Newton's laws of motion

Fluency

An object continues at rest or at constant velocity unless an unbalanced force acts on it.
$F_{\text{net}} = ma$ , where $F$ is in newtons (N), $m$ in kilograms (kg), $a$ in m/s $^2$ .
Mass is the amount of matter (kg); weight is the gravitational force on that mass (N). Weight depends on $g$ ; mass does not.
For every action there is an equal and opposite reaction; the forces act on two different objects.
The tendency of an object to resist changes in its motion; it depends on mass.

Fluency

$a = \dfrac{20}{4} = 5 \text{ m/s}^2$ .
$F = 1\,500 \times 2.5 = 3\,750 \text{ N}$ .
$m = \dfrac{50}{10} = 5 \text{ kg}$ .
$W = 90 \times 9.8 = 882 \text{ N}$ .
$m = \dfrac{196}{9.8} = 20 \text{ kg}$ .
$a = \dfrac{2 - 0}{0.8} = 2.5 \text{ m/s}^2$ . $F = 0.4 \times 2.5 = 1 \text{ N}$ .

Fluency

Examples: jumping (legs push on ground; ground pushes back on legs). Bat hits ball (bat on ball; ball on bat).
The water pushes the swimmer forwards with an equal and opposite force.
Friction is needed for the ground to push back on your foot; on frictionless ice, there is no reaction force to accelerate you forward.
$200$ N (Newton’s third law).
The forces act on different objects. The horse’s forward motion comes from the ground pushing back on the horse’s hooves (larger than the cart’s pull); that net forward force on the horse-cart system causes motion.

Reasoning

FBD: push $40$ N right, friction $15$ N left, normal up, weight down. $F_{\text{net}} = 40 - 15 = 25 \text{ N}$ . $a = \dfrac{25}{10} = 2.5 \text{ m/s}^2$ .
At terminal velocity, acceleration is zero, so net force is zero. Air resistance $= W = 80 \times 9.8 = 784 \text{ N}$ (upward).
Tension $T - W = ma$ , so $T = m(g + a) = 5(9.8 + 1) = 54 \text{ N}$ .
$a = \dfrac{0 - 15}{3} = -5 \text{ m/s}^2$ . $F = 2\,000 \times 5 = 10\,000 \text{ N}$ (opposing motion).
$F_{\text{net}} = ma = 1.2 \times 0.5 = 0.6$ N. Friction $= 5 - 0.6 = 4.4$ N.

Problem solving

Weight $= 70 \times 9.8 = 686$ N. Net force $= 686 - 400 = 286$ N downward. $a = \dfrac{286}{70} \approx 4.1 \text{ m/s}^2$ downward.
Net force $= 65 - 50 = 15$ N (in direction of $65$ N push). $a = \dfrac{15}{30} = 0.5 \text{ m/s}^2$ .
Weight $= 800 \times 9.8 = 7\,840$ N. $T - W = ma$ , so $T = 7\,840 + 800 \times 1.5 = 9\,040$ N.
$a = \dfrac{25}{8} = 3.125 \text{ m/s}^2$ . Net force $= 1\,300 \times 3.125 = 4\,062.5$ N. Thrust $= 4\,062.5 + 500 = 4\,562.5$ N.

Reasoning

(a) Moon: $W = 80 \times 1.6 = 128$ N. Earth: $W = 80 \times 9.8 = 784$ N. (b) Equally hard in terms of $F = ma$ — horizontal acceleration depends on mass, not weight. The Moon’s lower gravity only affects vertical forces like weight.
(a) $a = \dfrac{30}{0.1} = 300 \text{ m/s}^2$ . $F = 75 \times 300 = 22\,500$ N. (b) $a = \dfrac{30}{0.5} = 60 \text{ m/s}^2$ . $F = 75 \times 60 = 4\,500$ N — a fivefold reduction, crucial for survival.
Heavier mass ( $5$ kg) accelerates down; lighter ( $3$ kg) accelerates up. $a = \dfrac{(5 - 3) \times 9.8}{5 + 3} = \dfrac{19.6}{8} = 2.45 \text{ m/s}^2$ . Tension: for lighter mass, $T - mg = ma \Rightarrow T = 3(9.8 + 2.45) = 36.75$ N.
Rocket weight $= 2\,000 \times 9.8 = 19\,600$ N. (a) Thrust ( $25\,000$ N) exceeds weight, so yes — it lifts off with $F_{\text{net}} = 25\,000 - 19\,600 = 5\,400$ N giving $a = 2.7 \text{ m/s}^2$ . (b) Thrust $= m(g + a) = 2\,000 \times (9.8 + 3) = 25\,600$ N.

Year 10 answers