Linear momentum p = mv: Exercises — Physics deep dive

Before we start, a one-line reminder of the tools we will lean on, so no symbol appears unearned:

Definition The toolkit for this page

Momentum $p = m v$ : mass (a positive number, kilograms) times velocity (a vector — a number with a direction/sign). Unit $kg\cdotm/s$ .
Arrow vs no arrow: $p$ (with the arrow) is the full vector — it carries a direction. Plain $p$ (no arrow) means its magnitude $∣ p ∣$ — a non-negative number, the "length" of the arrow. So $K E = \frac{p ^{2}}{2 m}$ uses the magnitude, and a 1D signed value like $- 4.0$ is really the single component of $p$ along our chosen $+$ axis. We keep the arrow when direction matters and drop it when only size matters.
Force $F$ : the push or pull on an object. Newton stated his second law as "force is the rate at which momentum changes," written $F = \frac{d p}{d t}$ . Here $t$ is time (seconds), and $\frac{d p}{d t}$ is calculus shorthand for "how much $p$ changes per second" — the rate of change of momentum. Unit $N = kg\cdotm/s^{2}$ . We only need this for the L1 units trap.
In 1D, "direction" is just a $+$ or $-$ sign. We pick one direction (say right) to be $+$ .
In 2D, we split into $p_{x} = m v_{x}$ and $p_{y} = m v_{y}$ , then the length is $∣ p ∣ = p_{x}^{2} + p_{y}^{2}$ (Pythagoras on the right triangle the two components form).
Kinetic energy $K E = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ : a single number (scalar), never negative. The $p$ here is the magnitude $∣ p ∣$ .

Level 1 — Recognition

(Can you pick out $m$ and $v$ and multiply?)

Recall Solution

WHAT we do: plug the magnitudes into $p = m v$ . WHY: the object has one mass and one speed, nothing to add. $p = m v = 12 \times 1.5 = 18 kg\cdotm/s$ Direction (state it explicitly): momentum always points along $v$ because $m > 0$ never flips direction. In 1D sign notation, if we take the rolling direction as $+$ , the momentum is $p = + 18 kg\cdotm/s$ (a positive value = pointing the way it rolls).

Recall Solution

WHY rearrange: we know $p$ and $m$ , want $v$ , so invert $p = m v$ into $v = p / m$ . $v = \frac{p}{m} = \frac{6 \times 1 0 ^{- 6}}{2 \times 1 0 ^{- 6}} = 3 m/s$

Level 2 — Application

(One direct use of the formula, maybe with a sign or a rearrangement.)

Recall Solution

WHY the sign: velocity is a vector; "left" while right is $+$ means $v = - 25 m/s$ . $p = m v = 0.16 \times (- 25) = - 4.0 kg\cdotm/s$ The minus sign is the direction — this signed value is the component of $p$ along the $+$ axis. In 1D, that is all a vector needs.

Recall Solution

WHY rearrange: we are given $p$ and $v$ but want $m$ , so we invert the definition $p = m v$ by dividing both sides by $v$ , giving $m = p / v$ . $m = \frac{p}{v} = \frac{2.4}{8} = 0.30 kg$

Recall Solution

WHICH tool and WHY: we're given $K E$ and want $p$ (the magnitude), and the bridge that skips computing $v$ is $K E = \frac{p ^{2}}{2 m}$ . Solve for $p$ : $p^{2} = 2 m K E ⟹ p = 2 m K E = 2 \times 0.5 \times 16 = 16 = 4 kg\cdotm/s$ WHY only the positive root: solving $p^{2} = 16$ gives $p = \pm 4$ , but $p$ here is a magnitude — the length of the momentum arrow — and lengths are never negative. So we discard $- 4$ and keep $p = + 4 kg\cdotm/s$ . (If a direction were asked, we'd attach it separately; the magnitude itself stays positive.)

Level 3 — Analysis

(Two objects, or 2D — you must combine correctly.)

Recall Solution

WHY add with signs: total momentum is the vector sum $p_{A} + p_{B}$ ; in 1D that's an ordinary sum once signs are set (right $= +$ ). $p_{A} = 3 (+ 4) = + 12, p_{B} = 5 (- 2) = - 10$ $p_{tot} = 12 + (- 10) = + 2 kg\cdotm/s (to the right)$

Recall Solution

Step 1 — components (WHY): $m$ is a positive scalar, so it scales each velocity component the same way. $p_{x} = 0.30 \times 6 = 1.8 kg\cdotm/s, p_{y} = 0.30 \times 8 = 2.4 kg\cdotm/s$ Step 2 — magnitude (WHY Pythagoras): $p_{x}$ and $p_{y}$ are perpendicular, so they form the two legs of a right triangle whose hypotenuse is $∣ p ∣$ (look at the red triangle in the figure). $∣ p ∣ = 1. 8^{2} + 2. 4^{2} = 3.24 + 5.76 = 9 = 3.0 kg\cdotm/s$ Step 3 — direction (WHY arctan): the angle's steepness is captured by $tan θ = \frac{opposite}{adjacent} = \frac{p _{y}}{p _{x}}$ ; to recover the angle itself we ask "which angle has this tan?" — that is $arctan$ . $θ = arctan (\frac{2.4}{1.8}) = arctan (1.333) \approx 53. 1^{\circ}$ Both components are positive → quadrant I, so the raw $arctan$ value is already correct.

Common mistake L3 trap — "

$p_{x} = 0$ ? just type $arctan (p_{y} /0)$ " Why it feels right: the formula $θ = arctan (p_{y} / p_{x})$ looks universal, so students plug in $p_{x} = 0$ and hit division by zero (or " $arctan (\infty)$ ") and freeze — or worse, they add the two magnitudes $1.8 + 2.4 = 4.2$ thinking perpendicular arrows add straight. The fix (two parts):

Magnitudes: perpendicular legs never add straight; use $p_{x}^{2} + p_{y}^{2} = 3.0$ , not $4.2$ .
Edge case $p_{x} = 0$ : a vector straight up the $y$ -axis has $tan θ = p_{y} /0$ , undefined. But geometrically the angle is obvious: $+ 9 0^{\circ}$ if $p_{y} > 0$ , $- 9 0^{\circ}$ if $p_{y} < 0$ , and undefined if both components are zero (no direction). This is exactly why physicists use the two-argument $atan2 (p_{y}, p_{x})$ : you feed it both components separately, so it handles $p_{x} = 0$ and picks the correct quadrant automatically instead of choking on a single ratio.

Level 4 — Synthesis

(Momentum + energy, or a rearrangement chain.)

Recall Solution

WHICH tool and WHY: we have $p$ (magnitude) and $m$ for each, and want $K E$ without finding $v$ separately — use $K E = \frac{p ^{2}}{2 m}$ . $K E_{bullet} = \frac{4 ^{2}}{2 \times 0.02} = \frac{16}{0.04} = 400 J$ $K E_{block} = \frac{4 ^{2}}{2 \times 2} = \frac{16}{4} = 4 J$ The bullet has more energy by a factor $\frac{400}{4} = 100$ . WHY the huge gap: at fixed $p$ , $K E = p^{2} /2 m \propto 1/ m$ . The bullet has $100 \times$ less mass, so $100 \times$ the energy. Equal "oomph," wildly unequal "damage capacity."

Recall Solution

$p = m v$ is linear in $v$ : triple $v$ → momentum $\times 3$ . $K E = \frac{1}{2} m v^{2}$ is quadratic in $v$ : triple $v$ → energy $\times 3^{2} = 9$ . So momentum grows $3 \times$ , kinetic energy grows $9 \times$ .

Level 5 — Mastery

(Multi-step 2D vector problem with a twist.)

Recall Solution

Step 1 — each object's momentum (WHY per-axis): A moves purely along $x$ , B purely along $y$ , so each contributes to only one axis. $p_{A} : p_{A x} = 2 \times 3 = 6, p_{A y} = 0$ $p_{B} : p_{B x} = 0, p_{B y} = 1 \times 8 = 8$ Step 2 — add component by component (WHY): vector sum means add $x$ 's together and $y$ 's together, separately. $p_{x} = 6 + 0 = 6, p_{y} = 0 + 8 = 8 kg\cdotm/s$ Step 3 — magnitude (right triangle): the two totals are perpendicular legs (see the teal and orange arrows in the figure combining into the plum resultant). $∣ p ∣ = 6^{2} + 8^{2} = 36 + 64 = 100 = 10 kg\cdotm/s$ Step 4 — direction: $θ = arctan \frac{p _{y}}{p _{x}} = arctan \frac{8}{6} = arctan (1.333) \approx 53. 1^{\circ}$ north of east. Both components positive → quadrant I, no correction needed.

Recall Solution

Magnitude: $∣ p ∣ = (- 3)^{2} + (- 4)^{2} = 9 + 16 = 25 = 5 kg\cdotm/s$ . Direction — WHY the naive $arctan$ fails here: $tan$ repeats every $18 0^{\circ}$ , so $arctan$ only ever returns an angle between $- 9 0^{\circ}$ and $+ 9 0^{\circ}$ (quadrants I and IV). Our vector points down-and-left → quadrant III. The raw computation gives: $arctan (\frac{- 4}{- 3}) = arctan (1.333) \approx 53. 1^{\circ}$ That points up-and-right — the wrong way. Because both components are negative (quadrant III), add $18 0^{\circ}$ : $θ = 53. 1^{\circ} + 18 0^{\circ} = 233. 1^{\circ}$ (Equivalently $- 126. 9^{\circ}$ .) The figure shows the fake $53. 1^{\circ}$ arrow versus the true $233. 1^{\circ}$ arrow — same tan, opposite physical direction.

Recall Quick self-quiz (cover the right side — each answer carries its WHY)

$p$ of $12 kg$ at $1.5 m/s$ ::: $+ 18 kg\cdotm/s$ — direct $p = m v$ ; positive because it points the $+$ way. $∣ p ∣$ from $p_{x} = 1.8, p_{y} = 2.4$ ::: $3.0 kg\cdotm/s$ — Pythagoras on perpendicular legs, not $1.8 + 2.4$ . $K E$ of $2 kg$ block with $p = 4$ ::: $4 J$ — use $K E = p^{2} /2 m$ so no need to find $v$ first. Direction of $(p_{x}, p_{y}) = (- 3, - 4)$ ::: $233. 1^{\circ}$ — quadrant III ( $p_{x} < 0$ ), so add $18 0^{\circ}$ to the raw arctan. Triple the speed → $p \times ?$ , $K E \times ?$ ::: $\times 3$ and $\times 9$ — $p$ is linear in $v$ , $K E$ is quadratic ( $v^{2}$ ).

Connections

Parent: Linear momentum $p=mv$ — the definitions used throughout.
Kinetic Energy — the $K E = p^{2} /2 m$ bridge in L2·Q3 and L4·Q1.
Conservation of Linear Momentum — where the L3/L5 vector totals get used in real crashes.
Impulse–Momentum Theorem — why $kg\cdotm/s = N\cdots$ (L1 trap).
Newton's Second Law — the source of $F = d p / d t$ referenced in the toolkit.