1.P — Worked Problems: Classical Mechanics

Sixteen problems covering every chapter in Part 1. They climb from the kind H. C. Verma asks at the start of a chapter to the kind JEE Advanced asks at the end of a paper, and several of them look like a standard problem and quietly are not. Every solution shows the arithmetic, including the boring parts, and most end with a check. Attempt each one before opening the solution.

Problem 1 — Reading a graph rather than a formula

A car's velocity–time graph is a straight line from (0\ \text{s},\ 0\ \text{m/s}) up to (10\ \text{s},\ 20\ \text{m/s}), then flat at 20 m/s until t = 30 s, then a straight line down to (40\ \text{s},\ 0\ \text{m/s}). Find the acceleration in each phase, the total distance, and the average speed.

Solution

Accelerations are slopes.

Phase 1: a = \dfrac{20-0}{10-0} = 2.0\ \text{m/s}^2.

Phase 2: the line is flat, so a = 0.

Phase 3: a = \dfrac{0-20}{40-30} = -2.0\ \text{m/s}^2. Negative, because the velocity is falling.

Distance is the area under the graph. The shape is a trapezium, but it is safer to take it in three pieces.

Phase 1 is a triangle: \tfrac12\times10\times20 = 100\ \text{m}.

Phase 2 is a rectangle: 20\times20 = 400\ \text{m}.

Phase 3 is a triangle: \tfrac12\times10\times20 = 100\ \text{m}.

s_{\text{total}} = 100+400+100 = 600\ \text{m}

Average speed:

\bar{v} = \frac{600}{40} = 15\ \text{m/s}

What to notice. The average speed is not the average of 0 and 20, and it is not 20 either. Averaging endpoints only works when the acceleration is constant throughout, and here it changes three times. Reading areas off the graph avoids that trap entirely, which is why the graph method is worth keeping even once you know the equations.

Problem 2 — The two-stone problem

A stone is dropped from a 100 m cliff. One second later a second stone is thrown down from the same point at 20\ \text{m/s}. Do they meet before the ground? If so, where?

Solution

Take down as positive, g = 9.81\ \text{m/s}^2, and let t be the time since the first stone was released. The second stone has then been falling for (t-1).

First stone:

s_1 = \tfrac12(9.81)t^2 = 4.905t^2

Second stone, thrown at 20 m/s, in flight for (t-1):

s_2 = 20(t-1) + 4.905(t-1)^2

They meet when s_1 = s_2:

4.905t^2 = 20t-20 + 4.905(t^2-2t+1)

Expand the right side:

4.905t^2 = 20t - 20 + 4.905t^2 - 9.81t + 4.905

The 4.905t^2 cancels on both sides — which is the whole point, because both stones share the same acceleration:

0 = 20t - 9.81t - 20 + 4.905 = 10.19t - 15.095

t = \frac{15.095}{10.19} = 1.481\ \text{s}

Where:

s_1 = 4.905\times(1.481)^2 = 4.905\times2.193 = 10.76\ \text{m}

Check it happens in the air. The first stone reaches the ground when 4.905t^2 = 100, so t = 4.52 s. The meeting at 1.481 s is well before that, and at 10.8 m the stones are still high up. Valid.

What to notice. The quadratic terms cancelling is not luck. Both stones fall under the same g, so the difference in their positions grows only through the difference in their initial velocities — which is linear in time. That is why the meeting time came out of a linear equation rather than a quadratic, and it is worth spotting early because it saves the whole expansion.

Problem 3 — Projectile onto a slope

A ball is thrown at 20\ \text{m/s} at 60° above the horizontal from the foot of a hill that rises at 30°. How far up the slope does it land, measured along the slope?

Solution

This is the standard trap: the landing point is not at y = 0, so the range formula does not apply. Work in ordinary x–y coordinates and impose the slope as a constraint.

Components:

u_x = 20\cos60° = 10.0\ \text{m/s}, \qquad u_y = 20\sin60° = 17.32\ \text{m/s}

Positions:

x = 10t, \qquad y = 17.32t - 4.905t^2

The slope constraint. Any point on the hill satisfies y = x\tan30° = 0.5774x. So:

17.32t - 4.905t^2 = 0.5774\times10t = 5.774t

Divide through by t (discarding the t=0 launch root):

17.32 - 4.905t = 5.774

4.905t = 11.546 \quad\Longrightarrow\quad t = 2.354\ \text{s}

Landing point:

x = 10\times2.354 = 23.54\ \text{m}, \qquad y = 0.5774\times23.54 = 13.59\ \text{m}

Distance along the slope:

L = \sqrt{23.54^2+13.59^2} = \sqrt{554.1+184.7} = \sqrt{738.8} = 27.2\ \text{m}

Equivalently, L = x/\cos30° = 23.54/0.866 = 27.2 m. ✓

What to notice. On flat ground, 60° would give R = 400\sin120°/9.81 = 35.3 m. The uphill slope cuts it to 27.2 m along the slope, and only 23.5 m horizontally. Also worth knowing: the optimum launch angle on a slope is not 45°. For an upslope of angle \phi it is 45° + \phi/2, so here 60° happens to be exactly optimal — which is why this particular problem is set with these particular numbers.

Problem 4 — Two blocks and friction

A 5.0 kg block sits on top of a 10.0 kg block, which rests on a frictionless floor. The coefficient of static friction between the two blocks is 0.40. What is the largest horizontal force that can be applied to the lower block without the upper one sliding off?

Solution

The upper block has only one horizontal force available to accelerate it: friction from the block below. So the maximum acceleration the pair can share is set by how much friction the top block can receive.

Maximum friction on the top block. Its normal force is its own weight:

N = m_{\text{top}}g = 5.0\times9.81 = 49.05\ \text{N}

f_{\max} = \mu_sN = 0.40\times49.05 = 19.62\ \text{N}

Maximum shared acceleration:

a_{\max} = \frac{f_{\max}}{m_{\text{top}}} = \frac{19.62}{5.0} = 3.924\ \text{m/s}^2

Note this equals \mu_sg, which it must — the mass cancels, exactly as in the slope problem of Chapter 1.5.

Force needed to accelerate the whole system at that rate. The floor is frictionless, so the applied force accelerates all 15 kg:

F_{\max} = (m_1+m_2)a_{\max} = 15.0\times3.924 = 58.9\ \text{N}

What to notice. Now try it the other way round: apply the force to the top block instead. The top block can now be pushed directly, but the bottom block can only be dragged along by friction — and the maximum friction available is still 19.62 N, acting on 10.0 kg, so a_{\max} = 1.962\ \text{m/s}^2 and F_{\max} = 15.0\times1.962 = 29.4 N. Half as much. The system is not symmetric, because friction has to move a different mass in each case, and the mass it has to move is what limits you.

Problem 5 — The banked bend with friction

A bend of radius 80 m is banked at 15°. The coefficient of static friction is 0.50. Find the maximum safe speed.

Solution

Both the normal force and friction now have components pointing towards the centre. At maximum speed the car is about to slide up the bank, so friction acts down the slope.

Resolve into horizontal and vertical, with \theta = 15°, \sin\theta = 0.2588, \cos\theta = 0.9659.

Vertical (no acceleration). The normal force has an upward component N\cos\theta; friction, pointing down the slope, has a downward component f\sin\theta:

N\cos\theta - f\sin\theta = mg

Horizontal (centripetal). Both contribute inwards:

N\sin\theta + f\cos\theta = \frac{mv^2}{r}

With f = \mu N at the limit, substitute into both:

N(\cos\theta - \mu\sin\theta) = mg

N(\sin\theta + \mu\cos\theta) = \frac{mv^2}{r}

Divide the second by the first. Both N and m cancel:

\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta} = \frac{v^2}{rg}

v_{\max} = \sqrt{rg\cdot\frac{\tan\theta+\mu}{1-\mu\tan\theta}}

(The second form comes from dividing top and bottom of the fraction by \cos\theta.)

Now the numbers. \tan15° = 0.2679:

\frac{0.2679+0.50}{1-0.50\times0.2679} = \frac{0.7679}{0.8661} = 0.8866

v_{\max} = \sqrt{80\times9.81\times0.8866} = \sqrt{695.8} = 26.4\ \text{m/s} = 95\ \text{km/h}

Compare with the alternatives. Flat road, same friction: v = \sqrt{\mu gr} = \sqrt{0.5\times9.81\times80} = 19.8 m/s = 71 km/h. Banked but frictionless: v = \sqrt{rg\tan\theta} = \sqrt{80\times9.81\times0.2679} = 14.5 m/s = 52 km/h.

What to notice. The bank and the friction cooperate, and the combination beats either alone by a wide margin. Watch the denominator 1-\mu\tan\theta: as the bank gets steeper it shrinks towards zero and v_{\max} runs away to infinity. That is the mathematics of a wall-of-death, where the track is vertical, \tan\theta is infinite, and there is no upper speed limit at all — only a lower one.

Problem 6 — Energy with friction on a slope

A 2.0 kg block is pushed against a spring of stiffness k = 500\ \text{N/m}, compressing it 0.20 m, on a 30° incline with \mu_k = 0.25. It is released. How far up the slope does it travel from the release point?

Solution

Let d be the distance travelled along the slope after leaving the spring, measured from the release point, so the total travel including the spring's push is d and the spring pushes over the first 0.20 m of it.

Energy released by the spring:

E_s = \tfrac12kx^2 = \tfrac12\times500\times0.04 = 10.0\ \text{J}

Energy spent climbing. Rising along the slope by d gains height d\sin30° = 0.5d:

\Delta U = mgd\sin30° = 2.0\times9.81\times0.5d = 9.81d

Energy lost to friction. The normal force is mg\cos30° = 2.0\times9.81\times0.866 = 16.99\ \text{N}:

W_f = \mu_kNd = 0.25\times16.99\times d = 4.248d

Energy balance. The block stops when all 10.0 J has gone into height and heat:

10.0 = 9.81d + 4.248d = 14.06d

d = 0.711\ \text{m}

Does it stay there? Check whether static friction can hold it. Gravity's pull down the slope is mg\sin30° = 9.81 N. Maximum static friction, taking \mu_s \approx \mu_k = 0.25, is 0.25\times16.99 = 4.25 N. That is less than 9.81 N, so no — it slides back down.

What to notice. The friction term appears with a plus sign in the energy balance on the way up and would appear with a plus sign again on the way down, because friction always removes energy regardless of direction. That asymmetry is why the block will not return to the spring with the 10.0 J it started with: on the return trip over 0.711 m it loses another 3.02 J, arriving with 6.98 - 3.02 = 3.96 J of kinetic energy rather than 10.0.

Problem 7 — Rocket thrust from momentum

A rocket burns fuel at 250 kg/s, ejecting it at 2800 m/s relative to the rocket. Its initial mass is 30 000 kg. Find the thrust, and the initial acceleration.

Solution

Thrust is the rate at which momentum is thrown backwards:

F = v_e\frac{dm}{dt} = 2800\times250 = 700\,000\ \text{N} = 700\ \text{kN}

Initial acceleration. The rocket must also fight its own weight:

W = mg = 30\,000\times9.81 = 294\,300\ \text{N}

F_{\text{net}} = 700\,000 - 294\,300 = 405\,700\ \text{N}

a = \frac{405\,700}{30\,000} = 13.5\ \text{m/s}^2

What to notice. The thrust-to-weight ratio here is 700/294 = 2.4, and it must exceed 1 or the rocket cannot leave the pad at all. As fuel burns the mass falls while the thrust stays roughly constant, so the acceleration rises steeply — which is why crewed launches throttle back partway up, to keep the acceleration within what a human body tolerates. Note also that g has stolen 9.81\ \text{m/s}^2 of the available 23.3; this is called gravity loss, and over a full ascent it costs about 1.5 km/s of the delta-v budget in Chapter 11.7.

Problem 8 — Newton's cradle, and why one ball comes out

In a Newton's cradle, one ball is lifted and released and strikes four identical stationary balls. Exactly one ball leaves the far end at the incoming speed. Prove that two balls leaving at half the speed is impossible.

Solution

Let each ball have mass m and the incoming ball have speed v. Assume the collisions are elastic (steel on steel is very close).

Before: p = mv and K = \tfrac12mv^2.

Proposed outcome: two balls leave at v/2.

p = 2m\times\frac{v}{2} = mv \quad\text{✓ momentum is conserved}

K = 2\times\tfrac12m\left(\frac{v}{2}\right)^2 = m\frac{v^2}{4} = \tfrac12mv^2\times\frac{1}{2}

Only half the kinetic energy. So this outcome conserves momentum but destroys half the energy, and for a nearly elastic collision that is forbidden.

Actual outcome: one ball leaves at v.

p = mv \quad\text{✓}\qquad K = \tfrac12mv^2 \quad\text{✓}

Both conserved. This is the only outcome that satisfies both laws.

The general statement. For n balls leaving at speed u, momentum requires nmu = mv, so u = v/n. Energy then gives \tfrac12nm(v/n)^2 = \tfrac12mv^2/n, which equals the original energy only when n = 1.

What to notice. This is the clearest demonstration in mechanics that momentum conservation alone does not determine a collision. It gives one equation; you need the second, and which second equation you use — energy conservation, or a coefficient of restitution, or "they stick together" — is a physical fact about the bodies, not something the mathematics supplies. Drop two balls in and exactly two come out, for the same reason.

Problem 9 — Rotational: the falling rod

A uniform rod of length 1.5 m is held vertically on the ground and released, pivoting about its bottom end without slipping. How fast is its top end moving when it hits the ground?

Solution

Moment of inertia about the end (from the parallel axis theorem or the table):

I = \tfrac13ML^2

Energy conservation. The rod's centre of mass starts at L/2 and ends at ground level, so it falls a height L/2:

Mg\frac{L}{2} = \tfrac12I\omega^2 = \tfrac12\left(\tfrac13ML^2\right)\omega^2 = \tfrac16ML^2\omega^2

The M cancels:

g\frac{L}{2} = \tfrac16L^2\omega^2 \quad\Longrightarrow\quad \omega^2 = \frac{3g}{L}

\omega = \sqrt{\frac{3\times9.81}{1.5}} = \sqrt{19.62} = 4.43\ \text{rad/s}

Speed of the top end, which is at distance L from the pivot:

v = \omega L = 4.43\times1.5 = 6.64\ \text{m/s}

Compare with free fall. A ball dropped from 1.5 m arrives at \sqrt{2\times9.81\times1.5} = 5.42\ \text{m/s}. The rod's tip is moving faster than a freely falling object dropped from the same height — 6.64 against 5.42.

What to notice. Nothing is violated. The tip is being whipped round by the rest of the rod, whose lower portions are falling more slowly and effectively lending their energy to the end. There is a classic demonstration: balance a ball on the end of a hinged board, release the board, and the board's end beats the ball to the ground, leaving the ball hanging in mid-air behind it. The condition for the tip to outrun free fall is that the board's angle to the vertical exceeds about 35°, which is where \tfrac32g\cos\theta first exceeds g.

Problem 10 — Angular momentum: the turntable

A 60 kg person stands at the edge of a stationary merry-go-round of radius 2.0 m and moment of inertia 500\ \text{kg·m}^2, mounted on a frictionless bearing. The person walks around the rim at 1.5\ \text{m/s} relative to the ground. How fast does the merry-go-round rotate?

Solution

No external torque acts about the vertical axis, so total angular momentum stays at zero.

The person's angular momentum:

L_p = m v r = 60\times1.5\times2.0 = 180\ \text{kg·m}^2\text{/s}

The platform must carry the opposite:

L_{\text{plat}} = -180 = I\omega \quad\Longrightarrow\quad \omega = -\frac{180}{500} = -0.36\ \text{rad/s}

The minus sign means it turns the other way. Its rim speed is 0.36\times2.0 = 0.72 m/s, backwards.

What to notice. The person walks forwards and the floor walks backwards, exactly as in Chapter 1.4's third-law account of walking — except that here the "Earth" is a 500 kg·m² platform rather than a 6\times10^{24} kg planet, so the recoil is visible. This is also why a person walking around a small boat sets it spinning, and why astronauts on a spacewalk are taught never to make a large uncompensated arm movement.

Problem 11 — Statics: the hinged sign

A uniform horizontal beam of mass 30 kg and length 3.0 m is hinged to a wall at one end. A cable runs from the far end of the beam up to the wall, making 40° with the beam. A 50 kg sign hangs from the far end. Find the cable tension and the force at the hinge.

Solution

Weights: beam 30\times9.81 = 294.3 N at 1.5 m; sign 50\times9.81 = 490.5 N at 3.0 m.

Torques about the hinge — chosen so the two unknown hinge components vanish.

The cable's vertical component is T\sin40° = 0.6428T, and it acts at 3.0 m. Only that vertical component has a moment arm about the hinge; the horizontal component's line of action passes through the beam and therefore through the hinge.

T\sin40°\times3.0 = 294.3\times1.5 + 490.5\times3.0

1.928T = 441.5 + 1471.5 = 1913

T = \frac{1913}{1.928} = 992\ \text{N}

Hinge forces. Horizontal balance: the cable pulls the beam towards the wall with T\cos40° = 992\times0.766 = 760 N, so the hinge must push outwards with 760 N.

Vertical balance:

H_y + T\sin40° = 294.3+490.5

H_y = 784.8 - 992\times0.6428 = 784.8 - 637.7 = 147\ \text{N upward}

Magnitude and direction of the hinge force:

H = \sqrt{760^2+147^2} = \sqrt{577\,600+21\,600} = 774\ \text{N} \quad\text{at}\quad \tan^{-1}\frac{147}{760} = 11.0°\ \text{above horizontal}

What to notice. The cable tension of 992 N is larger than the total weight of 785 N, and the hinge force is larger still. That is the price of the shallow 40° angle: only 64% of the tension is doing useful vertical work, and the other 77% is wasted pulling horizontally. Make the cable steeper and both forces fall sharply; make it shallower and they run away. This is why a shelf bracket's diagonal is at roughly 45° and never at 10°.

Problem 12 — Buoyancy with two fluids

A block of wood (\rho = 600\ \text{kg/m}^3) floats in water with oil (\rho = 800\ \text{kg/m}^3) poured on top, so that the block is partly in each. If the block is a 20 cm cube, how much of it is in the water?

Solution

Let h be the depth of the block sitting in water, and (0.20-h) the depth in oil. Cross-sectional area A = 0.20\times0.20 = 0.04\ \text{m}^2.

Weight of the block:

W = \rho_{\text{wood}}\times V\times g = 600\times(0.20)^3\times g = 600\times0.008\,g = 4.8g\ \text{N}

Buoyancy is the weight of both displaced fluids added together — each layer contributes according to how much of the block is in it:

F_B = \rho_wAhg + \rho_oA(0.20-h)g = g\times0.04\left[1000h + 800(0.20-h)\right]

Set F_B = W and cancel g:

0.04\left[1000h + 160 - 800h\right] = 4.8

0.04\left[200h+160\right] = 4.8

200h + 160 = 120

200h = -40 \quad\Longrightarrow\quad h = -0.20\ \text{m}

A negative depth is impossible, and the negative result is the answer: the block does not reach the water at all. Check why. Wood at 600 kg/m³ is less dense than oil at 800, so it floats in the oil layer alone, provided the oil is deep enough.

Floating in oil only:

\frac{V_{\text{sub}}}{V} = \frac{600}{800} = 0.75 \quad\Longrightarrow\quad h_{\text{oil}} = 0.75\times0.20 = 0.15\ \text{m}

So the block sits 15 cm into the oil with 5 cm proud, and touches the water only if the oil layer is thinner than 15 cm.

Redo it for a denser block. Take \rho = 900\ \text{kg/m}^3, denser than oil and lighter than water:

W = 900\times0.008g = 7.2g

0.04[200h+160] = 7.2 \quad\Longrightarrow\quad 200h+160 = 180 \quad\Longrightarrow\quad h = 0.10\ \text{m}

Half in the water, half in the oil.

What to notice. An impossible answer is information, not a mistake. A negative depth told us the physical setup we assumed was wrong, and pointed straight at which assumption to drop. Always sanity-check a result against the geometry that produced it.

Problem 13 — Bernoulli and the pitot tube

An aircraft's pitot tube measures a pressure difference of 4.5 kPa between a forward-facing opening and a side-facing one, at an altitude where the air density is 0.90\ \text{kg/m}^3. What is the airspeed?

Solution

The side-facing opening reads the ambient pressure of air flowing past at speed v. The forward-facing opening reads the pressure where the air has been brought to a complete stop, called the stagnation point.

Apply Bernoulli between the two, at the same height so the \rho gy terms cancel:

P_{\text{static}} + \tfrac12\rho v^2 = P_{\text{stagnation}} + 0

\Delta P = \tfrac12\rho v^2

v = \sqrt{\frac{2\Delta P}{\rho}} = \sqrt{\frac{2\times4500}{0.90}} = \sqrt{10\,000} = 100\ \text{m/s}

360 km/h.

What to notice, and it has killed people. A pitot tube measures \tfrac12\rho v^2, which is called dynamic pressure and is what the airframe and wings actually respond to. But the airspeed indicator is calibrated using sea-level density, so at altitude, where \rho is lower, it reads less than the true speed. At 10 km, \rho \approx 0.41\ \text{kg/m}^3, so the same 4.5 kPa corresponds to a true airspeed of \sqrt{2\times4500/0.41} = 148 m/s while the instrument still shows 100 m/s worth. Pilots fly by indicated airspeed deliberately, because it is the number the wing cares about; the true speed matters for navigation and is computed separately. And a pitot tube blocked by ice reports a pressure difference that no longer corresponds to any speed at all — which is what happened to Air France 447 in 2009.

Problem 14 — Gravitation: the geostationary orbit, from scratch

Find the altitude at which a satellite orbits the Earth once every 24 hours, given GM_\oplus = 3.986\times10^{14}\ \text{m}^3/\text{s}^2 and R_\oplus = 6371 km.

Solution

Gravity supplies the centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r} = m\omega^2r

The satellite's mass cancels:

\frac{GM}{r^2} = \omega^2 r \quad\Longrightarrow\quad r^3 = \frac{GM}{\omega^2}

Angular velocity. One rotation in a day. Strictly this must be the sidereal day, 23 h 56 min 4 s = 86 164 s, because the satellite must match the Earth's rotation against the fixed stars rather than against the Sun:

\omega = \frac{2\pi}{86\,164} = 7.2921\times10^{-5}\ \text{rad/s}

r^3 = \frac{3.986\times10^{14}}{(7.2921\times10^{-5})^2} = \frac{3.986\times10^{14}}{5.3175\times10^{-9}} = 7.496\times10^{22}\ \text{m}^3

Take the cube root. Write it as 74.96\times10^{21} so the exponent divides by three cleanly:

r = (74.96)^{1/3}\times10^{7} = 4.2166\times10^{7}\ \text{m} = 42\,166\ \text{km}

Altitude above the surface:

h = 42\,166 - 6371 = 35\,795\ \text{km}

Orbital speed:

v = \omega r = 7.2921\times10^{-5}\times4.2166\times10^{7} = 3075\ \text{m/s} = 3.07\ \text{km/s}

What to notice. Using the 86 400 s solar day instead of the sidereal day shifts the answer by about 80 km — small in percentage terms and enormous in station-keeping terms, since a satellite 80 km off will drift out of its slot and out of every dish pointed at it. Note also the speed: at 3.07 km/s a geostationary satellite is much slower than the ISS at 7.66 km/s, because higher orbits are slower. And there is only one geostationary radius, which makes that ring a finite, allocated, internationally regulated resource.

Problem 15 — Combined: the loop-the-loop with friction

A 0.50 kg ball is released from height h on a track that leads into a vertical circular loop of radius 0.30 m. Friction dissipates 0.40 J between the release point and the top of the loop. Find the minimum h for the ball to complete the loop. Treat the ball as sliding, not rolling.

Solution

Condition at the top. The ball only just makes it when the track pushes with zero force and gravity alone supplies the centripetal force:

mg = \frac{mv_{\text{top}}^2}{r} \quad\Longrightarrow\quad v_{\text{top}}^2 = gr = 9.81\times0.30 = 2.943\ \text{m}^2/\text{s}^2

Kinetic energy needed at the top:

K_{\text{top}} = \tfrac12mv_{\text{top}}^2 = \tfrac12\times0.50\times2.943 = 0.7358\ \text{J}

Potential energy at the top, which is at height 2r = 0.60 m:

U_{\text{top}} = mg(2r) = 0.50\times9.81\times0.60 = 2.943\ \text{J}

Energy accounting. The starting potential energy must cover the energy at the top plus the friction loss:

mgh = K_{\text{top}} + U_{\text{top}} + W_f

0.50\times9.81\times h = 0.7358 + 2.943 + 0.40 = 4.079

h = \frac{4.079}{4.905} = 0.832\ \text{m}

Compare with the frictionless case. Without the 0.40 J loss, h = 3.679/4.905 = 0.750 m — which is exactly 2.5r, the textbook result. Friction has added 8 cm.

What to notice. The famous h = 2.5r answer is worth deriving rather than remembering: mgh = \tfrac12m(gr) + mg(2r), and cancelling mg gives h = r/2 + 2r = 2.5r. If the ball rolls instead of sliding, some of the energy goes into spin and the required height rises to 2.7r for a solid sphere, which follows from Chapter 1.8's c = 0.4. Whether the problem says "ball" or "block" changes the answer by 8%, and reading that word carefully is often the whole difficulty.

Problem 16 — The tricky one: the chain sliding off a table

A uniform chain of length L and mass M lies on a frictionless table with a length x_0 hanging over the edge. Find the speed of the chain when the last link leaves the table.

Solution

The mass hanging over the edge changes as the chain slides, so the driving force is not constant and none of the four kinematic equations applies. Energy is the way in.

Set up the potential energy. Take the table top as U = 0 and measure downwards as negative. The chain's mass per unit length is \lambda = M/L.

When a length x hangs over, that hanging piece has mass \lambda x and its centre of mass is at depth x/2 below the table. So:

U(x) = -\lambda x\,g\,\frac{x}{2} = -\frac{\lambda gx^2}{2}

Energy conservation from x_0 to L. The chain starts at rest:

\tfrac12Mv^2 = U(x_0) - U(L) = -\frac{\lambda gx_0^2}{2} + \frac{\lambda gL^2}{2} = \frac{\lambda g}{2}(L^2-x_0^2)

Substitute \lambda = M/L and cancel the M:

\tfrac12v^2 = \frac{g}{2L}(L^2-x_0^2)

\boxed{v = \sqrt{\frac{g(L^2-x_0^2)}{L}}}

Numbers. For L = 2.0 m and x_0 = 0.50 m:

v = \sqrt{\frac{9.81\times(4.0-0.25)}{2.0}} = \sqrt{\frac{9.81\times3.75}{2.0}} = \sqrt{18.39} = 4.29\ \text{m/s}

Two checks. If the whole chain starts hanging, x_0 = L, and v = 0 — correct, since nothing moves. If x_0 \to 0, v \to \sqrt{gL} = \sqrt{9.81\times2} = 4.43 m/s, which is the maximum possible. Compare that with a single mass dropped through L, which arrives at \sqrt{2gL} = 6.26 m/s. The chain is slower because its centre of mass falls only L/2, not L.

What to notice. The trap here is treating the hanging part as a block of fixed mass. It is not — it grows as the chain slides, so the force grows too, and F = ma with constant a is simply the wrong tool. The energy method sidesteps the changing mass entirely by asking only where the centre of mass started and where it ended. Whenever a problem has a changing mass distribution and asks for a speed rather than a time, reach for energy first.

A warning about the frictionless assumption. If the table has friction, this becomes genuinely hard, because the frictional force also changes as the chain slides — the weight on the table is shrinking. That version needs a differential equation, and it is a standard JEE Advanced escalation of this problem.