PMDC Unit 38 · 17 outcomes

Force and Motion

Unit 2 is how things move, and why. First the words (displacement, velocity, acceleration), then graphs, then a ball in the air, then Newton’s three laws and collisions.

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2.1

Displacement

PMDC: Describe displacement.

Displacement is the straight arrow from where you started to where you finished. It is a vector. Distance is the path you actually walked — a scalar, always positive.

Walk $3 m$ east then $4 m$ west: $distance = 7$ m, $d i s pl a ce m e n t = 1$ m west. A round trip has displacement zero.

Live graph — drag the sliders

∣ s ∣ = 3^{2} + 4^{2} = 5 m

$Δ s = s_{f} - s_{i}$ (vector)
distance ≥ |displacement|

Hold these

Displacement cares about start and end only.
It can be negative if you pick a positive direction and travel the other way.

2.2

Velocity

PMDC: Describe average velocity of objects.

Average velocity is displacement divided by the time it took. Average speed is distance divided by time. They match only when you never change direction.

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⟨ v ⟩ = Δ s /Δ t = 0

$v_{a v g} = Δ s /Δ t$
$s p ee d_{a} v g = distance /Δ t$

Hold these

Velocity is a vector. Speed is a scalar.
Zero displacement in a time interval means zero average velocity, even if you ran.

2.3

Displacement–time graph

PMDC: Interpret a displacement–time graph of objects moving along the same straight line.

On a displacement–time graph the slope is velocity. A straight line means constant velocity. A curve that gets steeper means speeding up. A flat line means at rest. Crossing the time axis means you passed the origin.

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v = d s / d t = 2 m/s

slope of s– $t = v$
area under v– $t = Δ s$

Hold these

Steeper s–t means faster.
A negative slope means motion in the negative direction.

2.4

Acceleration

PMDC: Describe acceleration.

Acceleration is how fast velocity changes. Speeding up, slowing down, or turning — all of those are acceleration, because velocity is a vector.

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a = co n s t an t

(straight v–t)

$a_{a v g} = Δ v /Δ t$
$a = d v / d t (in s t an t an eo u s)$

Hold these

Units of a: m s⁻².
If v and a point the same way, the object speeds up. Opposite ways, it slows down.

2.5

Uniform and variable acceleration

PMDC: Distinguish between uniform and variable acceleration.

Uniform acceleration: a is constant. The v–t graph is a straight line. The suvat equations work. Free fall near the ground (no air) is uniform, with $a = g$ downward.

Variable acceleration: a itself changes. A rocket that burns fuel, or a spring, or a car that eases off the pedal. You cannot drop in suvat blindly.

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s = \frac{1}{2} (4) (5)^{2} = 50 m

$u ni f or ma : v = u + a t$
$s = u t + \frac{1}{2} a t^{2}$
$v^{2} = u^{2} + 2 a s$
$s = (u + v) t /2$

Hold these

suvat needs constant a.

2.6

Projectile motion

PMDC: Explain that projectile motion is two-dimensional motion in a vertical plane.

Throw a ball. It moves across and up-and-down at the same time. That pair of motions, in a vertical plane, is projectile motion. The path is a parabola if air is ignored.

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R = u^{2} sin 2 θ / g = 40 m

two independent motions: horizontal + vertical

Hold these

Treat x and y separately. Time is the only shared number.

2.7

Ideal projectile

PMDC: Communicate the ideas of a projectile in the absence of air resistance.

Ideal means no air. Gravity is the only force after the throw, so the only acceleration is g downward. Nothing pushes sideways, so the horizontal velocity never changes.

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R = u^{2} sin 2 θ / g = 40 m

$a_{x} = 0 a_{y} = - g$

Hold these

No $ai r \Rightarrow n o$ horizontal $f or ce \Rightarrow v_{x}$ stays put.

2.8

Horizontal velocity is constant

PMDC: Explain that the horizontal component (v_H) of velocity is constant.

Split the launch velocity u at angle $θ : u_{x}$ = u cos $θ, u_{y}$ = u sin $θ$ . With no air, $u_{x}$ is the same at every point of the flight, including the top.

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R = u^{2} sin 2 θ / g = 40 m

$v_{x} = u cos θ (co n s t an t)$
$x = (u cos θ) t$

Hold these

At the top, $v_{x}$ is still u cos $θ$ . Only $v_{y}$ is zero there.

2.9

Vertical acceleration is free fall

PMDC: Acceleration is in the vertical direction and is the same as that of a vertically free-falling object.

The vertical motion is exactly free fall: $a_{y}$ = −g. Same as dropping a stone. The ball does not ‘run out of force’ at the top — g is still there, turning $v_{y}$ around.

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a = g

for both

$v_{y} = u sin θ - g t$
$y = (u sin θ) t - \frac{1}{2} g t^{2}$
$v_{y}^{2} = (u sin θ)^{2} - 2 g y$

Hold these

g never switches off during the flight.

2.10

Horizontal vs vertical motion

PMDC: Differentiate between the characteristics of horizontal motion and vertical motion.

Horizontal: constant velocity, zero acceleration, equal spaces in equal times. Vertical: changing velocity, acceleration g, the spaces stretch then shrink. They share time only. One does not wait for the other.

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H = u^{2} sin^{2} θ /2 g = 5.0 m

x-motion: $a = 0$ y-motion: $a = - g$ shared: t

Hold these

A bullet dropped and a bullet fired horizontally hit the ground together (same height, no air).

2.11

Height, range, time of flight, and equal-range angles

PMDC: For a frictionless projectile launched from ground level: how much higher, how far, where after a given time, how long in the air, the launch angle for maximum range, and the pair of angles that share a range.

Launch from the ground, land at the same height. Four numbers fall out of suvat. Maximum height H is where $v_{y}$ = 0. Time of flight T is the time until $y = 0$ again. Range R is how far along the ground in that time.

R is largest at $4 5^{\circ}$ . Two complementary angles, $θ$ and $9 0^{\circ} - θ$ , give the same range (one is a high lob, the other a flat throw).

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T = 2 u sin θ / g = 2.0 s

$H = (u sin θ)^{2} / (2 g)$
$T = 2 u sin θ / g$
$R = u^{2} sin 2 θ / g$
$R_{m a x}$ = u² / g at $θ = 4 5^{\circ}$
same range: $θ$ and ( $9 0^{\circ} - θ$ )

Hold these

Time to the top is T/2.
sin 2 $θ$ is largest when $2 θ = 9 0^{\circ}$ , so $θ = 4 5^{\circ}$ .
These four formulae need a flat landing at the launch height.

2.12

Newton’s laws of motion

PMDC: Apply Newton’s laws to explain the motion of objects in a variety of contexts.

First law: if the net force is zero, velocity stays the same — rest stays rest, steady motion stays steady. Inertia is that stubbornness.

Second law: net force equals mass times acceleration. Bigger F, bigger a. Bigger m, smaller a. Force and acceleration point the same way.

Third law: forces come in pairs. If A pushes B, B pushes A equally, opposite, on a different body. The pair never cancel on the same object.

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a = F / m = 6/2 = 3 m/s^{2}

Law I: $Σ F = 0 \Rightarrow v$ constant
$L a w I I : Σ F = ma$
Law III: $F_{A B} = - F_{B A}$

Hold these

Net force, not ‘a force’, decides a.
The third-law partner lives on the other object.

2.13

Newton’s second law and linear momentum

PMDC: Describe Newton’s second law of motion as rate of change of momentum.

Momentum $p = m$ v. Newton’s second law in its original form: net force is how fast momentum changes. If mass is constant that is just m a. If mass changes (a conveyor, a rocket) you must use dp/dt.

Impulse is force × time, and it equals the change in momentum. A longer collision (airbag) means a smaller force for the same $Δ p$ .

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J = m (- v - u) = - 10 N s

$p = m v$
$Σ F = d p / d t$
$im p u l se J = F Δ t = Δ p$

Hold these

Units of p: kg m s⁻¹, also N s.
Same $Δ p$ in more $t im e \Rightarrow s ma l l er$ F.

2.14

Newton’s third law and conservation of momentum

PMDC: Correlate Newton’s third law of motion and conservation of momentum.

In a collision, A hits B. The force of A on B and of B on A are equal and opposite (third law) and last for the same time. So the impulses cancel. The total momentum of A plus B does not change — if no outside force acts on the pair.

That is conservation of momentum. The third law is why it is true.

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a_{1} / a_{2} = m_{2} / m_{1} = 2

no external $f or ce \Rightarrow p_{t o t a l}$ constant
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Hold these

Always check: is the system isolated? Friction and the ground are outside forces.
Momentum is a vector. Pick a + direction and keep signs.

2.15

Elastic and inelastic collisions (1-D)

PMDC: Solve problems of elastic and inelastic collisions between two bodies in one dimension using conservation of momentum.

Every isolated collision keeps momentum. Kinetic energy is a different question.

Elastic: KE is also conserved (ideal billiard balls). Inelastic: some KE becomes heat and sound. Perfectly inelastic: the two bodies stick and move as one — maximum KE loss, momentum still conserved.

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v_{1} = 0 v_{2} = 6 m/s

always (isolated): $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
elastic also: ½ $m_{1} u_{1}$ ² + ½ $m_{2} u_{2}$ ² = ½ $m_{1} v_{1}$ ² + ½ $m_{2} v_{2}$ ²
perfectly inelastic: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}$ ) v

Hold these

Stuck $t o g e t h er \Rightarrow p er f ec tl y$ inelastic. Use one v after.
Do not use the KE equation unless the stem says elastic.

2.16

Momentum and explosive forces

PMDC: Describe that momentum is conserved in explosion situations.

A firecracker at rest splits. The bits fly apart. Internal explosive forces are third-law pairs, so total momentum stays zero if it started at rest. Recoil of a gun is the same idea: bullet one way, gun the other, momenta cancel.

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p = 4 \times 5 = 20 kg m/s

at rest then explode: $0 = m_{1} v_{1} + m_{2} v_{2}$
$v_{2} = - (m_{1} / m_{2}) v_{1}$

Hold these

Explosions add chemical energy (KE rises) but still keep momentum if isolated.

2.17

Perfectly elastic collision in one dimension

PMDC: Identify that for a perfectly elastic collision, the relative speed of approach equals the relative speed of separation.

For two bodies in a 1-D perfectly elastic collision, they close the gap at some relative speed, then open the gap at the same relative speed. That is the extra equation that, with momentum, solves for both final velocities.

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v_{1} = 0 v_{2} = 6 m/s

$e l a s t i c 1 - D : u_{1} - u_{2} = v_{2} - v_{1}$
(relative speed of $a pp r o a c h = relative$ speed of separation)

Hold these

This is not true for inelastic collisions.
If equal masses collide elastically head-on, they swap velocities.

Unit 38 formula sheet

Write these until they are cheaper than thinking, then open the quiz.

$Δ s = s_{f} - s_{i}$ (vector)
distance ≥ |displacement|
$v_{a v g} = Δ s /Δ t$
$s p ee d_{a} v g = distance /Δ t$
slope of s– $t = v$
area under v– $t = Δ s$
$a_{a v g} = Δ v /Δ t$
$a = d v / d t (in s t an t an eo u s)$
$u ni f or ma : v = u + a t$
$s = u t + \frac{1}{2} a t^{2}$
$v^{2} = u^{2} + 2 a s$
$s = (u + v) t /2$
two independent motions: horizontal + vertical
$a_{x} = 0 a_{y} = - g$
$v_{x} = u cos θ (co n s t an t)$
$x = (u cos θ) t$
$v_{y} = u sin θ - g t$
$y = (u sin θ) t - \frac{1}{2} g t^{2}$
$v_{y}^{2} = (u sin θ)^{2} - 2 g y$
x-motion: $a = 0$ y-motion: $a = - g$ shared: t
$H = (u sin θ)^{2} / (2 g)$
$T = 2 u sin θ / g$
$R = u^{2} sin 2 θ / g$
$R_{m a x}$ = u² / g at $θ = 4 5^{\circ}$
same range: $θ$ and ( $9 0^{\circ} - θ$ )
Law I: $Σ F = 0 \Rightarrow v$ constant
$L a w I I : Σ F = ma$
Law III: $F_{A B} = - F_{B A}$
$p = m v$
$Σ F = d p / d t$
$im p u l se J = F Δ t = Δ p$
no external $f or ce \Rightarrow p_{t o t a l}$ constant
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
always (isolated): $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
elastic also: ½ $m_{1} u_{1}$ ² + ½ $m_{2} u_{2}$ ² = ½ $m_{1} v_{1}$ ² + ½ $m_{2} v_{2}$ ²
perfectly inelastic: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}$ ) v
at rest then explode: $0 = m_{1} v_{1} + m_{2} v_{2}$
$v_{2} = - (m_{1} / m_{2}) v_{1}$
$e l a s t i c 1 - D : u_{1} - u_{2} = v_{2} - v_{1}$
(relative speed of $a pp r o a c h = relative$ speed of separation)

Open the Unit 38 quiz