Advanced Mechanics — Grade 11–12

A self-study guide for a harder mechanics problem set: full theory, methods, and step-by-step solutions to all 20 problems.

Written in English; Lithuanian physics terms are given in parentheses, e.g. momentum (judesio kiekis / impulsas), so the topic is always clear.

Theory + methods 20 worked solutions Equations rendered with MathJax Printable

How to use this guide

Each of the 20 problems is grouped under one of eight theory topics. Read the relevant theory first, attempt the problem on your own, then open the solution to check your method and answer. Click Expand all to reveal every solution, or print the page (solutions auto-expand for the PDF).

Convention. We take the free-fall acceleration (laisvojo kritimo pagreitis) $g = 10\ \mathrm{m/s^2}$ unless noted, and Earth's radius $R \approx 6.4\times10^{6}\ \mathrm{m}$. Where $g=9.8$ changes a number noticeably, both are given.

1. Work, energy & centre of mass (darbas, energija, masės centras)

Work (darbas): $A=\int\vec F\cdot d\vec s$; for a constant force $A=Fs\cos\alpha$.
Kinetic energy (kinetinė energija): $E_k=\tfrac12 mv^2$; work–energy theorem $A_{\text{net}}=\Delta E_k$.
Gravitational PE (potencinė energija): $E_p=mgh$, with $h$ = height of the centre of mass.
Energy conservation: $\tfrac12 mv^2+mgh=\text{const}$ when only gravity (and other conservative forces) do work.

Energy methods are usually the fastest route when you only care about speeds and positions, not the forces or the time taken. They fail the moment a non-conservative force (friction, a sudden impact) removes mechanical energy — then track that loss separately.

Why PE uses the centre of mass

The gravitational PE of an extended body is $\sum m_i g h_i = g\sum m_i h_i = Mg\,h_{\text{CM}}$, because the centre of mass (masės centras) is defined by $h_{\text{CM}}=\dfrac{\sum m_i h_i}{\sum m_i}$. So a rigid body behaves, for gravity, as if all its mass sat at the CM. For a uniform body the CM is at its geometric centre; for composite shapes, split into parts and combine: $x_{\text{CM}}=\dfrac{\sum m_i x_i}{M}$.

Tipping a body over an edge (kūno vertimas per briauną)

To tip a block you rotate it about a bottom edge. The CM then moves on a circular arc of radius $d$ = (edge→CM distance). Its height is highest exactly when the CM passes directly above the edge, where it equals $d$. Past that point the block is beyond its balance point and falls into place on its own. Therefore, if that peak height exceeds the final resting height, the work you must supply is only

$A_{\min}=mg\,(d-h_{\text{start}}),\qquad d=\text{edge→CM distance}.$

Intuition: you are only paying to lift the CM to the unstable equilibrium on top of the edge; gravity does the rest.

Centre-of-mass conservation (masės centro tvermė)

From $\vec F_{\text{ext}}=M\vec a_{\text{CM}}$: if there is no external horizontal force, the CM does not accelerate, and if it starts at rest it stays put horizontally: $\sum m_i\,\Delta x_i=0$. This single statement solves "boat", "recoil" and "sliding wedge" problems without any forces.

Pitfalls: measuring the CM distance to the face-centre instead of the edge; forgetting that the block free-falls after the peak (so you don't pay for the full final height); using $mgh$ with $h$ at the wrong reference point.

Used in: problems 1, 5, 9, 10.

2. Momentum & impulse (judesio kiekis ir jėgos impulsas)

Momentum (judesio kiekis / impulsas): $\vec p=m\vec v$ (a vector).
Newton's 2nd law, general form: $\vec F=\dfrac{d\vec p}{dt}$.
Impulse of a force (jėgos impulsas): $\vec J=\displaystyle\int \vec F\,dt=\Delta\vec p$.

Impulse is the tool for large, brief forces (impacts, kicks, bounces): you rarely know how the force varies in time, but you do know the momentum before and after, and $\vec J=\Delta\vec p$ connects them. If the contact time $\Delta t$ is known, the average force is $\bar F=J/\Delta t$.

Elastic bounce off a wall

Resolve the velocity into a part parallel to the wall and a part perpendicular. A smooth elastic wall leaves the parallel part unchanged and reverses the perpendicular part. If the velocity makes angle $\alpha$ with the wall surface, the perpendicular component is $v\sin\alpha$, so the change in momentum — equal and opposite to the impulse the wall receives — is

$J=2mv\sin\alpha\quad(\perp\text{ to the wall}).$

Pitfall: mixing up the angle. Measured from the wall you get $\sin\alpha$; measured from the normal you would get $\cos\alpha$. Read the figure carefully.

Used in: problems 2, 6, 14.

3. Collisions & explosions (smūgiai ir sprogimai)

Conservation of momentum (judesio kiekio tvermės dėsnis): in a closed system $\sum\vec p=\text{const}$, applied independently per axis ($x$, $y$). During an impact internal forces are equal-and-opposite, so they cancel in the total — momentum is conserved in every collision, elastic or not.

The three cases

Elastic (tamprusis smūgis) — momentum and kinetic energy conserved. Solving both for a 1-D hit on a stationary target gives $$v_1'=\frac{m_1-m_2}{m_1+m_2}v,\qquad v_2'=\frac{2m_1}{m_1+m_2}v.$$ Special cases worth memorising: equal masses exchange velocities; a very heavy target makes $v_1'\to-v$ (bounce back); a very light target gets $v_2'\to 2v$.
Perfectly inelastic (visiškai netamprusis) — bodies stick: $v'=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$. The lost kinetic energy is $\Delta E=\tfrac12\dfrac{m_1m_2}{m_1+m_2}(v_1-v_2)^2$.
Pass-through (bullet pierces a body) — momentum conserved; kinetic energy is not. The deficit $\Delta E=E_{k,\text{before}}-E_{k,\text{after}}$ becomes internal (heat/deformation).

Explosions are inelastic collisions run backwards: if the system is at rest at the instant it bursts (e.g. a shell at the top of its flight, where $v=0$), then $\sum\vec p_{\text{after}}=0$, and the fragment momenta close into a vector polygon.

Rise after impact: a body launched upward or swinging on a string climbs to $h=\dfrac{v'^2}{2g}$.

Pitfall: never apply energy conservation to an inelastic or pass-through collision — mechanical energy is lost there. Use momentum for the collision, then energy for the smooth motion before/after.

Used in: problems 3, 13, 17, 19, 20.

4. Newton's laws & pulleys (Niutono dėsniai, blokai)

For an ideal (massless, frictionless) pulley and an inextensible string, the two connected masses share the same acceleration magnitude and the same tension throughout the string. Write $\sum F=ma$ for each mass along its direction of motion and solve simultaneously.

For masses $m$ and $Nm$ (with $Nm>m$), the heavier side falls:

$a=\dfrac{(N-1)g}{N+1},\qquad T=\dfrac{2Nmg}{N+1}.$

Derivation: $Nmg-T=Nma$ and $T-mg=ma$; add to eliminate $T$.

Key subtlety — work in phases. A constraint can change mid-problem. When the heavy mass lands, the string goes slack: tension vanishes and the light mass becomes a free projectile, coasting up an extra $v^2/2g$ before stopping. Re-draw the forces every time the situation changes.

Used in: problem 4.

5. Variable-mass systems (kintamos masės sistemos)

When a body gains or loses mass, the general (Meshchersky) equation is

$m\dfrac{d\vec v}{dt}=\vec F_{\text{ext}}+\vec u_{\text{rel}}\dfrac{dm}{dt},$

where $\vec u_{\text{rel}}$ is the velocity of the entering/leaving mass relative to the body. The second term is the thrust. A rocket has $\vec u_{\text{rel}}\neq 0$ (gas shot out fast) and so gets thrust; but if mass simply drops away carrying the body's own velocity, then $u_{\text{rel}}=0$ and there is no thrust.

Sand cart (smėlio vežimėlis)

Sand leaking through the floor leaves with the cart's horizontal velocity ($u_{\text{rel}}=0$), so only the applied force acts: $m(t)\,\dfrac{dv}{dt}=F$ with $m(t)=M-\mu t$. Hence

$a(t)=\dfrac{F}{M-\mu t},\qquad v(t)=\dfrac{F}{\mu}\ln\dfrac{M}{M-\mu t}.$

Ropes and chains (lynas, grandinė)

Treat the rope by its linear density $\lambda=m/l$; a hanging length $y$ has mass $\lambda y$ with its CM at half its length. Energy conservation through the CM is usually cleanest; sometimes Newton's law with a variable hanging length is needed.

Pitfall: blindly writing $F=\frac{d(mv)}{dt}=m\dot v+v\dot m$ adds a fake "$v\dot m$" thrust. That extra term is only real when the lost mass leaves at a different velocity than the body. For drop-out sand it is zero.

Used in: problems 14, 15.

6. Circular motion (judėjimas apskritimu)

Centripetal acceleration (įcentrinis pagreitis): $a_c=\dfrac{v^2}{r}$, pointing to the centre. The net radial force supplies it: $F_{\text{net,radial}}=\dfrac{mv^2}{r}$. Forces along the path (tangential) change the speed; radial forces change only the direction.

Centripetal force is not a new force — it is whatever combination of tension, gravity and normal reaction happens to point inward. Always start from a free-body diagram and project onto the radial direction.

Vertical circle / loop

Bottom: $T_{\text{bot}}-mg=\dfrac{mv_{\text{bot}}^2}{r}$. Top: $T_{\text{top}}+mg=\dfrac{mv_{\text{top}}^2}{r}$.
The string/track can only push or pull one way, so contact is kept while $T\ge 0$. The critical point is the top: minimum there is $v_{\text{top,min}}=\sqrt{gr}$ (gravity alone supplies the centripetal force).
Energy conservation between top and bottom ($\Delta h=2r$) then gives $v_{\text{bot}}^2=v_{\text{top}}^2+4gr$, so the minimum bottom speed to complete a loop is $\sqrt{5gr}$.

Circle on an inclined plane (nuožulnioji plokštuma)

If the circle lies in an inclined plane of angle $\alpha$, split gravity into a component $g\cos\alpha$ pressing into the plane (giving the normal reaction $N=mg\cos\alpha$ and thus a constant-magnitude friction $f=\mu mg\cos\alpha$) and a component $g\sin\alpha$ acting within the plane — this is the "effective gravity" that plays the role of $g$ for the in-plane circular motion. Height changes in energy balances are $\Delta h=(\text{distance along the plane})\times\sin\alpha$.

Pitfall: using the full $g$ instead of $g\sin\alpha$ for motion in the plane, or forgetting that the radial component of gravity changes around the circle.

Used in: problems 12, 16, 18, 20.

7. Gravitation & orbits (gravitacija, kosminiai greičiai)

$F=\dfrac{GMm}{r^2}$, $E_p=-\dfrac{GMm}{r}$ (zero at infinity), at the surface $g=\dfrac{GM}{R^2}\Rightarrow GM=gR^2$.

Near the surface $E_p=mgh$ is fine, but for large heights you must use $-GMm/r$, because gravity weakens with distance. Total energy $E=\tfrac12 mv^2-\dfrac{GMm}{r}$ is conserved.

Cosmic speeds (kosminiai greičiai)

First cosmic (orbital) speed (pirmasis kosminis greitis) $v_{1k}=\sqrt{gR}\approx 7.9\ \mathrm{km/s}$ — the speed for a circular low orbit, from $\dfrac{mv^2}{R}=mg$.
Escape speed (antrasis kosminis greitis) $v_{2k}=\sqrt{2gR}=\sqrt2\,v_{1k}\approx 11.2\ \mathrm{km/s}$ — set total energy to zero: just barely reaches infinity and never returns.

For a vertically launched rocket, energy conservation $\tfrac12 v^2=\dfrac{GM\,h}{R(R+h)}$ rearranges to

$h=\dfrac{v^2R}{2gR-v^2}.$

Notice it blows up exactly when $v^2\to 2gR$ — the escape speed.

Orbits

Angular momentum (judesio kiekio momentas) is conserved, so at the nearest and farthest points (where $\vec v\perp\vec r$) $v_1R_1=v_2R_2$. Kepler's third law ties the period to the size: $T^2=\dfrac{4\pi^2 a^3}{GM}$ with semi-major axis $a=\dfrac{R_1+R_2}{2}$.

Pitfall: using $mgh$ for satellite-scale heights, or forgetting that orbital speed is largest at perigee and smallest at apogee.

Used in: problems 7, 11.

8. Potential energy & equilibrium (potencinė energija ir pusiausvyra)

Force from a potential: $F=-\dfrac{dU}{dr}$. Equilibrium (pusiausvyra) where $F=0$, i.e. $\dfrac{dU}{dr}=0$. It is stable if $\dfrac{d^2U}{dr^2}>0$ (a minimum of $U$) and unstable if $<0$ (a maximum).

Physically, a particle sits where the net force vanishes; it is stable if any small displacement produces a restoring force pushing it back — that is exactly the condition that $U$ curves upward (a valley). Method: differentiate $U$, set to zero to find $r_0$, then check the sign of the second derivative.

Pitfall: confusing "force is zero" (equilibrium) with "energy is zero". Equilibrium is about the slope of $U$, not its value.

Used in: problem 8.

Problems 1–20 — full solutions (uždavinių sprendimai)

Try each problem first, then open its solution. ★ = harder.

1Standing a brick upright — minimum workWork · centre of mass · tipping (darbas, masės centras)★★

A uniform brick (density $\rho$, dimensions $a\times a\times 2a$) lies on its side. Find the minimum work needed to stand it upright.

Fig. 1 — Tipping the brick about its bottom edge

Mass $M=\rho\,(a\cdot a\cdot 2a)=2\rho a^3$. Lying down, the cross-section in the tipping plane is a rectangle $2a$ (wide) $\times\,a$ (tall); the CM is at height $h_{\text{start}}=a/2$.

Tipping about the bottom edge, the CM rises to its peak when directly above the edge, at distance

$$d=\sqrt{a^2+\left(\tfrac a2\right)^2}=\frac{a\sqrt5}{2}\approx1.118\,a.$$

The final upright CM height is $a$ (half of $2a$). Since $1.118a>a$, the brick topples into place by itself after the peak, so the work is only needed up to the peak:

$$A_{\min}=Mg\left(\frac{a\sqrt5}{2}-\frac a2\right)=2\rho a^3 g\cdot\frac a2(\sqrt5-1)=\rho g a^4(\sqrt5-1).$$

Answer: $A_{\min}=\rho g a^4(\sqrt5-1)\approx1.24\,\rho g a^4.$

2Nitrogen molecule striking a wallImpulse · elastic bounce (jėgos impulsas)★

An $\mathrm{N_2}$ molecule ($v=600$ m/s) bounces elastically off a wall; the angle between its path and the wall is $\alpha=25^\circ$. Find the impulse delivered to the wall.

Fig. 2 — Elastic bounce; angle measured from the wall

Molecule mass: $m=\dfrac{0.028}{6.02\times10^{23}}\approx4.65\times10^{-26}\ \mathrm{kg}$.

Only the component perpendicular to the wall, $v\sin\alpha$, reverses; the parallel one is unchanged. The momentum change (and the impulse on the wall) is

$$J=2mv\sin\alpha=2\cdot4.65\times10^{-26}\cdot600\cdot\sin25^\circ\approx2.36\times10^{-23}\ \mathrm{kg\cdot m/s}.$$

Answer: $J\approx2.4\times10^{-23}\ \mathrm{N\cdot s}$, directed perpendicular to the wall.

3Shell exploding into three fragmentsVector momentum conservation (impulso tvermė)★★

At the top of its flight a shell explodes. $m_1=9$ kg, $v_1=60$ m/s and $m_2=18$ kg, $v_2=40$ m/s fly off horizontally at right angles; $v_3=300$ m/s. Find the third fragment's mass and direction.

Fig. 3 — Momentum triangle of the three fragments

At the highest point the velocity is zero, so $\sum\vec p=0$. The two momenta are perpendicular:

$$p_1=m_1v_1=540,\quad p_2=m_2v_2=720\ \mathrm{(kg\,m/s)},\quad |\vec p_1+\vec p_2|=\sqrt{540^2+720^2}=900.$$

The third fragment must carry $\vec p_3=-(\vec p_1+\vec p_2)$, so $p_3=900$ and

$$m_3=\frac{p_3}{v_3}=\frac{900}{300}=3\ \mathrm{kg}.$$

Direction: horizontal, opposite to the resultant of the first two, at $\tan\theta=720/540\Rightarrow\theta\approx53^\circ$ from the $\vec v_1$ line.

Answer: $m_3=3$ kg, flying horizontally opposite to the combined momentum of the first two (about $53^\circ$ from $\vec v_1$).

4Pulley with masses $m$ and $nm$Newton's laws · free rise (Niutono dėsniai)★★

A string over a light pulley carries $m$ (on the floor) and $nm$ (at height $h$). To what maximum height does $m$ rise?

Fig. 4 — Atwood-type pulley; m rises, nm falls

Phase 1 (string taut), acceleration $a=\dfrac{(n-1)g}{n+1}$. When $nm$ falls $h$, mass $m$ rises $h$ and reaches

$$v^2=2ah=\frac{2(n-1)gh}{n+1}.$$

Phase 2: $nm$ hits the floor, the string slackens, and $m$ rises freely by $\dfrac{v^2}{2g}=\dfrac{(n-1)h}{n+1}$. Total:

$$H=h+\frac{(n-1)h}{n+1}=\frac{2nh}{n+1}.$$

Answer: $H_{\max}=\dfrac{2nh}{n+1}.$

5Lifting a leaking bucketWork · variable mass (darbas, kintama masė)★★

A bucket is raised at constant speed from depth $H=20$ m; water leaks out at a constant rate. Full at the bottom ($V=15$ L), $z=\tfrac23$ of the water remains at the top. Bucket mass $m=2$ kg. Find the work.

Fig. 5 — Bucket rising while water leaks back

Initial water $15$ kg, final $\tfrac23\cdot15=10$ kg, lost $5$ kg. With constant speed and constant leak rate, the water mass falls linearly with height: $m_w(y)=15-5\,y/H$. Total lifted mass $=17-5y/H$.

$$A=g\!\int_0^H\!\Big(17-5\tfrac yH\Big)dy=g\big(17H-\tfrac52H\big)=14.5\,gH=14.5\cdot10\cdot20=2900\ \mathrm{J}.$$

(Equivalently, average load $=\tfrac{15+10}{2}+2=14.5$ kg.)

Answer: $A=14.5\,gH\approx2.9\ \mathrm{kJ}$ (≈2.84 kJ with $g=9.8$).

6Recoil of a skater on iceMomentum + friction (impulso tvermė, trintis)★

A person $M=70$ kg throws a stone $m=3$ kg horizontally at $v=8$ m/s. How far do they slide if $\mu=0.02$?

Fig. 6 — Throw and recoil on ice

Momentum: $V=\dfrac{mv}{M}=\dfrac{3\cdot8}{70}=0.343$ m/s. Friction then stops them over

$$d=\frac{V^2}{2\mu g}=\frac{0.343^2}{2\cdot0.02\cdot10}\approx0.29\ \mathrm{m}.$$

Answer: $d\approx0.29$ m (≈0.3 m).

7Rocket height & cosmic speedsGravitation · escape velocity (kosminiai greičiai)★★★

To what height $h$ does a vertically launched rocket rise with initial speed $v$? Evaluate $h_1$ for $v=v_{1k}$ (first cosmic speed) and $h_2$ for $v=\sqrt2\,v_{1k}$.

Fig. 7 — Rocket rising height h above radius R

Energy conservation with variable gravity, using $GM=gR^2$:

$$\tfrac12 v^2=\frac{gR^2 h}{R(R+h)}\ \Rightarrow\ h=\frac{v^2R}{2gR-v^2}.$$

If $v=v_{1k}=\sqrt{gR}$ ($v^2=gR$): $h_1=\dfrac{gR\cdot R}{2gR-gR}=R\approx6400$ km.

If $v=\sqrt2\,v_{1k}=\sqrt{2gR}$: the denominator $2gR-v^2=0$, so $h_2\to\infty$.

Answer: $h=\dfrac{v^2R}{2gR-v^2}$; $\ h_1=R$ (≈6400 km); $\ h_2\to\infty$. This is the escape (second cosmic) speed — the rocket leaves Earth's gravity and never returns.

8Equilibrium distance from $U(r)$Potential energy · equilibrium (pusiausvyra)★

$U(r)=\dfrac{a}{r^2}-\dfrac br$ with $a,b>0$. At what distance is the particle in (stable) equilibrium?

Fig. 8 — Potential U(r) with its minimum at r₀

Set $\dfrac{dU}{dr}=-\dfrac{2a}{r^3}+\dfrac{b}{r^2}=0\Rightarrow r_0=\dfrac{2a}{b}$.

Check: $\dfrac{d^2U}{dr^2}\big|_{r_0}=\dfrac{b}{r_0^3}>0$ — a minimum, so equilibrium is stable.

Answer: $r_0=\dfrac{2a}{b}$ (stable).

9Boat shift when two people swapCentre-of-mass conservation (masės centro tvermė)★★

A boat (length $L$, mass $M$) carries people $m_1$ and $m_2$ at its ends. Which way and how far does the boat move when they swap places?

Fig. 9 — Two people swapping ends of the boat

No external horizontal force ⇒ the CM stays fixed. Let the boat move $D$ (positive = right). Person 1 (left→right) moves $L+D$ over the ground; person 2 moves $-L+D$:

$$m_1(L+D)+m_2(-L+D)+MD=0\ \Rightarrow\ D=\frac{L(m_2-m_1)}{M+m_1+m_2}.$$

Answer: the boat shifts by $\dfrac{L\,|m_1-m_2|}{M+m_1+m_2}$ toward the side the lighter person moves to (opposite the heavier person's motion). If the masses are equal, it does not move.

10Two sliding wedges and a blockMomentum + energy · moving support★★★

Two identical smooth wedges (mass $M$ each) sit on a frictionless floor. A block $m$ is released from height $h$ on the first. (a) How high does it rise on the second? (b) The velocities of all three bodies at that instant?

Fig. 10 — Block sliding between two free wedges

Sliding off wedge 1. Momentum and energy conservation give the block's speed at the bottom and wedge 1's recoil speed:

$$v_m=\sqrt{\frac{2ghM}{M+m}},\qquad V_1=\frac mM v_m\ \text{(backward)}.$$

(a) Rise on wedge 2. At maximum height the block and wedge 2 move together at $u=\dfrac{mv_m}{m+M}$. Energy conservation gives

$$h'=\frac{Mv_m^2}{2(m+M)g}=h\left(\frac{M}{M+m}\right)^2.$$

(a) $h'=h\,\dfrac{M^2}{(M+m)^2}.$

(b) Velocities at that instant:

Wedge 1 (already separated) keeps moving backward at $V_1=\dfrac mM v_m=m\sqrt{\dfrac{2gh}{M(M+m)}}.$
Block $m$ and wedge 2 move together forward at $u=\dfrac{mv_m}{m+M}$, with $v_m=\sqrt{\dfrac{2ghM}{M+m}}.$

(b) Wedge 1: backward at $V_1=\tfrac mM v_m$; block + wedge 2: forward together at $u=\tfrac{mv_m}{m+M}$.

11Two satellites in the same orbitOrbits · angular momentum · Kepler★★★

Two satellites share one orbit; their separation varies between $L_1$ (at distance $R_1$ from Earth's centre) and $L_2$ (at $R_2$). The period is $T$. Find $R_1$ and $R_2$.

Fig. 11 — Elliptical orbit: apogee R₁, perigee R₂

The satellites follow the same ellipse a fixed time $\Delta t$ apart, so their separation is $L\approx v\,\Delta t$ — proportional to the local speed. Hence $\dfrac{v_1}{v_2}=\dfrac{L_1}{L_2}$ (here $v_1$ is the apogee speed at $R_1$).

Angular-momentum conservation $v_1R_1=v_2R_2$ gives $\dfrac{R_1}{R_2}=\dfrac{v_2}{v_1}=\dfrac{L_2}{L_1}$. Kepler's third law gives the semi-major axis $a=\dfrac{R_1+R_2}{2}=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}$. Combining:

$$R_1=\frac{2L_2}{L_1+L_2}\sqrt[3]{\frac{GMT^2}{4\pi^2}},\qquad R_2=\frac{2L_1}{L_1+L_2}\sqrt[3]{\frac{GMT^2}{4\pi^2}}.$$

Answer: max $R_1=\dfrac{2L_2}{L_1+L_2}\sqrt[3]{\tfrac{GMT^2}{4\pi^2}}$, min $R_2=\dfrac{2L_1}{L_1+L_2}\sqrt[3]{\tfrac{GMT^2}{4\pi^2}}$ (with $GM=gR_0^2$ for Earth).

12Body on a string around a poleCircular motion · projectile (apskritimas)★★★

A string (length $l$) is tied to the top $A$ of a vertical pole, with a body at the other end. Released from horizontal, the string breaks when its tension is $k$ times the body's weight, and the body lands at the pole's base $B$. Find the pole's height.

Fig. 12 — String breaks; body becomes a projectile

At angle $\theta$ below horizontal, $v^2=2gl\sin\theta$ and the tension is

$$T=mg\sin\theta+\frac{mv^2}{l}=3mg\sin\theta.$$

It breaks when $T=kmg\Rightarrow\sin\theta=\dfrac k3,\ \cos\theta=\dfrac{\sqrt{9-k^2}}{3}$. At that point (from $A$): $x_0=l\cos\theta$, $y_0=l\sin\theta$, velocity perpendicular to the string with $v_x=-v\sin\theta$ (toward the pole), $v_y=v\cos\theta$ (down). Time to reach the pole ($x=0$): $t=\dfrac{l\cos\theta}{v\sin\theta}$. Then $H=y_0+v\cos\theta\,t+\tfrac12 gt^2$; simplifying (using $\tfrac k3+\tfrac{9-k^2}{3k}=\tfrac3k$):

$$H=\frac{9l\,(k^2+3)}{4k^3}.$$

Check: $k=3$ (string breaks at the bottom) gives $H=l$. ✓

Answer: $H=\dfrac{9l(k^2+3)}{4k^3}$ (valid for $0

13Foam cube pierced by a bulletMomentum (pass-through) · rise★★

A foam cube $M=100$ g (height $a=10$ cm) sits on a surface. A bullet $m=10$ g pierces it vertically from below, entering at $v_1=100$ m/s and leaving at $v_2=95$ m/s. Does the cube jump? How high?

Fig. 13 — Bullet passing up through the foam cube

The bullet gives the cube an upward impulse (gravity's impulse during the brief passage is negligible):

$$J=m(v_1-v_2)=0.01\cdot5=0.05\ \mathrm{kg\,m/s}\ \Rightarrow\ V=\frac JM=0.5\ \mathrm{m/s}.$$

The average force ($\sim J/\Delta t$ with $\Delta t\approx a/\bar v\approx10^{-3}$ s, i.e. $\sim50$ N) hugely exceeds the cube's weight ($\sim1$ N), so it leaves the surface. Rise:

$$h=\frac{V^2}{2g}=\frac{0.25}{20}=0.0125\ \mathrm{m}.$$

Answer: yes, it jumps; $h\approx1.25$ cm (≈1.3 cm with $g=9.8$).

14Cart leaking sand under a forceVariable mass (kintamos masės sistema)★★★

Sand leaks through the floor of a cart pushed by a horizontal force $\vec F$; the rate is $\Delta m$ per unit time. At $t=0$ the cart is at rest with mass $M$. Find $a(t)$ and $v(t)$.

Fig. 14 — Cart pushed by F while sand leaks out

The leaking sand carries the cart's horizontal velocity, so there is no thrust. With $m(t)=M-\Delta m\,t$:

$$a(t)=\frac{F}{M-\Delta m\,t},\qquad v(t)=\int_0^t\frac{F\,dt'}{M-\Delta m\,t'}=\frac{F}{\Delta m}\ln\frac{M}{M-\Delta m\,t}.$$

Answer: $a(t)=\dfrac{F}{M-\Delta m\,t}$, $\ v(t)=\dfrac{F}{\Delta m}\ln\dfrac{M}{M-\Delta m\,t}$ (while sand remains).

15Rope sliding off a fenceChain energy · free fall (lynas, grandinė)★★★

A rope (length $l

Fig. 15 — Rope sliding over a frictionless fence

Speed as it leaves. By CM energy conservation: initially the CM is at $h-l/4$; with all the rope on one side the CM is at $h-l/2$, so it drops $l/4$:

$$\tfrac12 v^2=g\frac l4\ \Rightarrow\ v=\sqrt{\frac{gl}{2}}.$$

Free-fall phase. At that instant the top end is at height $h$ moving down at $v$, and the whole rope falls rigidly. "All landed" = the top end reaches the ground: $h=vt+\tfrac12 gt^2$, giving

$$t=\frac{-v+\sqrt{v^2+2gh}}{g}=\frac{\sqrt{4h+l}-\sqrt{l}}{\sqrt{2g}}.$$

Answer: $t=\dfrac{\sqrt{4h+l}-\sqrt{l}}{\sqrt{2g}}.$

16Circular motion on an inclined plane (a–c)Inclined circle · tension · friction★★★

Incline $\alpha=30^\circ$; a cord $l=20$ cm holds a body moving in a circle in the plane. At the highest point $v=3$ m/s. Find (a) $v_0$ if the top tension is half the bottom tension; (b) $\mu$; (c) the path length $x$ until it stops if, after $5/4$ turns, the cord breaks.

Fig. 16 — Circle in the inclined plane (g·sinα in-plane)

In the plane the effective gravity along it is $g\sin\alpha$; the normal reaction $N=mg\cos\alpha$ gives constant friction $f=\mu mg\cos\alpha$.

(a) Bottom $T_{\text{bot}}=mg\sin\alpha+\dfrac{mv_0^2}{l}$, top $T_{\text{top}}=\dfrac{mv^2}{l}-mg\sin\alpha$. With $T_{\text{bot}}=2T_{\text{top}}$:

$$v_0^2=2v^2-3gl\sin\alpha=18-3=15\ \Rightarrow\ v_0=\sqrt{15}\approx3.9\ \mathrm{m/s}.$$

(a) $v_0=\sqrt{15}\approx3.9$ m/s.

(b) Energy bottom→top (path = half-circle $\pi l$, height $2l\sin\alpha$):

$$\frac{v_0^2-v^2}{2}=2gl\sin\alpha+\mu g\cos\alpha\,\pi l\ \Rightarrow\ 3=2+\mu\cdot5.44\ \Rightarrow\ \mu\approx0.18.$$

(b) $\mu\approx0.18$ (≈0.20 with $g=9.8$).

(c) After $5/4$ turns the body is at a side point (a distance $u=l$ up the plane) moving up the slope — that is the only way it can stop, since $\mu<\tan\alpha$. Circular path so far: $s_1=\tfrac52\pi l$. Its speed there, from energy:

$$\tfrac12 v_{\text{br}}^2=\tfrac12 v_0^2-g\sin\alpha\,l-\mu g\cos\alpha\,\tfrac52\pi l=7.5-1-2.5=4\ \Rightarrow\ v_{\text{br}}^2=8.$$

After the break it slides up, braked by gravity and friction, a distance $d=\dfrac{v_{\text{br}}^2}{2g(\sin\alpha+\mu\cos\alpha)}\approx0.61$ m. Total path:

$$x=\tfrac52\pi l+d=1.57+0.61\approx2.18\ \mathrm{m}.$$

(c) $x\approx2.2$ m.

17Two balls — elastic collisionElastic collision · rise (tamprusis smūgis)★★

Balls $m_1=0.3$ kg and $m_2=0.2$ kg hang on equal strings. The first is raised to $H=7$ cm and released. To what heights $h_1,h_2$ do they rise after the elastic head-on collision?

Fig. 17 — Two balls on strings before the elastic hit

Before impact $v=\sqrt{2gH}$. Elastic-collision velocities:

$$v_1'=\frac{m_1-m_2}{m_1+m_2}v=0.2v,\qquad v_2'=\frac{2m_1}{m_1+m_2}v=1.2v.$$

Heights $h=(v'/v)^2H$:

$$h_1=0.2^2\cdot7=0.28\ \mathrm{cm},\qquad h_2=1.2^2\cdot7=10.08\ \mathrm{cm}.$$

Answer: $h_1=0.28$ cm, $h_2=10.08$ cm.

18Track with a half-loopLoop condition · projectile (kilpa)★★★

A frictionless track = a straight section + a half death-loop of radius $r$. An object given some initial speed leaves the top of the loop and lands back at the start. Find the required speed $v$ and the minimum straight length $x$.

Fig. 18 — Straight section + half-loop, then projectile

At the top (height $2r$) the object leaves horizontally, heading back toward the start, at $v_{\text{top}}$. To keep contact in the loop, $v_{\text{top}}^2\ge gr$; for the minimum $x$ take $v_{\text{top}}=\sqrt{gr}$.

Falling from $2r$: $t=2\sqrt{r/g}$. The horizontal distance back to the start equals $x$:

$$x=v_{\text{top}}\,t=\sqrt{gr}\cdot2\sqrt{\tfrac rg}=2r.$$

Energy conservation start→top: $v^2=v_{\text{top}}^2+4gr=5gr$, so $v=\sqrt{5gr}$.

Answer: $v=\sqrt{5gr}$, minimum straight length $x=2r$.

19Bullet through a ball on a stand (a, b)Momentum · projectile · energy loss★★★

A ball $M=200$ g rests on a stand at $h=5$ m. A bullet $m=10$ g, $v=500$ m/s, passes horizontally through its centre. (a) Where does the bullet land if the ball falls $l=20$ m from the stand? (b) What fraction $\beta$ of the bullet's kinetic energy becomes internal energy?

Fig. 19 — Bullet through a ball: two projectile ranges

Fall time from $h=5$ m: $t=\sqrt{2h/g}=1$ s, so the ball's speed is $V=l/t=20$ m/s.

Momentum: $u=v-\dfrac Mm V=500-20\cdot20=100$ m/s. The bullet also falls for $1$ s, so

$$L=u\,t=100\ \mathrm{m}.$$

(a) $L=100$ m.

(b) Initial KE $=\tfrac12 mv^2=1250$ J; after, $\tfrac12 mu^2+\tfrac12 MV^2=50+40=90$ J, so the loss is $1160$ J:

$$\beta=\frac{1250-90}{1250}=0.928.$$

(b) $\beta\approx0.93$ (≈93 %).

20Hoop hitting a stepAngular-momentum impact · rolling★★★

A hoop (ring) of radius $R$ rolling at $v$ hits a step of height $h$ ($h\ll R$). Find its speed after climbing onto the step, and the minimum $v_{\min}$ to climb it. The impact is inelastic and the hoop does not slip.

Fig. 20 — Hoop pivoting about the step corner P

During the impact the impulsive force acts at the step edge $P$, so angular momentum about $P$ is conserved. For a ring $I_c=MR^2$, $\omega=v/R$.

Before: $L=I_c\omega+Mv(R-h)=Mv(2R-h)$. After, the hoop pivots about $P$ with $I_P=I_c+MR^2=2MR^2$:

$$2MR^2\omega'=Mv(2R-h)\ \Rightarrow\ v'=\omega'R=v\,\frac{2R-h}{2R}=v\Big(1-\frac{h}{2R}\Big).$$

To climb, the CM must rise by $h$: $\tfrac12 I_P\omega'^2\ge Mgh\Rightarrow v'^2\ge gh$. Hence

$$v_{\min}=\frac{2R\sqrt{gh}}{2R-h}\approx\sqrt{gh}\quad(h\ll R).$$

Answer: $v'=v\dfrac{2R-h}{2R}$; $v_{\min}=\dfrac{2R\sqrt{gh}}{2R-h}\approx\sqrt{gh}.$

All numeric answers were checked computationally.