This MCQ module is based on: NCERT Exercises and Solutions: Laws of Motion
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NCERT Exercises and Solutions: Laws of Motion
🎓 Class 11
Physics
CBSE
Theory
Ch 4 – Laws of Motion
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Laws of Motion
📋 Chapter Summary
Newton's Laws of Motion — Key Takeaways
- Aristotle's fallacy: The notion that force is needed to maintain motion is wrong. Force is needed to change motion (to accelerate).
- Inertia: The natural property of objects to resist changes in their state of motion. Mass measures inertia.
- Newton's First Law: An object stays at rest or in uniform motion unless acted on by a net external force. Defines an inertial frame of reference.
- Linear momentum: \(\vec{p} = m\vec{v}\) — vector, in direction of velocity. SI unit: kg·m/s.
- Newton's Second Law: \(\vec{F} = d\vec{p}/dt = m\vec{a}\) (for constant mass). 1 N = 1 kg·m/s².
- Impulse: \(\vec{J} = \vec{F}\,\Delta t = \Delta\vec{p}\). Useful for forces over short times.
- Newton's Third Law: Action and reaction are equal in magnitude and opposite in direction, but act on DIFFERENT bodies (so they don't cancel).
- Conservation of Momentum: If \(F_{ext} = 0\), total momentum of an isolated system is conserved.
- Equilibrium: Net force = 0 (Newton's First Law). For three forces, they form a closed triangle.
- Friction: Static (\(0 \le f_s \le \mu_s N\)), kinetic (\(f_k = \mu_k N < f_{s,max}\)), rolling (much smaller). Angle of repose: tan θ = μ_s.
- Circular motion: Centripetal force \(F_c = mv^2/r\) towards center. On flat curves: \(v_{max} = \sqrt{\mu_s g R}\). On banked road: \(v_{opt} = \sqrt{Rg\tan\theta}\).
🔑 Key Terms & Formulas
InertiaResistance to change in motion. Measured by mass.
Momentum pp = mv (vector, kg·m/s)
Newton's 2nd LawF = ma = dp/dt
Impulse JJ = F·Δt = Δp
Newton's 3rd LawF_AB = −F_BA (different bodies)
Static friction0 ≤ f_s ≤ μ_s N (limiting)
Kinetic frictionf_k = μ_k N (μ_k < μ_s)
Angle of reposeμ_s = tan(θ_critical)
Centripetal forceF_c = mv²/r towards center
Banked road (no friction)v_opt = √(Rg tan θ)
Conservation of pIf F_ext = 0, p_total = const
EquilibriumΣ F_x = Σ F_y = Σ F_z = 0
📝 NCERT Exercises (Worked Solutions)
Exercise 4.1 — Net force in various scenarios
Give the magnitude and direction of the net force acting on:
(a) a drop of rain falling down with constant speed;
(b) a cork of mass 10 g floating on water;
(c) a kite skillfully held stationary in the sky;
(d) a car moving with constant velocity 30 km/h on rough road;
(e) a high-speed electron in space far from gravitating bodies and free of electric and magnetic fields.
In each case the body is either in uniform motion or at rest. By Newton's First Law, net force = 0.
(a) Rain drop at constant speed: net force = 0 (gravity + air drag balance).
(b) Floating cork: net force = 0 (gravity + buoyancy balance).
(c) Stationary kite: net force = 0 (gravity + tension + air force balance).
(d) Car at constant velocity: net force = 0 (engine thrust + friction balance).
(e) Free electron in deep space: net force = 0 (no fields, no gravitating bodies).
Insight: Constant velocity (or rest) ⇔ net force = zero (regardless of how complex the situation looks).
(a) Rain drop at constant speed: net force = 0 (gravity + air drag balance).
(b) Floating cork: net force = 0 (gravity + buoyancy balance).
(c) Stationary kite: net force = 0 (gravity + tension + air force balance).
(d) Car at constant velocity: net force = 0 (engine thrust + friction balance).
(e) Free electron in deep space: net force = 0 (no fields, no gravitating bodies).
Insight: Constant velocity (or rest) ⇔ net force = zero (regardless of how complex the situation looks).
Exercise 4.2 — Pebble in motion
A pebble of mass 0.05 kg is thrown vertically upwards. 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 your answers change if the pebble were thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
Throughout, the only force acting is gravity = mg downward.
\[F = mg = 0.05 \times 9.8 = \boxed{0.49 \text{ N, directed downward (vertically)}}\] This applies in all three cases (a, b, c) — direction does not change at the highest point.
Does it change for 45° throw? NO! Gravity always acts vertically downward regardless of launch angle. The motion path differs, but the only force is still 0.49 N down. (Without air resistance, motion is purely a parabola; with air, drag would be opposite to motion.)
\[F = mg = 0.05 \times 9.8 = \boxed{0.49 \text{ N, directed downward (vertically)}}\] This applies in all three cases (a, b, c) — direction does not change at the highest point.
Does it change for 45° throw? NO! Gravity always acts vertically downward regardless of launch angle. The motion path differs, but the only force is still 0.49 N down. (Without air resistance, motion is purely a parabola; with air, drag would be opposite to motion.)
Exercise 4.3 — Force on stone in lift / train
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg:
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at constant velocity 36 km/h,
(c) just after it is dropped from the window of a train accelerating with 1 m/s²,
(d) lying on the floor of a train accelerating with 1 m/s², the stone being at rest relative to the train.
(a) Stationary train: only gravity. F = mg = 0.1 × 9.8 = 0.98 N downward.
(b) Train at constant velocity (inertial frame): only gravity. F = 0.98 N downward.
(c) Train accelerating, stone DROPPED (no longer in contact): only gravity. F = 0.98 N downward. (Once free, the stone follows projectile motion in the inertial ground frame.)
(d) Stone on the floor, MOVING WITH the train (a = 1 m/s² horizontally): Net horizontal force = ma = 0.1 × 1 = 0.1 N in the direction of train's acceleration. (Vertical: gravity balances normal force, no net vertical force.)
Note: In (c), the stone has been released; the train's acceleration cannot affect it. In (d), friction from floor accelerates the stone with the train.
(b) Train at constant velocity (inertial frame): only gravity. F = 0.98 N downward.
(c) Train accelerating, stone DROPPED (no longer in contact): only gravity. F = 0.98 N downward. (Once free, the stone follows projectile motion in the inertial ground frame.)
(d) Stone on the floor, MOVING WITH the train (a = 1 m/s² horizontally): Net horizontal force = ma = 0.1 × 1 = 0.1 N in the direction of train's acceleration. (Vertical: gravity balances normal force, no net vertical force.)
Note: In (c), the stone has been released; the train's acceleration cannot affect it. In (d), friction from floor accelerates the stone with the train.
Exercise 4.4 — Particle on a string
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) equals: (i) T, (ii) T − mv²/l, (iii) T + mv²/l, (iv) 0. T is the tension in the string. Choose the correct alternative.
Answer: (i) T
The net inward (centripetal) force equals T (the only horizontal force pulling toward the centre). It must satisfy: T = mv²/l. The other options conflate the equation T = mv²/l with the net force itself.
The net inward (centripetal) force equals T (the only horizontal force pulling toward the centre). It must satisfy: T = mv²/l. The other options conflate the equation T = mv²/l with the net force itself.
Exercise 4.5 — Constant retardation
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body take to stop?
Given: F = -50 N, m = 20 kg, u = 15 m/s, v = 0.
\[a = F/m = -50/20 = -2.5 \text{ m/s}^2\] Using \(v = u + at\): \[0 = 15 + (-2.5)t \Rightarrow t = \boxed{6 \text{ s}}\]
\[a = F/m = -50/20 = -2.5 \text{ m/s}^2\] Using \(v = u + at\): \[0 = 15 + (-2.5)t \Rightarrow t = \boxed{6 \text{ s}}\]
Exercise 4.6 — Body acted on by force
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m/s to 3.5 m/s in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
\[a = (v - u)/t = (3.5 - 2.0)/25 = 0.06 \text{ m/s}^2\]
\[F = ma = 3.0 \times 0.06 = \boxed{0.18 \text{ N, in direction of motion}}\]
Exercise 4.7 — Two forces on a body
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Resultant force:
\[F_R = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \text{ N}\]
\[a = F_R/m = 10/5 = \boxed{2 \text{ m/s}^2}\]
Direction: \(\tan\theta = 6/8 = 0.75 \Rightarrow \theta = 36.87°\) with the 8 N force.
Exercise 4.8 — Driver applying brakes
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Total mass = 400 + 65 = 465 kg. Initial velocity = 36 km/h = 10 m/s.
\[a = (0 - 10)/4 = -2.5 \text{ m/s}^2\] \[F = ma = 465 \times (-2.5) = \boxed{-1162.5 \text{ N}}\] Magnitude ≈ 1163 N, opposite to direction of motion (retarding).
\[a = (0 - 10)/4 = -2.5 \text{ m/s}^2\] \[F = ma = 465 \times (-2.5) = \boxed{-1162.5 \text{ N}}\] Magnitude ≈ 1163 N, opposite to direction of motion (retarding).
Exercise 4.9 — Rocket equation
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m/s². Calculate the initial thrust (force) of the blast.
At lift-off: Thrust T − mg = ma, so T = m(g + a).
\[T = 20000 \times (9.8 + 5.0) = 20000 \times 14.8 = \boxed{2.96 \times 10^5 \text{ N}}\]
\[T = 20000 \times (9.8 + 5.0) = 20000 \times 14.8 = \boxed{2.96 \times 10^5 \text{ N}}\]
Exercise 4.10 — Body acted on by horizontal force
A body of mass 0.40 kg moving initially with constant speed of 10 m/s to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position at that instant to be x = 0. Predict its position at t = −5 s, 25 s, 100 s.
Setup: Take north as +x. m = 0.40 kg, u = +10 m/s, F = -8 N for 0 ≤ t ≤ 30 s.
Acceleration during force: a = -8/0.40 = -20 m/s².
At t = -5 s: No force, uniform motion. Position = -10 × 5 = -50 m (50 m south of origin).
At t = 25 s (within force interval): \[x = ut + \frac{1}{2}at^2 = 10(25) + 0.5(-20)(625) = 250 - 6250 = \boxed{-6000 \text{ m} = -6 \text{ km}}\]
At t = 100 s: Force acts only for 0 to 30 s. After that, uniform motion.
At t = 30 s: \(v = 10 + (-20)(30) = -590\) m/s. \(x(30) = 10(30) + 0.5(-20)(900) = 300 - 9000 = -8700\) m.
For 30 to 100 s: Δt = 70 s, no force. Δx = -590 × 70 = -41300 m.
Total: x(100) = -8700 + (-41300) = -50,000 m = -50 km.
Acceleration during force: a = -8/0.40 = -20 m/s².
At t = -5 s: No force, uniform motion. Position = -10 × 5 = -50 m (50 m south of origin).
At t = 25 s (within force interval): \[x = ut + \frac{1}{2}at^2 = 10(25) + 0.5(-20)(625) = 250 - 6250 = \boxed{-6000 \text{ m} = -6 \text{ km}}\]
At t = 100 s: Force acts only for 0 to 30 s. After that, uniform motion.
At t = 30 s: \(v = 10 + (-20)(30) = -590\) m/s. \(x(30) = 10(30) + 0.5(-20)(900) = 300 - 9000 = -8700\) m.
For 30 to 100 s: Δt = 70 s, no force. Δx = -590 × 70 = -41300 m.
Total: x(100) = -8700 + (-41300) = -50,000 m = -50 km.
Exercise 4.11 — Truck velocity time graph
A truck starts from rest and accelerates uniformly at 2.0 m/s². At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
At t = 10 s, truck (and stone) horizontal velocity: v_x = 0 + 2 × 10 = 20 m/s.
After dropping at t = 10 s, the stone has: - Horizontal velocity 20 m/s (constant, no air drag) - Initial vertical velocity 0, then falls under gravity.
At t = 11 s (i.e., 1 s after release):
v_y = 0 + g × 1 = 9.8 m/s downward.
v_x = 20 m/s (unchanged).
(a) Velocity: \[|\vec{v}| = \sqrt{20^2 + 9.8^2} = \sqrt{496} \approx \boxed{22.27 \text{ m/s}}\] Angle: tan θ = 9.8/20 ≈ 0.49 ⇒ θ ≈ 26.1° below horizontal.
(b) Acceleration: Once dropped, the only force on stone is gravity. So \[a = \boxed{9.8 \text{ m/s}^2 \text{ downward}}\] The truck's horizontal acceleration is irrelevant after stone leaves contact.
After dropping at t = 10 s, the stone has: - Horizontal velocity 20 m/s (constant, no air drag) - Initial vertical velocity 0, then falls under gravity.
At t = 11 s (i.e., 1 s after release):
v_y = 0 + g × 1 = 9.8 m/s downward.
v_x = 20 m/s (unchanged).
(a) Velocity: \[|\vec{v}| = \sqrt{20^2 + 9.8^2} = \sqrt{496} \approx \boxed{22.27 \text{ m/s}}\] Angle: tan θ = 9.8/20 ≈ 0.49 ⇒ θ ≈ 26.1° below horizontal.
(b) Acceleration: Once dropped, the only force on stone is gravity. So \[a = \boxed{9.8 \text{ m/s}^2 \text{ downward}}\] The truck's horizontal acceleration is irrelevant after stone leaves contact.
Exercise 4.12 — Bob in trolley
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m/s. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position?
(a) At extreme position: Bob momentarily at rest. After cutting string, it falls vertically under gravity. Trajectory: vertical straight line (free fall).
(b) At mean position: Bob has horizontal velocity 1 m/s. After cutting, it has horizontal velocity + downward gravity. Trajectory: parabolic path (projectile motion).
(b) At mean position: Bob has horizontal velocity 1 m/s. After cutting, it has horizontal velocity + downward gravity. Trajectory: parabolic path (projectile motion).
Exercise 4.13 — Man inside lift
A man of mass 70 kg stands on a weighing scale in a lift which is moving:
(a) upwards with a uniform speed of 10 m/s;
(b) downwards with a uniform acceleration of 5 m/s²;
(c) upwards with a uniform acceleration of 5 m/s².
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
The scale reads the normal force N (apparent weight). Take g = 10 m/s².
(a) Constant speed: a = 0. N = mg = 70 × 10 = 700 N.
(b) Downward accel 5: N = m(g − a) = 70 × 5 = 350 N (feels lighter).
(c) Upward accel 5: N = m(g + a) = 70 × 15 = 1050 N (feels heavier).
(d) Free fall: a = g downward. N = m(g − g) = 0 N (weightlessness — astronauts in orbit experience this).
(a) Constant speed: a = 0. N = mg = 70 × 10 = 700 N.
(b) Downward accel 5: N = m(g − a) = 70 × 5 = 350 N (feels lighter).
(c) Upward accel 5: N = m(g + a) = 70 × 15 = 1050 N (feels heavier).
(d) Free fall: a = g downward. N = m(g − g) = 0 N (weightlessness — astronauts in orbit experience this).
Exercise 4.14 — Particle position graph
The position-time (x-t) graph of a particle of mass 4 kg is shown to be a piecewise linear graph (constant velocity in segments). (i) What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s? Take x = 0, t = 0 as origin.
For piecewise linear x-t (constant slopes), velocity is constant in each segment, so acceleration = 0. Hence force = 0 in each segment.
For t < 0: F = 0 (uniform motion or rest).
For 0 < t < 4 s: F = 0.
For t > 4 s: F = 0.
At t = 0 and t = 4 s (slope changes), there must be impulsive forces (instantaneous Δp). The impulse equals m × (Δv) at each break point.
For t < 0: F = 0 (uniform motion or rest).
For 0 < t < 4 s: F = 0.
For t > 4 s: F = 0.
At t = 0 and t = 4 s (slope changes), there must be impulsive forces (instantaneous Δp). The impulse equals m × (Δv) at each break point.
Exercise 4.15 — Two bodies linked, force on one
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Total mass = 30 kg. System acceleration: a = F/M = 600/30 = 20 m/s².
(i) F applied to 10-kg body A (B trails): Tension T pulls B forward. \[T = m_B \times a = 20 \times 20 = \boxed{400 \text{ N}}\]
(ii) F applied to 20-kg body B (A trails): Tension T pulls A forward. \[T = m_A \times a = 10 \times 20 = \boxed{200 \text{ N}}\]
Tension differs because in each case the tension only has to accelerate the trailing mass.
(i) F applied to 10-kg body A (B trails): Tension T pulls B forward. \[T = m_B \times a = 20 \times 20 = \boxed{400 \text{ N}}\]
(ii) F applied to 20-kg body B (A trails): Tension T pulls A forward. \[T = m_A \times a = 10 \times 20 = \boxed{200 \text{ N}}\]
Tension differs because in each case the tension only has to accelerate the trailing mass.
Exercise 4.16 — Two masses suspended over pulley
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
For the heavier mass m₂ = 12 kg (going down): m₂g − T = m₂a
For the lighter mass m₁ = 8 kg (going up): T − m₁g = m₁a
Adding: \((m_2 - m_1)g = (m_1 + m_2)a\) \[a = \frac{(12-8)\times 9.8}{12+8} = \frac{4 \times 9.8}{20} = \boxed{1.96 \text{ m/s}^2}\]
For tension: \[T = m_1(g + a) = 8 \times (9.8 + 1.96) = 8 \times 11.76 = \boxed{94.1 \text{ N}}\]
For the lighter mass m₁ = 8 kg (going up): T − m₁g = m₁a
Adding: \((m_2 - m_1)g = (m_1 + m_2)a\) \[a = \frac{(12-8)\times 9.8}{12+8} = \frac{4 \times 9.8}{20} = \boxed{1.96 \text{ m/s}^2}\]
For tension: \[T = m_1(g + a) = 8 \times (9.8 + 1.96) = 8 \times 11.76 = \boxed{94.1 \text{ N}}\]
Exercise 4.17 — Decay of nucleus
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.
Initial momentum = 0 (nucleus at rest). By conservation of momentum, final total momentum = 0.
\[m_1 \vec{v}_1 + m_2 \vec{v}_2 = 0\] \[\vec{v}_2 = -\frac{m_1}{m_2}\vec{v}_1\] The negative sign shows \(\vec{v}_2\) is opposite to \(\vec{v}_1\). ∎
This is the basis for nuclear fission products always recoiling in opposite directions — a direct consequence of momentum conservation.
\[m_1 \vec{v}_1 + m_2 \vec{v}_2 = 0\] \[\vec{v}_2 = -\frac{m_1}{m_2}\vec{v}_1\] The negative sign shows \(\vec{v}_2\) is opposite to \(\vec{v}_1\). ∎
This is the basis for nuclear fission products always recoiling in opposite directions — a direct consequence of momentum conservation.
Exercise 4.18 — Two billiard balls collision
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
For a single ball, initial velocity = +6 m/s, final = -6 m/s.
Impulse on this ball = Δp = m(v_f − v_i) = 0.05 × (-6 − 6) = -0.6 kg·m/s.
Magnitude = 0.6 kg·m/s. The other ball receives equal and opposite impulse: +0.6 kg·m/s. By Newton's Third Law, both balls receive equal magnitude impulses.
Impulse on this ball = Δp = m(v_f − v_i) = 0.05 × (-6 − 6) = -0.6 kg·m/s.
Magnitude = 0.6 kg·m/s. The other ball receives equal and opposite impulse: +0.6 kg·m/s. By Newton's Third Law, both balls receive equal magnitude impulses.
Exercise 4.19 — Shell explosion mid-flight
A shell of mass 0.020 kg is fired by 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?
Conservation of momentum (initially at rest):
\[m_s v_s + M_g V_g = 0\]
\[V_g = -\frac{m_s v_s}{M_g} = -\frac{0.020 \times 80}{100} = \boxed{-0.016 \text{ m/s}}\]
Magnitude = 1.6 cm/s, direction opposite to shell.
Exercise 4.20 — Batsman strikes ball
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Speed: 54 km/h = 15 m/s. The ball deflects by 45° — this means each velocity component (incident and reflected) makes 22.5° with the bat's normal.
Method: Take the bat's normal as the x-axis. By symmetry (elastic deflection):
- Incident: v at angle 22.5° to normal, speed 15 m/s.
- Reflected: v at angle 22.5° on other side of normal.
- v_x flips sign; v_y unchanged.
Change in v_x = 2 × 15 × cos(22.5°) = 30 × 0.924 = 27.72 m/s.
Impulse magnitude = m × Δv_x = 0.15 × 27.72 = 4.16 N·s.
The direction of impulse is along the bat's normal (i.e., perpendicular to the deflecting surface).
Method: Take the bat's normal as the x-axis. By symmetry (elastic deflection):
- Incident: v at angle 22.5° to normal, speed 15 m/s.
- Reflected: v at angle 22.5° on other side of normal.
- v_x flips sign; v_y unchanged.
Change in v_x = 2 × 15 × cos(22.5°) = 30 × 0.924 = 27.72 m/s.
Impulse magnitude = m × Δv_x = 0.15 × 27.72 = 4.16 N·s.
The direction of impulse is along the bat's normal (i.e., perpendicular to the deflecting surface).
Exercise 4.21 — Stone whirled in horizontal circle
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40/π rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Frequency: f = (40/π)/60 = 40/(60π) = 2/(3π) rev/s.
Angular speed: ω = 2πf = 2π × 2/(3π) = 4/3 rad/s.
(a) Tension at given speed: \[T = m\omega^2 r = 0.25 \times (4/3)^2 \times 1.5 = 0.25 \times 16/9 \times 1.5 = \boxed{0.667 \text{ N}}\]
(b) Max speed with T_max = 200 N: \[T_{max} = \frac{m v_{max}^2}{r}\] \[v_{max} = \sqrt{\frac{T_{max} \cdot r}{m}} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{1200} \approx \boxed{34.64 \text{ m/s}}\]
Angular speed: ω = 2πf = 2π × 2/(3π) = 4/3 rad/s.
(a) Tension at given speed: \[T = m\omega^2 r = 0.25 \times (4/3)^2 \times 1.5 = 0.25 \times 16/9 \times 1.5 = \boxed{0.667 \text{ N}}\]
(b) Max speed with T_max = 200 N: \[T_{max} = \frac{m v_{max}^2}{r}\] \[v_{max} = \sqrt{\frac{T_{max} \cdot r}{m}} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{1200} \approx \boxed{34.64 \text{ m/s}}\]
Exercise 4.22 — Stone after string breaks
If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Answer: (b) The stone flies off tangentially at the instant the string breaks.
Explanation: The instantaneous velocity of an object in circular motion is always tangent to the circle. Once the centripetal force (string tension) vanishes, no force changes the direction. So the stone flies off in a straight line, in the direction of its instantaneous velocity — i.e., along the tangent.
This is also why a hammer thrower releases the hammer at the optimal moment to send it in the desired direction (tangent at release point).
Explanation: The instantaneous velocity of an object in circular motion is always tangent to the circle. Once the centripetal force (string tension) vanishes, no force changes the direction. So the stone flies off in a straight line, in the direction of its instantaneous velocity — i.e., along the tangent.
This is also why a hammer thrower releases the hammer at the optimal moment to send it in the desired direction (tangent at release point).
Exercise 4.23 — Reaction forces in special situations
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.
(a) A horse pulls a cart by pushing the ground backward via its hooves; the ground pushes back (friction) which propels horse + cart. In empty space, there's no surface to push against, so no propulsion.
(b) Inertia of motion. The passenger's body shares the bus's velocity. When bus brakes, the bus slows but the body tends to continue at original velocity → throws forward.
(c) When pulling, the applied force has an upward component that REDUCES the normal force on the ground, hence reducing friction. When pushing, the applied force pushes the mower DOWN, increasing normal force and friction. So pulling needs less force.
(d) Moving hands backward increases the time interval Δt over which the ball decelerates. Since impulse Δp is fixed, the average force F = Δp/Δt decreases — gentler impact, less pain. Same physics as airbags.
(b) Inertia of motion. The passenger's body shares the bus's velocity. When bus brakes, the bus slows but the body tends to continue at original velocity → throws forward.
(c) When pulling, the applied force has an upward component that REDUCES the normal force on the ground, hence reducing friction. When pushing, the applied force pushes the mower DOWN, increasing normal force and friction. So pulling needs less force.
(d) Moving hands backward increases the time interval Δt over which the ball decelerates. Since impulse Δp is fixed, the average force F = Δp/Δt decreases — gentler impact, less pain. Same physics as airbags.
🎓 Chapter 4 Complete! You've now mastered Newton's three laws — the foundation of classical mechanics. These principles govern everything from a falling apple to satellite orbits. The next chapter (Work, Energy and Power) builds on these laws to introduce powerful concepts of energy and conservation. Keep practicing problems and visualising forces!
Frequently Asked Questions - NCERT Exercises and Solutions: Laws of Motion
What are the key NCERT exercise types in Chapter 4 Laws of Motion?
NCERT Class 11 Physics Chapter 4 Laws of Motion exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Laws of Motion?
For numerical problems in NCERT Class 11 Physics Chapter 4 Laws of Motion: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 4?
From NCERT Class 11 Physics Chapter 4 (Laws of Motion), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 4 Laws of Motion problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 4 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 4 Laws of Motion exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 4 Laws of Motion solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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