Solved Examples

How to Use This Section

Sections 1 to 8 taught you the physics. This section is where you find out whether you actually own it.

What follows is 42 fresh worked problems — new numbers, new framings, and combinations the theory sections deliberately did not attempt. They are ordered roughly easy to hard, and they are grouped by idea so you can attack a weak area directly:

Examples	Topic	Sections to revise if you get stuck
1 to 4	First law, inertia, inertial frames	Section 1
5 to 13	Second law, impulse, force-time graphs, variable force	Section 2
14 to 19	Third law, recoil, collisions, explosions, mass flow	Section 3
20 to 23	Equilibrium, Lami's theorem, multi-string FBDs	Sections 4 and 5
24 to 29	Friction, angle of repose, inclines, stopping distances	Section 6
30 to 34	Circular dynamics — level road, banked road, conical pendulum	Section 7
35 to 39	Connected bodies, pulleys, Atwood, lifts, ropes with mass	Section 8
40 to 42	Genuinely hard, two or three skills chained together	everything

How to work them

Read the problem, close the page, and solve it yourself. Then compare. Reading a worked solution feels like learning and is not; the only thing that transfers to an exam hall is having done it with your own pen.

Key Point: Three habits appear in nearly every solution below, and they are worth more than any formula in this chapter. (1) Draw the free-body diagram before writing a single equation. (2) For any friction problem, test whether the body moves — compare the driving force with $(f_s)_{max} = \mu_s N$ — before you decide whether to use $f_s$ or $f_k$. (3) After you get an answer, check it against an equation you did not use.

[Board Important] Every example states the value of $g$ it uses. Some problems say "take $g = 10$ m/s^2", while others use 9.8. Both appear below, always stated, and never mixed inside one problem. In an exam, use whatever the question says; if it says nothing, say what you assumed.

Right. Pen, paper, and let's go.

Solved Examples

Example 1: The pebble thrown upwards

A pebble of mass 0.05 kg is thrown vertically upwards. Take $g = 10$ m/s^2 and ignore air resistance. Give the magnitude and direction of the net force on the pebble (a) during its upward motion, (b) during its downward motion, (c) at the highest point, where it is momentarily at rest. Do the answers change if it is thrown at $45°$ to the horizontal instead?

Solution:

1. List the forces, not the motions. Once the pebble has left the hand, only one force acts on it: gravity. There is no "force of throw" travelling with it — that force stopped existing the moment contact ended. Air resistance is ignored by instruction.

2. So in every one of the three cases: $$F_{net} = mg = 0.05 \times 10 = 0.5\ \text{N, vertically downward}$$

3. (a) Going up. The velocity is upward, the force is downward. Nothing wrong with that — a force opposite to the velocity is what slows a body. $F_{net} = 0.5$ N down.

4. (b) Coming down. Velocity down, force down, so it speeds up. $F_{net} = 0.5$ N down.

5. (c) At the top. Here $v = 0$, and this is where students go wrong. Zero velocity does not mean zero force. If the force were zero at the top, the first law says the pebble would stay there forever. It does not. $F_{net} = 0.5$ N down.

6. Thrown at $45°$? Gravity does not know or care about the launch angle. The path becomes a parabola, the horizontal velocity stays constant throughout, but the net force is still $mg = 0.5$ N vertically down at every instant, including the top of the flight (where the velocity is not zero now — it is horizontal).

Final Answer: 0.5 N vertically downward in all three cases, and unchanged for a $45°$ launch.

Takeaway: The single most examined confusion in this whole chapter is between velocity and force. They are independent: a body can have zero velocity and a huge force (the pebble at the top), or a huge velocity and zero force (a spacecraft coasting). The first law connects force to change of velocity, never to velocity itself.

Example 2: Where the string is cut

A bob of mass 0.1 kg hangs from the ceiling by a string 2 m long and is set oscillating. Its speed at the mean (lowest) position is 1 m/s. Take $g = 10$ m/s^2. What is the trajectory of the bob if the string is cut (a) at an extreme position, (b) at the mean position?

Solution:

1. The principle. The instant the string is cut, tension vanishes. Only gravity remains, so the bob becomes a projectile with whatever velocity it happened to have at that instant. The whole question is therefore: what is the velocity vector at the moment of cutting?

2. (a) At an extreme position. At the extremes, the bob is momentarily at rest — $v = 0$. A projectile launched with zero velocity simply falls. Trajectory: a vertical straight line, straight down.

3. (b) At the mean position. Here the speed is 1 m/s and the velocity is horizontal (the string is vertical, and the velocity of a pendulum bob is always perpendicular to the string). A projectile launched horizontally under gravity follows a parabola. Trajectory: a parabolic arc.

4. Putting a number on (b). Suppose the bob is 1.8 m above the floor when the string is cut at the mean position. The fall time is $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.8}{10}} = \sqrt{0.36} = 0.6\ \text{s}$$ and in that time it travels horizontally $$x = vt = 1 \times 0.6 = 0.6\ \text{m}$$

5. A bonus check on the tension just before the cut. At the lowest point the bob is moving on a circle of radius 2 m, so the net upward force must supply $\frac{mv^2}{r}$: $$T - mg = \frac{mv^2}{r} \quad\Longrightarrow\quad T = 0.1 \times 10 + \frac{0.1 \times 1^2}{2} = 1 + 0.05 = 1.05\ \text{N}$$

Final Answer: (a) a vertical straight line; (b) a parabola — landing 0.6 m away if cut 1.8 m above the floor.

Takeaway: "Cut the string" problems are always answered the same way: find the velocity at that instant, then treat it as a projectile. The bob does not fly outward along the string, and it does not stop — inertia carries it off along the tangent. [NEET Important] Notice how part (a) and part (b) differ only in when you cut, and give completely different paths.

Example 3: The braking bus, and the box that keeps going

A bus travelling at 15 m/s brakes uniformly and comes to rest in 5 s. A smooth (effectively frictionless) box sits on the floor of the bus, 2.0 m behind the partition at the front. Does it hit the partition? If so, when, and at what relative speed?

Solution:

1. The bus first. $$a_{bus} = \frac{0 - 15}{5} = -3.0\ \text{m/s}^2$$

2. Now the box — and here is the whole point. The floor is frictionless, so there is no horizontal force on the box at all. By the first law it keeps moving at a constant 15 m/s. The bus slows; the box does not.

3. Work in the ground frame (an inertial frame, so Newton's laws are valid there without corrections). Measuring both from the same starting line, $$x_{box} = 15t, \qquad x_{bus} = 15t - 1.5t^2$$ so the box gains on the front of the bus by $$\Delta x = x_{box} - x_{bus} = 1.5t^2$$

4. It hits when $\Delta x = 2.0$ m: $$1.5t^2 = 2.0 \quad\Longrightarrow\quad t = \sqrt{\frac{2.0}{1.5}} = 1.15\ \text{s}$$ The bus is still moving at $15 - 3(1.15) = 11.54$ m/s at that moment, so this happens early in the braking.

5. Relative speed at impact: $$v_{rel} = 15 - (15 - 3 \times 1.15) = 3 \times 1.15 = 3.46\ \text{m/s}$$

Final Answer: Yes — it strikes the partition after 1.15 s, at a relative speed of 3.46 m/s.

Takeaway: Notice what we did not say: we never claimed a force threw the box forward. Nothing pushed it. The bus was taken out from under it. [JEE Tip] If you insist on working in the bus's frame, you must add a pseudo-force of $ma = 3m$ N forward, because the bus frame is non-inertial — that machinery is Section 10's job. The ground frame needs no such thing, which is exactly why it is the safer choice in an exam.

Example 4: Four forces and a constant velocity

A body moves with constant velocity while four coplanar forces act on it. Three of them are $$\vec{F}_1 = (4\hat{i} + 2\hat{j})\ \text{N}, \qquad \vec{F}_2 = (-3\hat{i} + 5\hat{j})\ \text{N}, \qquad \vec{F}_3 = (2\hat{i} - 6\hat{j})\ \text{N}$$ Find the fourth force, its magnitude and its direction.

Solution:

1. Read "constant velocity" correctly. Constant velocity means zero acceleration, which by the second law means zero net force. It does not mean zero force on the body — it means the forces cancel. $$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0$$

2. Add the three known forces component by component: $$\sum F_x = 4 - 3 + 2 = 3\ \text{N}, \qquad \sum F_y = 2 + 5 - 6 = 1\ \text{N}$$

3. The fourth force must cancel that: $$\vec{F}_4 = -(3\hat{i} + \hat{j}) = (-3\hat{i} - \hat{j})\ \text{N}$$

4. Magnitude and direction: $$|\vec{F}_4| = \sqrt{3^2 + 1^2} = \sqrt{10} = 3.16\ \text{N}$$ Both components are negative, so it points into the third quadrant, at $$\tan^{-1}\left(\frac{1}{3}\right) = 18.4°$$ below the negative $x$-axis.

5. Check: $(4-3+2-3)\hat{i} + (2+5-6-1)\hat{j} = 0\hat{i} + 0\hat{j}$. Clean.

Final Answer: $\vec{F}_4 = (-3\hat{i} - \hat{j})$ N, magnitude 3.16 N, at $18.4°$ below the negative $x$-axis.

Takeaway: The whole problem is one line of physics — "constant velocity $\Rightarrow$ $\sum\vec{F} = 0$" — and three lines of arithmetic. [JEE Tip] Also note what is not needed: the mass of the body, the actual value of the velocity, and the direction of motion. If a question hands you numbers you never use, that is often deliberate.

Example 5: Two quick second-law drills

(a) A constant retarding force of 50 N acts on a 20 kg body moving initially at 15 m/s. How long does it take to stop, and how far does it travel? (b) A constant force acting on a 3.0 kg body changes its speed from 2.0 m/s to 3.5 m/s in 25 s, without changing direction. Find the magnitude and direction of the force.

Solution:

1. (a) Acceleration first: $$a = \frac{F}{m} = \frac{-50}{20} = -2.5\ \text{m/s}^2$$

2. Time to stop, from $v = u + at$ with $v = 0$: $$0 = 15 - 2.5t \quad\Longrightarrow\quad t = \frac{15}{2.5} = 6.0\ \text{s}$$

3. Distance, from $v^2 = u^2 + 2as$: $$0 = 225 - 2(2.5)s \quad\Longrightarrow\quad s = \frac{225}{5} = 45\ \text{m}$$

4. (b) Acceleration from the speed change: $$a = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} = 0.06\ \text{m/s}^2$$

5. Force: $$F = ma = 3.0 \times 0.06 = 0.18\ \text{N}$$ The speed increased along an unchanged direction, so the force is along the direction of motion.

Final Answer: (a) 6.0 s, travelling 45 m; (b) 0.18 N, in the direction of motion.

Takeaway: Part (b) is a useful reality check on how small forces can be. A force of 0.18 N — about the weight of a 18 g biscuit — is enough to change a 3 kg body's speed by 1.5 m/s, provided you give it 25 seconds. Force sets the rate of change of momentum, not the change itself; a long enough time turns any force into a large impulse.

Example 6: Two perpendicular forces

A body of mass 5 kg is acted upon by two perpendicular forces of 8 N and 6 N. Find the magnitude and direction of its acceleration.

Solution:

1. Set up axes along the forces. Put the 8 N force along $+x$ and the 6 N force along $+y$ (they are perpendicular, so this is allowed and it makes the arithmetic trivial): $$\vec{F}_{net} = (8\hat{i} + 6\hat{j})\ \text{N}$$

2. Resultant magnitude — the familiar 3-4-5 triangle, scaled: $$|\vec{F}_{net}| = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{N}$$

3. Acceleration: $$a = \frac{F_{net}}{m} = \frac{10}{5} = 2.0\ \text{m/s}^2$$

4. Direction. The acceleration is parallel to the net force (the second law is a vector equation, so $\vec{a}$ and $\vec{F}$ always point the same way): $$\theta = \tan^{-1}\left(\frac{6}{8}\right) = 36.9°\ \text{from the 8 N force}$$

5. Component check: $a_x = 8/5 = 1.6$ m/s^2 and $a_y = 6/5 = 1.2$ m/s^2, and $\sqrt{1.6^2 + 1.2^2} = 2.0$ m/s^2. Consistent.

Final Answer: $a = 2.0$ m/s^2, at $36.9°$ from the 8 N force (that is, $53.1°$ from the 6 N force).

Takeaway: Two things worth locking in. First, add the forces as vectors, then divide by $m$ — never add magnitudes ($8 + 6 = 14$ N is wrong and gives 2.8 m/s^2). Second, always state the direction relative to something named, because "$36.9°$" on its own means nothing to an examiner.

Example 7: The three-wheeler and the child

The driver of a three-wheeler moving at 36 km/h sees a child in the road and brings the vehicle to rest in 4.0 s, just in time. The three-wheeler has mass 400 kg and the driver 65 kg. What is the average retarding force on the vehicle?

Solution:

1. Convert to SI first, every time: $$36\ \text{km/h} = 36 \times \frac{1000}{3600} = 10\ \text{m/s}$$

2. Retardation: $$a = \frac{v - u}{t} = \frac{0 - 10}{4.0} = -2.5\ \text{m/s}^2$$

3. What mass is being retarded? The driver is decelerating too, and the force that does it comes from the vehicle. So the retarding force from the road must slow both: $$m_{total} = 400 + 65 = 465\ \text{kg}$$

4. Second law: $$F = ma = 465 \times (-2.5) = -1162.5\ \text{N}$$ The minus sign says "opposite to the motion". Magnitude 1162.5 N.

5. Cross-check by impulse. $\Delta p = 465 \times (0 - 10) = -4650$ kg m/s, delivered in 4.0 s, so $F = -4650/4.0 = -1162.5$ N. Same.

Final Answer: 1162.5 N, directed opposite to the motion.

Takeaway: The driver's 65 kg is not decoration — leaving it out gives 1000 N and loses the mark. Ask yourself every time: what exactly is the system this force is accelerating? Also note the word average: the real braking force varies from instant to instant, and $F = \Delta p/\Delta t$ gives you its time-average, which is all the data allows.

Example 8: The rocket's initial thrust

A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 m/s^2. Take $g = 10$ m/s^2. Calculate the initial thrust of the blast.

Solution:

1. Free-body diagram of the rocket at lift-off. Two vertical forces: the thrust $F$ upward from the exhaust gases, and the weight $Mg$ downward.

2. Weight: $$Mg = 20000 \times 10 = 2.0 \times 10^5\ \text{N}$$

3. Second law, taking up as positive: $$F - Mg = Ma$$ $$F = M(g + a) = 20000(10 + 5) = 20000 \times 15 = 3.0 \times 10^5\ \text{N}$$

4. Sanity check. The thrust must exceed the weight, or the rocket would not leave the pad — and indeed $3.0 \times 10^5 > 2.0 \times 10^5$. The surplus, $1.0 \times 10^5$ N, is the net force, and $10^5/20000 = 5$ m/s^2 as required.

Final Answer: $F = 3.0 \times 10^5$ N.

Takeaway: The pattern $F = M(g+a)$ is the same one you meet in the lift (Example 37) and in every "lifting something with acceleration" problem. [NEET Important] The commonest error is writing $F = Ma = 10^5$ N and forgetting that the thrust also has to hold up the weight. If a body accelerates upward, the upward force is always bigger than $mg$.

Example 9: A force with an expiry date

A body of mass 0.40 kg is moving at a constant 10 m/s to the north. From $t = 0$ it is subject to a constant force of 8.0 N directed south, for 30 s. Take $x = 0$ at $t = 0$. Predict its position at $t = -5$ s, $t = 25$ s and $t = 100$ s.

Solution:

1. Set up. Take north as positive $x$. While the force acts, $$a = \frac{F}{m} = \frac{-8.0}{0.40} = -20\ \text{m/s}^2$$

2. At $t = -5$ s the force has not started yet. The body has been coasting at a constant 10 m/s, so $$x = ut = 10 \times (-5) = -50\ \text{m}$$ That is, 50 m south of the origin — which makes sense: five seconds before it reached the origin, it was 50 m short of it.

3. At $t = 25$ s the force is still acting (it runs from 0 to 30 s), so use the uniform-acceleration formula: $$x = ut + \tfrac{1}{2}at^2 = 10(25) + \tfrac{1}{2}(-20)(25)^2 = 250 - 6250 = -6000\ \text{m}$$ The body reversed long ago — it turned around at $t = u/|a| = 0.5$ s — and is now 6 km south.

4. At $t = 100$ s. Here is the trap. The force stops at $t = 30$ s. After that the body coasts. So do it in two stages. $$x(30) = 10(30) + \tfrac{1}{2}(-20)(900) = 300 - 9000 = -8700\ \text{m}$$ $$v(30) = 10 + (-20)(30) = -590\ \text{m/s}$$ From $t = 30$ s to $t = 100$ s (70 s) there is no force, so the velocity stays at $-590$ m/s: $$x(100) = -8700 + (-590)(70) = -8700 - 41300 = -50000\ \text{m}$$

Final Answer: $x = -50$ m at $t = -5$ s; $x = -6000$ m ($-6$ km) at $t = 25$ s; $x = -50000$ m ($-50$ km) at $t = 100$ s.

Takeaway: Two lessons in one problem. (1) Kinematic formulas only apply over intervals in which the acceleration is actually constant — the moment the force switches off, you start a new stage with the old stage's final velocity as the new initial velocity. (2) Applying $x = ut + \frac{1}{2}at^2$ blindly all the way to $t = 100$ s gives $-99000$ m, and it is wrong by almost a factor of two.

Example 10: Dropped from an accelerating truck

A truck starts from rest and accelerates uniformly at 2.0 m/s^2. At $t = 10$ s, a person standing on top of the truck, 6 m above the ground, drops a stone. Take $g = 10$ m/s^2 and neglect air resistance. Find (a) the velocity and (b) the acceleration of the stone at $t = 11$ s.

Solution:

1. The truck at the instant of release. $$v_{truck} = 0 + 2.0 \times 10 = 20\ \text{m/s, horizontally}$$ The stone was travelling with the truck, so at the moment it is released it has a horizontal velocity of 20 m/s.

2. What forces act on the stone after release? Only gravity — vertical, downward. There is no horizontal force, so the stone's horizontal velocity stays at 20 m/s for ever after. The truck keeps accelerating; the stone does not. (You met exactly this point in Section 1, in a new coat.)

3. One second after release ($t = 11$ s): $$v_x = 20\ \text{m/s (unchanged)}, \qquad v_y = 0 + g(1) = 10\ \text{m/s downward}$$

4. Combine: $$v = \sqrt{20^2 + 10^2} = \sqrt{500} = 22.4\ \text{m/s}$$ at an angle below the horizontal of $$\tan^{-1}\left(\frac{10}{20}\right) = 26.6°$$

5. Check it is still in the air. Falling 6 m takes $t = \sqrt{2h/g} = \sqrt{1.2} = 1.10$ s, and $1.10 > 1.00$ s, so at $t = 11$ s the stone has not landed. Good — the question is answerable.

6. (b) The acceleration is just $g$: 10 m/s^2, vertically downward. The truck's 2.0 m/s^2 is irrelevant to the stone the moment contact ends.

Final Answer: (a) 22.4 m/s at $26.6°$ below the horizontal; (b) 10 m/s^2 vertically downward.

Takeaway: Step 5 is the habit to steal from this problem. Before you compute a projectile's state at some time, check it is still in flight. And note the split personality of the answer: the velocity remembers the truck (through $v_x = 20$), while the acceleration has forgotten it completely.

Example 11: Force and impulse from a position-time graph

The graph shows the position-time record of a particle of mass 4 kg: it sits at $x = 0$ for all $t < 0$, moves along a straight line reaching $x = 3$ m at $t = 4$ s, and then stays at $x = 3$ m. Find (a) the force on the particle for $t < 0$, for $t > 4$ s and for $0 < t < 4$ s; (b) the impulse at $t = 0$ and at $t = 4$ s.

Solution:

1. Turn the $x$-$t$ graph into velocities. Velocity is the slope.

• For $t < 0$: the graph is flat, so $v = 0$.
• For $0 < t < 4$ s: a straight line from $(0, 0)$ to $(4, 3)$, so $v = \frac{3 - 0}{4 - 0} = 0.75$ m/s, constant.
• For $t > 4$ s: flat again, so $v = 0$.

1. (a) The force. In each of the three intervals the velocity is constant, so $a = 0$, so $$F = ma = 0 \quad \text{in all three intervals}$$ Yes — zero force even while it is moving at 0.75 m/s. That is the first law, and it is exactly the point of the question.

2. (b) The impulse at $t = 0$. The velocity jumps from 0 to 0.75 m/s: $$J = \Delta p = m\,\Delta v = 4(0.75 - 0) = +3\ \text{kg m/s}$$

3. The impulse at $t = 4$ s. The velocity jumps from 0.75 m/s back to 0: $$J = 4(0 - 0.75) = -3\ \text{kg m/s}$$

4. Why "impulse" and not "force" at the corners? At a sharp corner of the $x$-$t$ graph, the velocity changes in zero time, so the force is infinite for an instant. The force is unanswerable; the impulse — force integrated over time — is perfectly finite and equals the momentum change. That is precisely when the impulse-momentum theorem earns its keep.

Final Answer: (a) $F = 0$ in all three intervals; (b) $+3$ kg m/s at $t = 0$ and $-3$ kg m/s at $t = 4$ s.

Takeaway: Read graphs in this order: position $\to$ slope gives velocity $\to$ slope of that gives acceleration $\to$ multiply by $m$ for force. A straight $x$-$t$ segment always means zero force, however steep it is. And a kink in the $x$-$t$ graph always means an impulse. The two impulses here are equal and opposite, so the total momentum change over the whole record is zero — which it must be, since the particle starts and ends at rest.

Example 12: A force-time pulse, and the area under it

A 4 kg block rests on a frictionless horizontal surface. A horizontal force is applied: it grows linearly from 0 to 8 N over the first 2 s, stays at 8 N from $t = 2$ s to $t = 5$ s, then falls linearly back to zero by $t = 6$ s. Find (a) the total impulse, (b) the final speed, (c) the maximum acceleration, and (d) the average force over the 6 s.

Solution:

1. The shape is a trapezium, so the impulse — which is the area under the $F$-$t$ graph — is best found piece by piece.

2. (a) Piecewise areas: $$\text{rise (triangle)} = \tfrac{1}{2}(2)(8) = 8\ \text{N s}$$ $$\text{flat top (rectangle)} = (3)(8) = 24\ \text{N s}$$ $$\text{fall (triangle)} = \tfrac{1}{2}(1)(8) = 4\ \text{N s}$$ $$J = 8 + 24 + 4 = 36\ \text{N s}$$

3. (b) Final speed. By the impulse-momentum theorem, and since the surface is frictionless so no other horizontal force acts, $$J = m v_f - m v_i = 4v_f - 0 \quad\Longrightarrow\quad v_f = \frac{36}{4} = 9.0\ \text{m/s}$$

4. (c) Maximum acceleration happens when the force is largest, at 8 N: $$a_{max} = \frac{8}{4} = 2.0\ \text{m/s}^2$$

5. (d) Average force over the whole 6 s: $$\bar{F} = \frac{J}{\Delta t} = \frac{36}{6} = 6.0\ \text{N}$$ Check: a constant 6 N for 6 s would also produce $36/4 = 9$ m/s. That is what "average force" means.

6. Intermediate speeds, for practice. Up to $t = 2$ s the impulse is 8 N s, so $v = 2$ m/s; up to $t = 5$ s it is 32 N s, so $v = 8$ m/s. The last second adds only 1 m/s, because the force is dying away.

Final Answer: (a) 36 N s; (b) 9.0 m/s; (c) 2.0 m/s^2; (d) 6.0 N.

Takeaway: When a force varies with time, stop looking for $a$ and start looking for area. Impulse handles any shape of $F(t)$ — triangles, trapezia, curves — with no calculus beyond finding an area, and it delivers the velocity change directly. [JEE Tip] Whenever a graph of force against time appears in a paper, the question is almost certainly about impulse.

Example 13: A force that dies away

A 2 kg body at rest on a frictionless surface is acted on by a horizontal force $F = (12 - 3t)$ N, where $t$ is in seconds. Find (a) the initial acceleration, (b) the instant the force vanishes, (c) the speed at that instant, and (d) the speed at $t = 8$ s.

Solution:

1. (a) Initial acceleration. At $t = 0$, $F = 12$ N, so $$a_0 = \frac{12}{2} = 6.0\ \text{m/s}^2$$ This is an instantaneous value — the second law is a relation between force and acceleration at the same moment, and both are changing here.

2. (b) The force vanishes when $12 - 3t = 0$: $$t = 4.0\ \text{s}$$ After that, $F$ becomes negative — the force reverses and starts retarding the body.

3. (c) Speed at $t = 4$ s. The force is not constant, so use impulse (the area under the $F$-$t$ line, which is a triangle of base 4 s and height 12 N): $$J = \int_0^4 (12 - 3t)\,dt = \left[12t - \tfrac{3}{2}t^2\right]_0^4 = 48 - 24 = 24\ \text{N s}$$ $$v = \frac{J}{m} = \frac{24}{2} = 12\ \text{m/s}$$ That is the maximum speed, because the force turns negative afterwards.

4. (d) Speed at $t = 8$ s. From 4 s to 8 s the force is negative, and by symmetry that second triangle has exactly the same area, $-24$ N s. So $$J_{total}(0 \to 8) = 24 - 24 = 0 \quad\Longrightarrow\quad v(8) = 0$$ The body accelerates for 4 s, decelerates for 4 s, and ends at rest — but far from where it started: 32 m by the time it reaches top speed, and 64 m by the time it stops.

5. A speed on the way up, for practice. At $t = 2$ s, $J = 24 - 6 = 18$ N s, so $v = 9$ m/s.

Final Answer: (a) 6.0 m/s^2; (b) $t = 4.0$ s; (c) 12 m/s; (d) 0.

Takeaway: A force of zero at $t = 4$ s does not mean the body is at rest there — it means the body has stopped changing its velocity, so that is where the speed peaks. [JEE Tip] "Maximum velocity" almost always means "where the acceleration crosses zero", and finding that instant is usually the first line of the solution.

Example 14: The recoil of the gun

A shell of mass 0.020 kg is fired from a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun?

Solution:

1. Choose the system: gun + shell together. The explosion forces are internal to that system, and the external forces (gravity, the normal reaction from the ground) are vertical, so horizontal momentum is conserved.

2. Before firing, everything is at rest: $$p_{before} = 0$$

3. After firing, taking the shell's direction as positive: $$p_{after} = m_s v_s - m_g v_g = 0.020(80) - 100 v_g$$

4. Set them equal: $$0.020 \times 80 = 100 v_g \quad\Longrightarrow\quad 1.6 = 100 v_g \quad\Longrightarrow\quad v_g = 0.016\ \text{m/s}$$

5. Check the vector sum. Shell: $+1.6$ kg m/s. Gun: $100 \times 0.016 = 1.6$ kg m/s backwards, so $-1.6$. Total zero, as required. The speed ratio is 5000 to 1, the exact inverse of the mass ratio.

Final Answer: The gun recoils at 0.016 m/s, that is 1.6 cm/s, opposite to the shell.

Takeaway: Both objects receive the same magnitude of momentum (1.6 kg m/s), because they push each other with equal and opposite forces for the same length of time. It is the speeds that differ wildly, and only because the masses do. [NEET Important] In every recoil problem the pattern is $m_1v_1 = m_2v_2$ — a lighter body always ends up faster.

Example 15: The impulse each billiard ball receives

Two billiard balls, each of mass 0.05 kg, move towards each other at 6 m/s and rebound with the same speed. What impulse does each ball receive from the other?

Solution:

1. Take a direction. Let ball A move along $+x$ at 6 m/s and ball B along $-x$ at 6 m/s.

2. Momenta before: $$p_A = 0.05(+6) = +0.3\ \text{kg m/s}, \qquad p_B = 0.05(-6) = -0.3\ \text{kg m/s}$$ Total: 0.

3. Momenta after (each rebounds, so each reverses): $$p_A' = -0.3\ \text{kg m/s}, \qquad p_B' = +0.3\ \text{kg m/s}$$ Total: 0. Momentum is conserved, as it must be.

4. Impulse on A (impulse-momentum theorem): $$J_A = p_A' - p_A = -0.3 - (+0.3) = -0.6\ \text{kg m/s}$$

5. Impulse on B: $$J_B = +0.3 - (-0.3) = +0.6\ \text{kg m/s}$$

6. Third-law check: $J_A + J_B = 0$. The two impulses are equal in magnitude and opposite in direction, exactly as Newton's third law requires — the two balls push each other with equal and opposite forces for the identical contact time.

Final Answer: Each ball receives an impulse of magnitude 0.6 kg m/s, directed opposite to its own original motion.

Takeaway: The magnitude is $2mv$, not $mv$ — a reversal changes momentum by twice the original amount. Students routinely halve this answer. [JEE/NEET] And note that the total momentum was zero both before and after: conservation of momentum is satisfied trivially here, which is why the interesting quantity is the impulse on each ball individually.

Example 16: The batsman's deflection

A batsman deflects a ball by an angle of $45°$ without changing its speed, which is 54 km/h. The ball has mass 0.15 kg. What impulse is imparted to the ball?

Solution:

1. Convert: $$54\ \text{km/h} = 54 \times \frac{1000}{3600} = 15\ \text{m/s}$$ So $|\vec{p}_i| = |\vec{p}_f| = 0.15 \times 15 = 2.25$ kg m/s.

2. Get the geometry right — this is the whole problem. The ball arrives at the bat and is deflected, meaning it is sent back on the other side, making $45°$ with the line it came in on. So the incoming and outgoing velocity vectors are at $180° - 45° = 135°$ to each other, not $45°$.

3. Set it up in components. Let the incoming velocity be along $+x$; the outgoing velocity is then at $135°$: $$\vec{p}_i = 2.25\hat{i}, \qquad \vec{p}_f = 2.25(\cos 135°\,\hat{i} + \sin 135°\,\hat{j}) = (-1.591\hat{i} + 1.591\hat{j})$$

4. Impulse: $$\vec{J} = \vec{p}_f - \vec{p}_i = (-1.591 - 2.25)\hat{i} + 1.591\hat{j} = (-3.841\hat{i} + 1.591\hat{j})$$ $$|\vec{J}| = \sqrt{3.841^2 + 1.591^2} = \sqrt{14.75 + 2.53} = 4.16\ \text{kg m/s}$$

5. The one-line version. When the speed is unchanged and the two vectors are at angle $\phi$ to each other, $$|\vec{J}| = 2mv\sin\frac{\phi}{2} = 2(0.15)(15)\sin 67.5° = 4.5 \times 0.9239 = 4.16\ \text{kg m/s}$$ Equivalently, in terms of the stated deflection $\theta = 45°$: $|\vec{J}| = 2mv\cos\frac{\theta}{2} = 4.5\cos 22.5° = 4.16$ kg m/s.

6. Direction. $\vec{J}$ points along the bisector of the reversed incoming and the outgoing directions — that is, straight out from the bat face, which is exactly where you would expect the bat to push.

Final Answer: $|\vec{J}| = 4.16$ kg m/s, that is about 4.2 kg m/s, directed along the bisector of the two paths.

Takeaway: Everything hinges on step 2. If you had read "deflected by $45°$" as "the two velocity vectors are $45°$ apart", you would have got $2mv\sin 22.5° = 1.72$ kg m/s — a completely different answer for the same words. [JEE Tip] In any change-of-direction impulse problem, draw the two momentum vectors from a common point and measure the angle between them. Never work from a description alone, and remember that the reversal case ($\phi = 180°$) gives the maximum possible, $2mv$.

Example 17: Why the fragments must fly apart back to back

A nucleus is at rest in the laboratory frame. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.

Solution:

1. The system is isolated. Nuclear forces are internal to the parent nucleus, and during the split no external force acts. Therefore the total linear momentum is conserved as a vector.

2. Before: the nucleus is at rest, so $$\vec{p}_{total} = \vec{0}$$

3. After: two fragments of masses $m_1$ and $m_2$ with velocities $\vec{v}_1$ and $\vec{v}_2$: $$m_1\vec{v}_1 + m_2\vec{v}_2 = \vec{0}$$

4. Rearrange: $$m_1\vec{v}_1 = -\,m_2\vec{v}_2$$ Masses are positive scalars, so $\vec{v}_1$ and $\vec{v}_2$ must be antiparallel — exactly opposite in direction. That is the proof, and it is three lines long.

5. The speeds are not equal, though. Taking magnitudes, $$\frac{v_1}{v_2} = \frac{m_2}{m_1}$$ The lighter fragment goes faster, in exact inverse proportion.

6. A number, to make it concrete. A nucleus of mass number 220 at rest emits an alpha particle (mass number 4), leaving a daughter of mass number 216. If the alpha leaves at speed $v$, the daughter recoils at $$v_d = \frac{4}{216}v = \frac{v}{54}$$ in the opposite direction, and the two momenta are identical in magnitude.

Final Answer: The two fragments must move in exactly opposite directions, with speeds inversely proportional to their masses.

Takeaway: This is the recoil of a gun (Example 14), the explosion of a bomb, and a person stepping off a boat — the same one-line argument every time. [JEE/NEET] And note carefully what makes it work: the total momentum before was zero, so the total after must also be zero, and the only way two vectors add to zero is by being antiparallel. If the parent nucleus had been moving, the fragments would in general not be back to back in the lab frame.

Example 18: Sand onto a moving conveyor belt

Sand drops vertically at a steady rate of 5 kg/s onto a horizontal conveyor belt that is moving at a constant 2 m/s. What extra horizontal force must the belt motor supply to keep the belt moving at that speed, and what extra power does it need?

Solution:

1. What is the physics? The sand lands with zero horizontal velocity. The belt has to drag it up to 2 m/s. That change of horizontal momentum needs a horizontal force, and it is friction between the sand and the belt that provides it.

2. Momentum given to the sand per second. In one second, 5 kg of sand goes from 0 to 2 m/s horizontally: $$\frac{\Delta p}{\Delta t} = \frac{dm}{dt} \times v = 5 \times 2 = 10\ \text{N}$$

3. By the second law in its momentum form, that rate of change of momentum is the force on the sand: $$F = v\frac{dm}{dt} = 10\ \text{N}$$ By the third law, the sand drags backwards on the belt with 10 N, so the motor must supply an extra 10 N to hold the speed steady.

4. Extra power: $$P = Fv = 10 \times 2 = 20\ \text{W}$$

5. Cross-check by impulse over 4 s. In 4 s, 20 kg of sand is loaded and each kilogram ends at 2 m/s, so the momentum delivered is $20 \times 2 = 40$ kg m/s. The impulse from a steady 10 N over 4 s is $10 \times 4 = 40$ N s. Identical, as they must be.

Final Answer: An extra force of 10 N and an extra power of 20 W.

Takeaway: This is $\vec{F} = \frac{d\vec{p}}{dt}$ used in the form the $F = ma$ shortcut cannot handle: the mass is changing, not the velocity. The same formula $F = v\,\frac{dm}{dt}$ governs a water jet hitting a wall, a rocket burning fuel, and a hose filling a bucket. [JEE Tip] Whenever a problem gives you a rate in kg/s, reach for $\frac{dp}{dt}$, never for $ma$.

Example 19: Four "explain why" questions

Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch.

Solution:

1. (a) The horse in empty space. The horse pushes backwards on the ground; by the third law the ground pushes forwards on the horse, and that reaction is the force which drives horse and cart forward. In empty space there is no ground to push against, so no reaction is generated, and the horse's legs thrash uselessly. The tension in the traces is internal to the horse-plus-cart system and can never move it as a whole.

2. (b) Passengers thrown forward. They are not thrown — nothing pushes them. The seat and the floor decelerate with the bus; the passenger's upper body has no such force acting on it and, by the first law, keeps moving forward at the original speed. Relative to the bus, that looks like being flung forward. This is exactly Example 3, with a person instead of a box.

3. (c) Pull versus push, with numbers. Take a 20 kg mower, $\mu_s = 0.4$, a handle at $30°$ to the horizontal, and $g = 10$ m/s^2. The vertical component of the applied force changes the normal reaction, and friction follows $N$. Pulling (force at $30°$ above the horizontal): $N = mg - F\sin 30°$, and to just move it, $F\cos 30° = \mu_s N$. Solving the pair, $$F(\cos 30° + \mu_s\sin 30°) = \mu_s mg \quad\Longrightarrow\quad F = \frac{0.4 \times 200}{0.866 + 0.2} = 75.0\ \text{N}$$ with $N = 200 - 37.5 = 162.5$ N. Pushing (force at $30°$ below the horizontal): now $N = mg + F\sin 30°$, so $$F = \frac{0.4 \times 200}{0.866 - 0.2} = 120.1\ \text{N}$$ with $N = 260.1$ N. Pushing needs 1.6 times as much force.

4. (d) The cricketer's hands. The catch must destroy the same momentum $\Delta p$ either way, so $$F_{avg} = \frac{\Delta p}{\Delta t}$$ Drawing the hands back stretches $\Delta t$ — the only quantity he can control. Triple the stopping time and the average force on his hands falls to a third. The impulse is fixed; the force is negotiable.

Final Answer: (a) no ground means no reaction force; (b) inertia of the passenger, not a forward force; (c) pulling reduces $N$, so 75.0 N instead of 120.1 N; (d) a longer $\Delta t$ means a smaller $F$ for the same $\Delta p$.

Takeaway: These four look like essay questions and are actually four different laws in disguise: (a) the third law, (b) the first law, (c) friction with a variable normal reaction, (d) the impulse-momentum theorem. In a board exam, name the law you are using in the first sentence of each answer — that sentence usually carries the mark.

Example 20: A lamp on two chains

A lamp of mass 12 kg hangs from a hook in the ceiling. Two chains run from the hook to the ceiling, one making $30°$ with the ceiling and the other $45°$, on opposite sides. Take $g = 10$ m/s^2. Find the tension in each chain.

Solution:

1. Isolate the hook — the point where the three forces meet. The lamp's weight pulls straight down on it with $W = 12 \times 10 = 120$ N, and the two chains pull along their own lengths, up and outwards.

2. Choose axes and resolve. Let $T_1$ be the tension in the $30°$ chain (up and to the left) and $T_2$ in the $45°$ chain (up and to the right). Angles are measured from the horizontal ceiling. $$\sum F_x = 0: \qquad T_2\cos 45° - T_1\cos 30° = 0$$ $$\sum F_y = 0: \qquad T_1\sin 30° + T_2\sin 45° = 120$$

3. From the first equation: $$T_2 = T_1\frac{\cos 30°}{\cos 45°} = T_1\frac{0.8660}{0.7071} = 1.2247\,T_1$$

4. Substitute into the second: $$T_1(0.5) + 1.2247\,T_1(0.7071) = 120$$ $$0.5\,T_1 + 0.8660\,T_1 = 120 \quad\Longrightarrow\quad 1.3660\,T_1 = 120 \quad\Longrightarrow\quad T_1 = 87.8\ \text{N}$$ $$T_2 = 1.2247 \times 87.8 = 107.6\ \text{N}$$

5. Cross-check with Lami's theorem. Draw the three forces from the hook: $T_1$ at $150°$, $T_2$ at $45°$, $W$ at $270°$. The angles between successive pairs are $105°$ (between the chains), $135°$ ($W$ to $T_2$) and $120°$ ($W$ to $T_1$), which sum to $360°$ as they must. Then $$\frac{120}{\sin 105°} = \frac{T_1}{\sin 135°} = \frac{T_2}{\sin 120°} = 124.23$$ giving $T_1 = 124.23 \times 0.7071 = 87.8$ N and $T_2 = 124.23 \times 0.8660 = 107.6$ N. The two methods agree exactly.

Final Answer: $T_1 = 87.8$ N in the $30°$ chain; $T_2 = 107.6$ N in the $45°$ chain.

Takeaway: The steeper chain always carries the larger tension, because it is doing more of the vertical lifting — a two-second sanity check you can run on any answer of this type. [JEE Tip] Also notice that $T_1 + T_2 = 195.4$ N, far more than the 120 N being supported. Tensions do not add up to the weight unless both strings are vertical.

Example 21: A sphere against a smooth wall — Lami in two lines

A sphere of mass 10 kg rests against a smooth vertical wall. It is held by a light string attached to the wall above, and the string makes $30°$ with the wall. Take $g = 10$ m/s^2. Find the tension in the string and the reaction of the wall.

Solution:

1. Three forces act on the sphere, and they are concurrent (all lines pass through the centre): the weight $W = 100$ N vertically down; the normal reaction $N$ from the wall, horizontal (the wall is smooth, so there is no friction and the reaction is perpendicular to the wall); and the tension $T$ along the string, at $30°$ to the vertical.

2. By components. Taking $x$ horizontal (away from the wall) and $y$ vertical: $$\sum F_x = 0: \qquad N - T\sin 30° = 0$$ $$\sum F_y = 0: \qquad T\cos 30° - 100 = 0$$

3. From the vertical equation: $$T = \frac{100}{\cos 30°} = \frac{100}{0.8660} = 115.5\ \text{N}$$

4. From the horizontal equation: $$N = T\sin 30° = 115.5 \times 0.5 = 57.7\ \text{N}$$ Equivalently, $N = W\tan 30° = 100 \times 0.5774 = 57.7$ N.

5. Lami's theorem as a check. The angle between $N$ and $W$ is $90°$; between $W$ and $T$ is $150°$; between $T$ and $N$ is $120°$. So $$\frac{W}{\sin 120°} = \frac{T}{\sin 90°} = \frac{N}{\sin 150°}$$ $$\frac{100}{0.8660} = 115.5 = \frac{T}{1} \quad\Longrightarrow\quad T = 115.5\ \text{N}, \qquad N = 115.5 \times 0.5 = 57.7\ \text{N}$$

Final Answer: $T = 115.5$ N and $N = 57.7$ N.

Takeaway: Two standard results worth memorising for this configuration: $T = \dfrac{W}{\cos\theta}$ and $N = W\tan\theta$, with $\theta$ measured from the wall. Both grow without limit as the string is made more horizontal ($\theta \to 90°$), which is why a sphere resting against a wall is always hung from well above it. [NEET Important] $T$ is always greater than $W$ here — if your answer for a slanted supporting string comes out less than the weight, you have made a sign or angle error.

Example 22: Two masses balanced across a double incline

A wedge has two smooth faces sloping at $30°$ and $60°$, meeting at an apex where a light frictionless pulley is mounted. A 10 kg block on the $30°$ face is joined by a light string over the pulley to a block of mass $m$ on the $60°$ face. Take $g = 10$ m/s^2. Find $m$ for equilibrium, the tension, and the normal reaction on each face.

Solution:

1. Two free-body diagrams, two sets of axes. For each block choose axes along and perpendicular to its own face — the two blocks get different axes, and that is fine, because they are separate bodies.

2. The 10 kg block on the $30°$ face. Along its slope, at rest: $$T - m_1 g\sin 30° = 0 \quad\Longrightarrow\quad T = 10 \times 10 \times 0.5 = 50\ \text{N}$$

3. The block of mass $m$ on the $60°$ face. Same string, so the same tension (light string, frictionless pulley): $$T - mg\sin 60° = 0 \quad\Longrightarrow\quad 50 = m \times 10 \times 0.8660$$ $$m = \frac{50}{8.660} = 5.77\ \text{kg}$$

4. Normal reactions, from the perpendicular equations: $$N_1 = m_1 g\cos 30° = 100 \times 0.8660 = 86.6\ \text{N}$$ $$N_2 = mg\cos 60° = 57.7 \times 0.5 = 28.9\ \text{N}$$

5. The elegant form of the condition. Setting the two along-slope equations equal: $$m_1 g\sin\theta_1 = m_2 g\sin\theta_2 \quad\Longrightarrow\quad \frac{m_2}{m_1} = \frac{\sin\theta_1}{\sin\theta_2} = \frac{\sin 30°}{\sin 60°} = 0.577$$ Check: $5.77/10 = 0.577$. Consistent.

Final Answer: $m = 5.77$ kg, $T = 50$ N, $N_1 = 86.6$ N and $N_2 = 28.9$ N.

Takeaway: The balancing condition is $m_1\sin\theta_1 = m_2\sin\theta_2$, and it says something you can see: the steeper face needs the lighter block, because a steeper slope makes a larger fraction of the weight act along the string. The normal reactions never enter the balance at all — they are perpendicular to the string and do no competing.

Example 23: Held aside by a horizontal force

An 8 kg block hangs from the ceiling on a light string. A horizontal force $F$ pushes it sideways until the string makes $37°$ with the vertical, and holds it there. Take $g = 10$ m/s^2, $\sin 37° = 0.6$, $\cos 37° = 0.8$. (a) Find $F$ and the tension $T$. (b) What is the smallest force that could hold the block at that same $37°$, and in what direction must it act?

Solution:

1. (a) FBD of the block: weight $mg = 80$ N down; tension $T$ along the string, at $37°$ to the vertical; applied force $F$ horizontal. Three forces, in equilibrium. $$\sum F_y = 0: \qquad T\cos 37° - 80 = 0$$ $$\sum F_x = 0: \qquad T\sin 37° - F = 0$$

2. From the vertical equation: $$T = \frac{80}{0.8} = 100\ \text{N}$$

3. From the horizontal equation: $$F = T\sin 37° = 100 \times 0.6 = 60\ \text{N}$$ or in one step, $F = mg\tan 37° = 80 \times 0.75 = 60$ N.

4. (b) The minimum force. Here is the trick: whatever $F$ is, the tension can only act along the string, so $F$ must cancel the component of the weight perpendicular to the string. The most efficient way to do that is to make $F$ itself perpendicular to the string, so that none of it is wasted along the string direction. Resolving perpendicular to the string: $$F_{min} = mg\sin 37° = 80 \times 0.6 = 48\ \text{N}$$ and resolving along the string, $$T = mg\cos 37° = 80 \times 0.8 = 64\ \text{N}$$

5. Check part (b): the three forces $T = 64$ N, $F = 48$ N and $mg = 80$ N form a right-angled triangle, and $\sqrt{64^2 + 48^2} = \sqrt{4096 + 2304} = \sqrt{6400} = 80$. It closes.

Final Answer: (a) $T = 100$ N and $F = 60$ N; (b) $F_{min} = 48$ N applied perpendicular to the string, with $T$ then only 64 N.

Takeaway: The horizontal force is not the cheapest way to hold the block aside — 60 N horizontal against 48 N perpendicular to the string, a 20% saving. [JEE Tip] The general rule: to hold a body displaced against a string, the minimum force is always $mg\sin\theta$, applied at right angles to the string. Notice too that the horizontal push increases the tension to 100 N, while the minimum force decreases it to 64 N.

Example 24: The does-it-move test, four ways

A 25 kg crate sits on a horizontal floor with $\mu_s = 0.4$ and $\mu_k = 0.3$. Take $g = 9.8$ m/s^2. Find the friction force and the acceleration when the applied horizontal force is (a) 80 N, (b) 98 N, (c) 120 N. (d) Once it is sliding, what force keeps it moving at constant speed?

Solution:

1. Do the set-up once. Vertically there is no acceleration, so $$N = mg = 25 \times 9.8 = 245\ \text{N}$$ $$(f_s)_{max} = \mu_s N = 0.4 \times 245 = 98\ \text{N}, \qquad f_k = \mu_k N = 0.3 \times 245 = 73.5\ \text{N}$$ Those two numbers now decide everything.

2. (a) $F = 80$ N. Test: $80 < 98$, so the crate does not move. Static friction is self-adjusting and takes exactly the value needed for equilibrium: $$f_s = 80\ \text{N}, \qquad a = 0$$

3. (b) $F = 98$ N. Test: $98 = 98$ — it is exactly on the verge. Static friction is at its ceiling, and motion is impending but has not begun: $$f_s = 98\ \text{N}, \qquad a = 0$$

4. (c) $F = 120$ N. Test: $120 > 98$, so it breaks free and slides. Only now may you use the kinetic formula: $$f = f_k = 73.5\ \text{N}$$ $$a = \frac{120 - 73.5}{25} = \frac{46.5}{25} = 1.86\ \text{m/s}^2$$

5. (d) Constant speed once sliding. Constant velocity means $a = 0$, so the applied force must exactly balance kinetic friction: $$F = f_k = 73.5\ \text{N}$$

Final Answer: (a) $f = 80$ N, $a = 0$; (b) $f = 98$ N, $a = 0$; (c) $f = 73.5$ N, $a = 1.86$ m/s^2; (d) 73.5 N.

Takeaway: Part (d) contains the sentence that catches people out. It takes 98 N to start the crate and only 73.5 N to keep it going, because $\mu_k < \mu_s$. So a crate that has just been set moving will accelerate if you keep pushing with the 98 N that started it. [NEET Important] And burn in the sequence: compute $N$, compute $(f_s)_{max}$, compare, and only then choose your formula. Applying $f_k = \mu_k N$ to a stationary block is the single most common silent error in this chapter.

Example 25: From the angle of repose to a real slide

A block placed on a plank just begins to slide when the plank is tilted to $22°$. The plank is then raised to $35°$, where the coefficient of kinetic friction is 0.35. Take $g = 9.8$ m/s^2. Find (a) $\mu_s$, (b) the acceleration of the block down the $35°$ slope, (c) its speed after sliding 2.0 m from rest.

Solution:

1. (a) The angle of repose. At the angle at which sliding just begins, the down-slope weight component exactly equals the maximum static friction: $$mg\sin\theta_r = \mu_s\,mg\cos\theta_r \quad\Longrightarrow\quad \mu_s = \tan\theta_r = \tan 22° = 0.404$$ The mass cancels, which is why this is such a good way to measure $\mu_s$ — no force meter and no balance needed, just a protractor.

2. (b) First, the does-it-move test at $35°$. Compare the driving force with the maximum static friction, per kilogram: $$g\sin 35° = 9.8 \times 0.5736 = 5.62\ \text{N/kg} \quad\text{against}\quad \mu_s g\cos 35° = 0.404 \times 9.8 \times 0.8192 = 3.24\ \text{N/kg}$$ Since $5.62 > 3.24$ — equivalently $\tan 35° = 0.700 > \mu_s = 0.404$ — the block slides, and kinetic friction applies.

3. Now the second law along the slope, taking down-slope as positive: $$mg\sin 35° - \mu_k\,mg\cos 35° = ma$$ $$a = g(\sin 35° - \mu_k\cos 35°) = 9.8(0.5736 - 0.35 \times 0.8192)$$ $$a = 9.8(0.5736 - 0.2867) = 9.8 \times 0.2869 = 2.81\ \text{m/s}^2$$

4. (c) Speed after 2.0 m from rest, using $v^2 = u^2 + 2as$: $$v = \sqrt{2 \times 2.81 \times 2.0} = \sqrt{11.25} = 3.35\ \text{m/s}$$ (It takes $t = \sqrt{2s/a} = 1.19$ s to cover that distance.)

Final Answer: (a) $\mu_s = 0.404$; (b) $a = 2.81$ m/s^2 down the slope; (c) $v = 3.35$ m/s.

Takeaway: The angle-of-repose test $\tan\theta = \mu_s$ is the fastest experiment in the whole chapter. [JEE/NEET] Note the shortcut you now own for any slope: if $\tan\theta > \mu_s$ the block slides; if $\tan\theta \le \mu_s$ it stays put, regardless of its mass. Compare the tangent of the angle with $\mu_s$ and you have answered "does it move" in one line.

Example 26: Braking downhill, on the level, and uphill

A car travelling at 20 m/s brakes so hard that its wheels lock and it skids to a stop. The coefficient of kinetic friction is 0.6 and $g = 10$ m/s^2. Find the stopping distance (a) going down a $10°$ slope, (b) on level ground, (c) going up a $10°$ slope.

Solution:

1. Common set-up. Once the wheels lock, the car is a sliding block. On a slope of angle $\theta$, $N = mg\cos\theta$, so friction is $f_k = \mu_k mg\cos\theta$, always opposing the motion — that is, up the slope while the car moves down it. Gravity contributes $mg\sin\theta$ down the slope in every case. The mass will cancel throughout.

2. (a) Going downhill. Friction acts up the slope, gravity's component acts down it, so they fight: $$a = -g(\mu_k\cos 10° - \sin 10°) = -10(0.6 \times 0.9848 - 0.1736) = -10(0.5909 - 0.1736) = -4.17\ \text{m/s}^2$$ $$s = \frac{v^2}{2|a|} = \frac{400}{2 \times 4.17} = 47.9\ \text{m}$$

3. (b) On the level. No slope component at all: $$a = -\mu_k g = -6.0\ \text{m/s}^2, \qquad s = \frac{400}{12} = 33.3\ \text{m}$$

4. (c) Going uphill. Now friction acts down the slope (opposing the upward motion) and gravity's component also acts down the slope, so they cooperate: $$a = -g(\mu_k\cos 10° + \sin 10°) = -10(0.5909 + 0.1736) = -7.65\ \text{m/s}^2$$ $$s = \frac{400}{2 \times 7.65} = 26.2\ \text{m}$$

5. Read the three numbers together: 47.9 m, 33.3 m, 26.2 m. The downhill stopping distance is 1.8 times the uphill one, from a slope of just $10°$.

Final Answer: (a) 47.9 m; (b) 33.3 m; (c) 26.2 m.

Takeaway: Slope changes stopping distance far more than most people expect, and always in the dangerous direction going down. Note also that $s = \frac{v^2}{2\mu_k g}$ on the level has no mass in it — a loaded lorry and an empty car with the same tyres stop in the same distance — but it does have $v^2$, so doubling the speed quadruples the distance. [JEE Tip] Watch the sign of the $\sin\theta$ term: it subtracts from the retardation downhill and adds to it uphill. Getting that backwards is worth a wrong answer every time.

Example 27: The cheapest way to drag a crate

A 40 kg crate rests on a floor with $\mu_s = 0.5$. Take $g = 10$ m/s^2. It is to be dragged by a rope pulling at an angle $\theta$ above the horizontal. (a) What force is needed if you pull horizontally? (b) What angle needs the least force, and how much is it? (c) Compare with pulling at $45°$.

Solution:

1. Set up the general case. With the rope at $\theta$ above the horizontal, the upward component lightens the crate: $$N = mg - F\sin\theta$$ and on the verge of moving, the horizontal component equals the maximum static friction: $$F\cos\theta = \mu_s N = \mu_s(mg - F\sin\theta)$$

2. Solve for $F$: $$F(\cos\theta + \mu_s\sin\theta) = \mu_s mg \quad\Longrightarrow\quad \boxed{\,F = \frac{\mu_s mg}{\cos\theta + \mu_s\sin\theta}\,}$$

3. (a) Horizontal pull, $\theta = 0$: $$F = \frac{0.5 \times 400}{1} = 200\ \text{N}$$

4. (b) The best angle. $F$ is smallest when the denominator $\cos\theta + \mu_s\sin\theta$ is largest, and that expression peaks at $$\tan\theta = \mu_s \quad\Longrightarrow\quad \theta = \tan^{-1}(0.5) = 26.6°$$ which is the angle of friction. At that angle the denominator equals $\sqrt{1 + \mu_s^2}$, so $$F_{min} = \frac{\mu_s mg}{\sqrt{1 + \mu_s^2}} = \frac{200}{\sqrt{1.25}} = \frac{200}{1.1180} = 178.9\ \text{N}$$

5. (c) At $45°$: $$F = \frac{200}{0.7071 + 0.5 \times 0.7071} = \frac{200}{1.0607} = 188.6\ \text{N}$$ Worse than $26.6°$, better than horizontal.

6. The saving: $200 \to 178.9$ N is a reduction of 10.6%. Modest, but free.

Final Answer: (a) 200 N; (b) $\theta = 26.6°$ needing 178.9 N; (c) 188.6 N at $45°$.

Takeaway: The optimum angle is $\theta = \tan^{-1}\mu_s$, the angle of friction — the same angle that turns up as the angle of repose. That is not a coincidence: both mark the point where the total contact force (normal plus friction) is tilted as far from the vertical as the surfaces allow. [JEE Tip] The two results $F_{min} = \dfrac{\mu_s mg}{\sqrt{1+\mu_s^2}}$ and $\theta_{opt} = \tan^{-1}\mu_s$ are worth memorising outright; they are asked directly and they save a differentiation under exam pressure.