Exercises — Newton's second law — F = ma (net force), impulse-momentum form

WHAT we want: acceleration $a$ from a known net force and mass. WHY $F=ma$: the mass is constant (one solid crate), so the simple child-form applies. $a=\frac{F_{\text{net}}}{m}=\frac{20}{5}=4{\text{m/s}}^2\text{east}.$ The direction of $\vec{a}$ always matches the direction of ${\vec{F}}_{\text{net}}$.

Momentum is just mass times velocity: $p=mv=0.5\times 8=4\text{kg⋅m/s}.$ Units check: $\text{kg}\times \text{m/s}=\text{kg⋅m/s}$. Direction is the same as the velocity.

For a constant force the time-integral collapses to a product: $J=F\Delta t=6\times 3=18\text{N⋅s}.$ And $\text{N⋅s}=\text{kg⋅m/s}$ — the same units as momentum, because impulse is a change in momentum.

Step 1 — net force. Choose right = positive. Add the forces with signs: $F_{\text{net}}=+18-6=+12\text{N (rightward)}.$ Why signs? Forces are vectors; along one line a sign is the whole direction. Step 2 — divide by mass: $a=\frac{F_{\text{net}}}{m}=\frac{12}{3}=4{\text{m/s}}^2\text{right}.$

Step 1 — assign signs. Let the incoming direction be positive: $v_i=+15$, $v_f=-12$ (it now travels the other way). Step 2 — change in momentum: $\Delta p=m(v_f-v_i)=0.20(-12-15)=0.20\times (-27)=-5.4\text{kg⋅m/s}.$ Step 3 — average force from $J=F\Delta t=\Delta p$: $F=\frac{\Delta p}{\Delta t}=\frac{-5.4}{0.05}=-108\text{N}.$ The minus sign means the force points opposite to the ball's original motion (away from the wall). Magnitude: $108\text{N}$.

Step 1 — add the forces as vectors. They are perpendicular, so they form the two legs of a right triangle (see figure). The net force is the hypotenuse: $|{\vec{F}}_{\text{net}}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{N}.$ Why Pythagoras? Perpendicular vectors add tip-to-tail into a right triangle; the length of the diagonal is $\sqrt{a^2+b^2}$. Step 2 — divide by mass: $a=\frac{|{\vec{F}}_{\text{net}}|}{m}=\frac{10}{2}=5{\text{m/s}}^2,$ pointing along the same $6\text{-}8\text{-}10$ diagonal (about $53°$ north of east).

The momentum change is fixed — it does not care how you stop: $\Delta p=m(v_f-v_i)=60(0-20)=-1200\text{kg⋅m/s},|\Delta p|=1200.$ (a) $F=\frac{1200}{0.10}=12,000\text{N}.$ (b) $F=\frac{1200}{0.50}=2,400\text{N}.$ Ratio: $12000/2400=5$. Stretching the time by $5\times$ shrinks the force by $5\times$. That's the whole point of crumple zones.

WHY area? Impulse is $\int Fdt$ — the accumulated force over time — which is exactly the area under the $F$-vs-$t$ curve. The graph is a triangle with base $4\text{s}$ and height $8\text{N}$ (see figure): $J=\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(4)(8)=16\text{N⋅s}.$ Final speed from $J=\Delta p=m{v}_f-m{v}_i$ with $v_i=0$: $v_f=\frac{J}{m}=\frac{16}{2}=8\text{m/s}.$

Phase 1 (0–3 s): impulse $J_1=F\Delta t=10\times 3=30\text{N⋅s}$, so $v_3=\frac{J_1}{m}=\frac{30}{4}=7.5\text{m/s}.$ Phase 2 (3–5 s): no force ⇒ no impulse ⇒ momentum unchanged. By Newton's First Law (Inertia) the object coasts at constant velocity. $v_5=7.5\text{m/s}.$

WHY not $F=ma$? The mass changes, so we must use the momentum form. Horizontally there is no external force, so horizontal momentum is conserved (this is Conservation of Linear Momentum): $m_i{v}_i=m_f{v}_f.$ $200\times 6=(200+40)v_f\Rightarrow 1200=240v_f\Rightarrow v_f=5\text{m/s}.$ The car slows down even though nothing pushed it backward — the added mass shares the same fixed momentum. A naive "$a=0\Rightarrow v$ constant" would give $6\text{m/s}$ — wrong by the product-rule term $v\frac{dm}{dt}$.

Target impulse: $\Delta p=m\Delta v=1.5\times 12=18\text{N⋅s}$ (from rest). Stage 1 impulse: $J_1=9\times 1=9\text{N⋅s}$. Remaining impulse for stage 2: $J_2=18-9=9\text{N⋅s}$. Stage 2 time: $t_2=\frac{J_2}{F_2}=\frac{9}{3}=3\text{s}.$ Notice: same $9\text{N⋅s}$ each stage, but the weaker force needs $3\times$ longer — "small force, long time" buys the same momentum as "big force, short time."

The gases push the bullet forward and the bullet-gas pair push the rifle back with an equal-and-opposite force (Newton's Third Law), for the same contact time — so the impulses are equal and opposite, and total momentum stays zero: $0=m_b{v}_b+m_r{v}_r.$ $0=(0.01)(400)+(2)v_r\Rightarrow v_r=-\frac{4}{2}=-2\text{m/s}.$ The minus sign = backward. The rifle recoils at $2\text{m/s}$.

WHY vectors matter: the speed is unchanged ($10\to 10$), yet the momentum vector rotated $90°$, so $\Delta \vec{p}\ne 0$. Work component by component. Let east = $+x$, north = $+y$. ${\vec{p}}_i=(m\cdot 10,0)=(3,0)\text{kg⋅m/s},{\vec{p}}_f=(0,m\cdot 10)=(0,3).$ $\Delta \vec{p}={\vec{p}}_f-{\vec{p}}_i=(0-3,3-0)=(-3,+3).$ Magnitude: $|\vec{J}|=|\Delta \vec{p}|=\sqrt{(-3{)}^2+3^2}=\sqrt{18}\approx 4.24\text{N⋅s}.$ The impulse points north-west (up-and-to-the-left in the figure) — not along either the incoming or outgoing path.

(a) Impulse = integral of force over time. Because $F$ is not constant we must integrate (the area under $F=12t$, a triangle of height $24$ at $t=2$): $J={\int }_0^{2}12tdt={\left[6t^2\right]}_0^2=6(4)-0=24\text{N⋅s}.$ (b) $\Delta v=\frac{J}{m}=\frac{24}{3}=8\text{m/s}.$ (c) Average force = impulse divided by the time it acted: $\bar{F}=\frac{J}{\Delta t}=\frac{24}{2}=12\text{N}.$ Sanity check: $F$ ran from $0$ to $24\text{N}$ linearly, so its average is $\frac{0+24}{2}=12\text{N}$. ✓

(a) Work–energy (Work-Energy Theorem): the space-integral of force equals the change in kinetic energy. $W=Fd=5\times 4=20\text{J}=\frac{1}{2}m{v}_f^2-0.$ $v_f=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\times 20}{2}}=\sqrt{20}\approx 4.47\text{m/s}.$ (b) Impulse = change in momentum: $J=m{v}_f-0=2\times 4.472\approx 8.94\text{N⋅s}.$ Contact time from $J=F\Delta t$: $\Delta t=\frac{J}{F}=\frac{8.944}{5}\approx 1.79\text{s}.$ The lesson: impulse integrates force over time; work integrates force over distance. Same force, two different accumulations — one gives momentum, the other gives energy.