Laws of Motion — Easy Practice (Set 1)

Quick Recap — Newton's Laws & Momentum

• First law (inertia): a body stays at rest or in uniform motion unless acted on by a net external force.
• Second law: $\vec{F} = m\vec{a} = \dfrac{d\vec{p}}{dt}$; the SI unit of force is the newton (N).
• Third law: every action has an equal and opposite reaction, and the two act on different bodies.
• Momentum $\vec{p} = m\vec{v}$ (unit kg m/s); impulse $= \vec{F}\,\Delta t = \Delta\vec{p}$ (unit N s).
• Conservation of momentum: with no net external force, the total momentum stays constant.
• Friction: opposes relative motion; static $f_s \le \mu_s N$, kinetic $f_k = \mu_k N$.

Beyond-NCERT JEE Formulae

These extend the Newton's-law and momentum basics of the earlier sets and go straight to the tools JEE loves to test.

1. Friction: angle of friction, angle of repose, and the smart pull

The total contact force (normal $N$ plus friction $f$) tilts away from the surface normal by the angle of friction $\lambda$, with $\tan\lambda=\mu$. On an incline a resting block is on the verge of slipping at the angle of repose $\alpha$; setting $mg\sin\alpha=\mu mg\cos\alpha$ gives $\tan\alpha=\mu$. So the two are one and the same angle: $\alpha=\lambda=\tan^{-1}\mu$.

Block on a rough incline (angle $\theta$).

• Stays put with no applied force whenever $\theta\le\alpha$; then friction is self-adjusting, $f=mg\sin\theta$ (not $\mu mg\cos\theta$).
• Sliding freely down: $a=g(\sin\theta-\mu\cos\theta)$.
• Force applied ALONG the plane to drag it up: $F=mg(\sin\theta+\mu\cos\theta)$; to just stop it sliding down: $F=mg(\sin\theta-\mu\cos\theta)$.

Least force to pull a block along the floor. Pulling at angle $\theta$ above the horizontal, the force needed is $F=\dfrac{\mu mg}{\cos\theta+\mu\sin\theta}$. The denominator peaks at $\tan\theta=\mu$, so the pull is smallest at that angle, $\theta=\tan^{-1}\mu$ (exactly the angle of friction), where

$$F_{min}=\dfrac{\mu mg}{\sqrt{1+\mu^2}}.$$

[JEE Tip] Pull, do not push. A pull at angle $\theta$ eases the normal reaction to $N=mg-F\sin\theta$, whereas a push angled below the horizontal raises it to $N=mg+F\sin\theta$ and jams the block down harder. Always resolve the applied force into the normal direction before writing $f=\mu N$.

2. Pseudo force and apparent weight (non-inertial frames)

Sitting in a frame that accelerates with $\vec a_0$, tack a pseudo-force $-m\vec a_0$ onto every body and then solve as if the frame were at rest.

• Lift / apparent weight. The floor or spring balance reads $N=m(g+a)$ when accelerating up (or slowing while descending), $N=m(g-a)$ when accelerating down (or slowing while rising), and $N=0$ in free fall.
• Effective gravity. For a pendulum or a block on a smooth incline inside a lift accelerating up, simply swap $g$ for $g_{eff}=g+a_0$; the normal reaction, the sliding acceleration and the time period all scale with $g_{eff}$.

[JEE Tip] It is the direction of the ACCELERATION that decides heavy or light, never the direction of motion. A lift moving up but braking has $a$ pointing down, so you feel lighter, not heavier.

3. Banking of roads WITH friction

The frictionless design speed is $v_0=\sqrt{rg\tan\theta}$. Once friction $\mu$ is available the safe range widens on both sides:

$$v_{max}=\sqrt{\dfrac{rg(\tan\theta+\mu)}{1-\mu\tan\theta}},\qquad v_{min}=\sqrt{\dfrac{rg(\tan\theta-\mu)}{1+\mu\tan\theta}}.$$

Above $v_{max}$ the car tends to slide OUT and up the bank, so friction points down the slope; below $v_{min}$ it tends to slip in and down, so friction points up. A flat road ($\theta=0$) collapses this to $v_{max}=\sqrt{\mu rg}$.

[JEE Tip] If $\mu\tan\theta\ge1$ the denominator of $v_{max}$ vanishes: the bend is so steep and grippy that there is no upper speed limit at all. And if $\tan\theta\le\mu$ there is no lower limit either, so a car may even park on the slope without sliding down.

4. Constraint relations (string, pulley, wedge)

For a single inextensible string the velocity components of its two ends ALONG the string are equal:

$$v_1\cos\theta_1=v_2\cos\theta_2.$$

• Movable pulley. Two rope segments carry it, so the effort is halved while the free end moves twice as fast as the pulley: $a_{free}=2a_{pulley}$.
• Wedge and block. A block kept on the moving face of a wedge must share the wedge's motion perpendicular to that face, which ties the block's acceleration to the wedge's.

[JEE Tip] Never guess the acceleration link. Write 'total string length equals a constant' and differentiate twice; the relation that drops out is exact, even for slanted ropes or stacked pulleys.

5. Connected bodies over a pulley (Atwood and its cousins)

Ideal Atwood machine ($m_1>m_2$, light string, frictionless pulley):

$$a=\dfrac{(m_1-m_2)g}{m_1+m_2},\qquad T=\dfrac{2m_1m_2g}{m_1+m_2}.$$

• The hook holding the pulley feels $2T=\dfrac{4m_1m_2g}{m_1+m_2}$, which is LESS than the dead weight $(m_1+m_2)g$ — a suspended Atwood machine reads light while it runs.
• Block on a smooth table pulled by a hanging mass: $a=\dfrac{m_2g}{m_1+m_2}$, $T=\dfrac{m_1m_2g}{m_1+m_2}$.
• Rough table (coefficient $\mu$ under the table block $m_1$): $a=\dfrac{(m_2-\mu m_1)g}{m_1+m_2}$, and it moves only if $m_2>\mu m_1$.

[JEE Tip] In any Atwood set-up the tension is trapped between the two weights, $m_2g<T<m_1g$. If your $T$ escapes that band, a sign is wrong.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Least force to drag a block

A 10 kg crate rests on a floor with coefficient of friction $\mu=0.75$. Find the smallest force that can drag it along the floor and the angle at which it must act. Take $g=10$ m/s$^2$.

Set-up. A pull at angle $\theta$ above the horizontal needs $F=\dfrac{\mu mg}{\cos\theta+\mu\sin\theta}$, which is least when $\tan\theta=\mu$.

Working. The best angle is $\theta=\tan^{-1}0.75\approx37^\circ$. Then $F_{min}=\dfrac{\mu mg}{\sqrt{1+\mu^2}}=\dfrac{0.75\times10\times10}{\sqrt{1+0.5625}}=\dfrac{75}{1.25}=60$ N. A flat horizontal pull would instead have needed $\mu mg=75$ N, so slanting the rope to $37^\circ$ saves 15 N.

Answer. About 60 N, applied at roughly $37^\circ$ above the horizontal.

Example 2 — Banking with friction: fastest and slowest safe speeds

A bend of radius 30 m is banked at $45^\circ$, and the tyre-road friction coefficient is $\mu=0.5$. Take $g=10$ m/s$^2$. Find the greatest and least speeds for rounding it without skidding.

Set-up. With $\tan45^\circ=1$, use $v_{max}=\sqrt{\dfrac{rg(\tan\theta+\mu)}{1-\mu\tan\theta}}$ and $v_{min}=\sqrt{\dfrac{rg(\tan\theta-\mu)}{1+\mu\tan\theta}}$.

Working. $v_{max}=\sqrt{\dfrac{30\times10\times(1+0.5)}{1-0.5}}=\sqrt{\dfrac{300\times1.5}{0.5}}=\sqrt{900}=30$ m/s. Then $v_{min}=\sqrt{\dfrac{300\times(1-0.5)}{1+0.5}}=\sqrt{\dfrac{150}{1.5}}=\sqrt{100}=10$ m/s. The frictionless design speed $\sqrt{rg\tan\theta}=\sqrt{300}\approx17.3$ m/s lies neatly between the two.

Answer. $v_{max}=30$ m/s and $v_{min}=10$ m/s.

Example 3 — Apparent weight in a lift

A 60 kg passenger stands on a weighing scale inside a lift. Find the reading when the lift (a) accelerates upward at 2 m/s$^2$, (b) accelerates downward at 2 m/s$^2$, (c) falls freely after the cable snaps. Take $g=10$ m/s$^2$.

Set-up. With up taken positive, the scale reads $N$ where $N-mg=ma$, so $N=m(g+a)$.

Working. (a) Accelerating up, $N=60\times(10+2)=720$ N. (b) Accelerating down, $a=-2$, so $N=60\times(10-2)=480$ N. (c) Free fall means $a=-g$, giving $N=60\times(10-10)=0$ — the passenger is momentarily weightless.

Answer. 720 N, 480 N, and 0.

Example 4 — Atwood machine: acceleration, tension, and pull on the hook

Blocks of 6 kg and 4 kg hang from the ends of a light string over a frictionless pulley. Find the acceleration, the string tension, and the force on the hook that holds the pulley. Take $g=10$ m/s$^2$.

Set-up. Use $a=\dfrac{(m_1-m_2)g}{m_1+m_2}$ and $T=\dfrac{2m_1m_2g}{m_1+m_2}$; the hook carries both string segments, so it feels $2T$.

Working. $a=\dfrac{(6-4)\times10}{6+4}=2$ m/s$^2$ and $T=\dfrac{2\times6\times4\times10}{10}=48$ N. Cross-check: for the light block $T=m_2(g+a)=4\times12=48$ N, and for the heavy one $T=m_1(g-a)=6\times8=48$ N. The hook feels $2T=96$ N, less than the dead weight $(6+4)\times10=100$ N.

Answer. $a=2$ m/s$^2$, $T=48$ N, and the hook carries 96 N.

Example 5 — Constraint: velocity components along a rope

A boat is winched towards a jetty by a rope running over a bollard 4 m above the water. At the instant the straight rope from boat to bollard is 5 m long, it is being hauled in at 2 m/s. How fast is the boat then moving?

Set-up. Only the component of the boat's velocity ALONG the rope shortens it, so $v_{boat}\cos\theta=v_{rope}$, where $\theta$ is the angle the rope makes with the water.

Working. With the rope 5 m long and the bollard 4 m up, the horizontal reach is $\sqrt{5^2-4^2}=3$ m, so $\cos\theta=\dfrac{3}{5}=0.6$. Then $v_{boat}=\dfrac{v_{rope}}{\cos\theta}=\dfrac{2}{0.6}=\dfrac{10}{3}\approx3.33$ m/s. The boat outruns the rope, and it speeds up further as it nears the jetty and $\theta$ grows towards a right angle.

Answer. $\dfrac{10}{3}\approx3.33$ m/s.

Example 6 — Wedge constraint: pushing a wedge so its block does not slide

A block sits on the smooth $37^\circ$ face of a wedge. With what horizontal acceleration must the wedge be pushed so that the block neither climbs nor slips on the face? Also find the normal reaction on a 2 kg block. Take $g=10$ m/s$^2$, $\sin37^\circ=0.6$, $\cos37^\circ=0.8$.

Set-up. In the wedge's frame the block feels a backward pseudo-force $ma$. For no sliding, balance the forces along the incline: $ma\cos\theta=mg\sin\theta$, so $a=g\tan\theta$.

Working. $a=g\tan37^\circ=10\times0.75=7.5$ m/s$^2$. The normal reaction, perpendicular to the face, is $N=mg\cos\theta+ma\sin\theta=2\times(10\times0.8+7.5\times0.6)=2\times12.5=25$ N — matching the tidy form $N=\dfrac{mg}{\cos\theta}=\dfrac{20}{0.8}=25$ N.

Answer. Push the wedge at 7.5 m/s$^2$; the normal reaction is 25 N.

Example 7 — Dragging up versus holding on a rough incline

A 5 kg block lies on a rough incline of $37^\circ$ with $\mu=0.5$. Find the force along the plane needed to (a) drag it steadily up and (b) just prevent it sliding down, and the ratio of the two. Take $g=10$ m/s$^2$, $\sin37^\circ=0.6$, $\cos37^\circ=0.8$.

Set-up. Friction reverses between the cases: dragging up it points down the slope, holding it points up. So $F_1=mg(\sin\theta+\mu\cos\theta)$ and $F_2=mg(\sin\theta-\mu\cos\theta)$.

Working. Since $\tan37^\circ=0.75>\mu=0.5$, the block would indeed run down on its own, so it genuinely needs holding. With $mg=50$ N: $F_1=50\times(0.6+0.5\times0.8)=50\times1.0=50$ N and $F_2=50\times(0.6-0.5\times0.8)=50\times0.2=10$ N. Their ratio is $\dfrac{F_1}{F_2}=5$.

Answer. $F_1=50$ N up the plane, $F_2=10$ N, ratio 5.