: Revision notes

Mass $m$ (kg) is the amount of matter in a body and is the same everywhere. Weight $W$ (N) is the gravitational force on the body, acting vertically downwards: $W=mg.$ Here $g$ is the acceleration due to gravity. Unless a question says otherwise, use $g=9.8$ m s$^{-2}$; some questions specify another value, such as $10$ m s$^{-2}$, and you must use that value. The value of $g$ is not a universal constant: it varies with location, for example about $9.8$ m s$^{-2}$ on Earth's surface and $1.6$ m s$^{-2}$ on the Moon. At A Level you may assume that $g$ is constant over the heights in a problem, and you do not need the inverse square law of gravitation. Example: a $0.2$ kg stone weighs $0.2\times9.8=1.96$ N on Earth and $0.2\times1.6=0.32$ N on the Moon, but its mass is $0.2$ kg in both places.

A body moving freely under gravity has weight as the only force, so $a=g$ downwards: the same for every body, whatever its mass. Model it as a particle with no air resistance and constant $g$. Use the constant-acceleration equations with a sign convention. Taking up as positive, $a=-g$; taking down as positive, $a=+g$. State your choice at the start and keep to it. $v=u+at,\quad s=ut+\tfrac12at^2,\quad v^2=u^2+2as,\quad s=\tfrac12(u+v)t.$

For a body thrown upwards from ground level with speed $u$ (up positive, $a=-g$):

• At the greatest height $v=0$: $H=\dfrac{u^2}{2g}$ and the time to the top is $\dfrac{u}{g}$.
• It returns to the ground when $s=0$, after $T=\dfrac{2u}{g}$.
• It passes any given height with the same speed going up and going down. Example: $u=14$, $g=9.8$: $H=\frac{196}{19.6}=10$ m, $T=2.86$ s. At $4$ m on the way down, $v^2=196-2(9.8)(4)=117.6$, so the speed is $10.8$ m s$^{-1}$.

A body dropped from rest has $u=0$, so $s=\frac12gt^2$ and $v^2=2gs$. Example: a coin dropped $40$ m: $40=4.9t^2$ gives $t=2.86$ s, and $v=\sqrt{2(9.8)(40)}=28$ m s$^{-1}$. If a body is thrown from above the ground, the final displacement is negative (up positive). Ball thrown up at $15$ m s$^{-1}$ from $1.5$ m above the ground: reaches the ground when $-1.5=15t-4.9t^2$, i.e. $4.9t^2-15t-1.5=0$, so $t=3.16$ s after rejecting the negative root. For the last second of a fall, subtract the distance fallen in the first $(T-1)$ seconds from the total height: $40-4.9(1.857)^2=23.1$ m.

1. Draw a sketch with the positive direction marked.
2. List $s$, $u$, $v$, $a$, $t$ with signs, then choose the equation that has the unknown and three knowns.
3. Use the value of $g$ stated in the question; give answers to 3 significant figures (or to the accuracy implied by $g$).
4. Comment on the model when asked: ignoring air resistance makes the predicted times shorter and speeds larger than in reality; and use the idea that $g$ depends on location when comparing places.