Laws of Motion Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1(a) The raindrop falls at constant speed, so its acceleration is zero: net force = 0.
2. 2(b) The cork's weight downward is balanced by the upthrust of water: net force = 0.
3. 3(c) The kite is stationary (zero acceleration): net force = 0 by Newton's first law.
4. 4(d) Constant velocity means zero acceleration: net force = 0.
5. 5(e) Far from material objects and free of fields, nothing acts on the electron: net force = 0.

Step-by-step solution

1. 1The only force acting on the pebble is gravity, which acts vertically downward in all three cases.
2. 2F = ma = mg = 0.05 × 10 = 0.5 N, directed downward.
3. 3(a) During upward motion: 0.5 N downward.
4. 4(b) During downward motion: 0.5 N downward.
5. 5(c) At the highest point: still 0.5 N downward (acceleration due to gravity is constant).
6. 6If thrown at 45°, only the vertical component of velocity is zero at the top — gravity still acts downward with the same magnitude, so the answers do not change.

Step-by-step solution

1. 1(a) Just after release, only gravity acts: F = mg = 0.1 × 10 = 1 N, vertically downward.
2. 2(b) The train moves at constant velocity, so there is no horizontal force; once released, the stone's horizontal motion does not change. Net force = 1 N, vertically downward.
3. 3(c) The train accelerates at 1 m s⁻², but the force causing that acceleration acts between train and train — it is not transmitted to the released stone. After release, F = mg = 1 N, vertically downward.
4. 4(d) The stone at rest on the accelerating floor receives the full horizontal push: F = ma = 0.1 × 1 = 0.1 N in the direction of motion of the train (the vertical weight is balanced by the floor's normal reaction).

Step-by-step solution

1. 1On a smooth horizontal table, the weight and the normal reaction cancel vertically.
2. 2The only horizontal force on the particle is the tension T of the string.
3. 3This tension alone provides the centripetal force mv²/l towards the centre.
4. 4Hence the net force on the particle is T itself. (Option i).

Step-by-step solution

1. 1Retarding force F = −50 N (negative — opposing motion).
2. 2Using F = ma, the deceleration a = F/m = −50/20 = −2.5 m s⁻².
3. 3From v = u + at with v = 0 and u = 15 m s⁻¹: t = −u/a = −15/(−2.5) = 6 s.

Step-by-step solution

1. 1a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m s⁻².
2. 2F = ma = 3.0 × 0.06 = 0.18 N.
3. 3Since the direction of motion is unchanged, the force acts in the direction of motion.

Step-by-step solution

1. 1Resultant of the two perpendicular forces: R = √(8² + 6²) = √100 = 10 N.
2. 2Direction: θ = tan⁻¹(6/8) = 36.87° with the 8 N force.
3. 3Acceleration a = F/m = 10/5 = 2 m s⁻².
4. 4The acceleration acts along the resultant force, at 36.87° with the 8 N force.

Step-by-step solution

1. 1u = 36 km/h = 10 m s⁻¹; v = 0; total mass = 400 + 65 = 465 kg.
2. 2a = (v − u)/t = (0 − 10)/4 = −2.5 m s⁻².
3. 3F = ma = 465 × (−2.5) = −1162.5 N.
4. 4The average retarding force is 1162.5 N, opposing the motion.

Step-by-step solution

1. 1Upward equation of motion for the rocket: F − mg = ma.
2. 2F = m(g + a) = 20,000 × (10 + 5.0).
3. 3F = 20,000 × 15 = 3 × 10⁵ N.

Step-by-step solution

1. 1Acceleration due to the force: a = F/m = −8.0/0.40 = −20 m s⁻² (southward; north taken positive).
2. 2At t = −5 s (before the force is applied, a′ = 0): x = ut = 10 × (−5) = −50 m.
3. 3At t = 25 s: x = ut + ½at² = 10 × 25 + ½(−20)(25)² = 250 − 6250 = −6000 m.
4. 4From t = 0 to 30 s: x₁ = 10 × 30 + ½(−20)(30)² = 300 − 9000 = −8700 m.
5. 5Velocity at t = 30 s: v = u + at = 10 − 20 × 30 = −590 m s⁻¹.
6. 6From t = 30 s to 100 s (force ends, a′ = 0): x₂ = vt = −590 × 70 = −41300 m.
7. 7At t = 100 s: x = x₁ + x₂ = −8700 − 41300 = −50000 m.

Step-by-step solution

1. 1Velocity of truck at t = 10 s: v = u + at = 0 + 2 × 10 = 20 m s⁻¹ — this is the stone's horizontal velocity at release.
2. 2At t = 11 s the horizontal velocity is unchanged (no horizontal force): vₓ = 20 m s⁻¹.
3. 3Vertical velocity: vᵧ = uᵧ + aᵧ Δt = 0 + 10 × (1) = 10 m s⁻¹ downward.
4. 4Resultant speed: v = √(vₓ² + vᵧ²) = √(20² + 10²) = √500 ≈ 22.36 m s⁻¹.
5. 5Direction: θ = tan⁻¹(vᵧ/vₓ) = tan⁻¹(10/20) = 26.57° with the horizontal (direction of the truck's motion).
6. 6(b) After release only gravity acts: acceleration = g = 10 m s⁻² vertically downward.

Step-by-step solution

1. 1(a) At an extreme position the bob is momentarily at rest (speed = 0). If the string is cut, only gravity acts, so the bob falls vertically downward with acceleration g.
2. 2(b) At the mean position the bob moves horizontally at 1 m s⁻¹, tangential to its arc.
3. 3After the string is cut, it keeps that horizontal velocity and accelerates downward under gravity.
4. 4This combination (constant horizontal velocity + uniform vertical acceleration) gives a parabolic trajectory.

Step-by-step solution

1. 1(a) Uniform speed ⇒ a = 0: R = mg = 70 × 10 = 700 N; reading = 700/10 = 70 kg.
2. 2(b) Downward acceleration: R = m(g − a) = 70(10 − 5) = 350 N; reading = 35 kg.
3. 3(c) Upward acceleration: R = m(g + a) = 70(10 + 5) = 1050 N; reading = 105 kg.
4. 4(d) Free fall: a = g ⇒ R = m(g − g) = 0; reading = 0 (the man is weightless).

Step-by-step solution

1. 1(a) For t < 0: the displacement is zero (position coincides with the time axis), so the velocity and force are both zero.
2. 2For 0 < t < 4 s: the x-t graph is a straight line with constant slope — constant velocity, zero acceleration, hence zero force.
3. 3For t > 4 s: the position stays at 3 m (graph parallel to time axis) — the particle is at rest, so again zero force.
4. 4(b) At t = 0 the velocity jumps from 0 to the slope value 3/4 m s⁻¹: impulse = m(v − u) = 4(3/4 − 0) = 3 kg m s⁻¹.
5. 5At t = 4 s the velocity drops from 3/4 m s⁻¹ to 0: impulse = 4(0 − 3/4) = −3 kg m s⁻¹.

Step-by-step solution

1. 1Total mass m = m_A + m_B = 10 + 20 = 30 kg.
2. 2System acceleration a = F/m = 600/30 = 20 m s⁻².
3. 3(i) Force applied to A: F − T = m_A a ⇒ T = 600 − 10 × 20 = 400 N.
4. 4(ii) Force applied to B: F − T = m_B a ⇒ T = 600 − 20 × 20 = 200 N.

Step-by-step solution

1. 1For the 8 kg mass going up: T − m₁g = m₁a.
2. 2For the 12 kg mass going down: m₂g − T = m₂a.
3. 3Adding: (m₂ − m₁)g = (m₁ + m₂)a.
4. 4a = (12 − 8)/(12 + 8) × 10 = 4/20 × 10 = 2 m s⁻².
5. 5T = (2m₁m₂/(m₁ + m₂))g = (2 × 12 × 8)/20 × 10 = 96 N.

Step-by-step solution

1. 1The parent nucleus is at rest, so the initial linear momentum of the system is zero.
2. 2Let m₁, v₁ and m₂, v₂ be the masses and velocities of the two fragments.
3. 3Conservation of linear momentum: 0 = m₁v₁ + m₂v₂.
4. 4Hence v₁ = −(m₂/m₁)v₂.
5. 5The minus sign shows the two fragments move in opposite directions.

Step-by-step solution

1. 1Initial momentum of each ball: pᵢ = 0.05 × 6 = 0.3 kg m s⁻¹.
2. 2After collision, the direction reverses: p_f = −0.3 kg m s⁻¹ for each ball.
3. 3Impulse = change in momentum = p_f − pᵢ = −0.3 − 0.3 = −0.6 kg m s⁻¹.
4. 4Each ball receives an impulse of 0.6 kg m s⁻¹, directed opposite to its initial motion (by Newton's third law, equal and opposite).

Step-by-step solution

1. 1The gun and shell are initially at rest: total initial momentum = 0.
2. 2Final momentum = m_{shell}v − MV (gun recoils opposite to the shell).
3. 3Conservation of momentum: 0 = 0.020 × 80 − 100 × V.
4. 4V = (0.020 × 80)/100 = 1.6/100 = 0.016 m s⁻¹.

Step-by-step solution

1. 1v = 54 km/h = 15 m s⁻¹; mass m = 0.15 kg.
2. 2The deflection is 45°, so each velocity makes θ = 22.5° with the angle bisector.
3. 3The momentum component along the bisector reverses; the perpendicular component is unchanged.
4. 4Impulse = change in momentum = 2mv cos θ = 2 × 0.15 × 15 × cos 22.5°.
5. 5= 4.5 × 0.9239 ≈ 4.16 kg m s⁻¹.

Step-by-step solution

1. 1n = 40 rev/min = 40/60 = 2/3 rev s⁻¹; angular speed ω = 2πn.
2. 2Tension provides the centripetal force: T = m r ω² = m r (2πn)².
3. 3T = 0.25 × 1.5 × (2π × 2/3)² = 0.25 × 1.5 × (4.19)² ≈ 6.57 N.
4. 4Maximum speed: T_max = m v_max²/r ⇒ v_max = √(T_max r/m) = √(200 × 1.5/0.25).
5. 5v_max = √1200 ≈ 34.64 m s⁻¹.

Step-by-step solution

1. 1At the instant the string breaks, the centripetal force disappears and no net force acts on the stone.
2. 2By Newton's first law, the stone continues moving in the direction of its velocity at that instant.
3. 3The velocity is always tangential to the circular path at the break point.
4. 4Hence the stone flies off tangentially: option (b).

Step-by-step solution

1. 1(a) To move forward the horse pushes the ground backward; the ground's reaction throws the horse (and cart) forward. Empty space gives no such reaction force, so the horse cannot move forwards — nothing to push against.
2. 2(b) When the bus stops suddenly, the lower body (in contact with the seat) stops, but the upper body continues forward by inertia (Newton's first law) — passengers are thrown forward.
3. 3(c) Pulling at an angle θ: the vertical component of the force, F sin θ, acts upward and reduces the effective weight (mg − F sin θ). Pushing: the vertical component acts downward, increasing the effective weight (mg + F sin θ). A smaller effective weight during pulling makes it easier.
4. 4(d) By F = ma = mΔv/Δt, the stopping force is inversely proportional to impact time. Moving the hands backward increases Δt, decreasing the average force on the hands and preventing injury.