NCERT Solutions for Class 11 Physics Chapter 4: Laws of Motion

Concept. By Newton’s first law, a body at rest or moving with constant velocity (in a straight line, with constant speed) has zero net force on it. Equivalent statement from the second law: F = ma, and a = 0 when the velocity does not change. In every part below the body is either at rest or in uniform motion (in an inertial frame), so the net force is zero even though individual forces (weight, drag, buoyancy, tension) may be large.

1. In (a) the drop first accelerates; as speed increases the drag grows until it equals mg; from then on the speed is constant.
2. In (b) the weight is 0.01 kg × 10 m s⁻² = 0.1 N; buoyant force = weight of water displaced = 0.1 N.
3. In (c) three forces are in equilibrium (F_wind + T + W = 0, see Fig. 4.7).
4. In (d) the speed 30 km h⁻¹ = 8.3 m s⁻¹ does not matter: only the constancy of the velocity matters.
5. In (e) the electron moves in a straight line with constant velocity for ever (law of inertia); its speed may be nearly that of light but there is no net force.

The key idea: zero net force does not mean no force on the body; it means the forces cancel in vector sum. Motion with constant velocity requires no net force: Aristotle’s idea that a force is needed to keep a body moving is a fallacy (Section 4.2).

Extra depth. In (e), the relativistic correction (mass increase) does not change the conclusion: a free electron moves uniformly. If an electric or magnetic field were present there would be a force. Parts (a)–(d) are examples of dynamic equilibrium (balanced forces with constant velocity) and (b), (c) of static equilibrium (balanced forces at rest). In all of them an observer in an inertial frame (the ground) measures zero acceleration.

Conclusion: The net force is zero in all five cases (the forces present balance each other, and no force acts on the electron).

Concept. Once the pebble has left the hand the only force acting is gravity (air resistance is neglected), the weight W = mg = 0.05 kg × 10 m s⁻² = 0.5 N directed vertically downward. The net force depends on the forces present, not on the direction or magnitude of the velocity; the hand’s force acts only while the hand is in contact.

1. The net force F = mg is the same in all three stages: F = 0.05 × 10 = 0.5 N (taking g = 10 m s⁻²; 0.49 N with g = 9.8).
2. The acceleration of the pebble is g downward all through the motion; the velocity changes from upward to zero to downward.
3. Throw at 45°: the answers do not change. The net force is again 0.5 N vertically downward at every point of the parabolic path, including the highest point, because the horizontal component of velocity does not require any horizontal force (first law); only the vertical component of velocity is affected by gravity.

Contrast. At the highest point of the 45° throw the pebble is not at rest (it still has the horizontal velocity v cos45°) but the net force is still mg downward, which is why the path curves.

Extra depth. This exercise separates velocity from force: velocity can be up, zero or down while the force is always down. Such questions often ask what the force would be if air resistance were included: during ascent both weight and drag act downward (net force > mg), during descent drag acts upward (net force < mg), and at the top the drag is zero (net force = mg). Without air resistance, as here, the net force is mg throughout.

More to know — effect of air resistance. With air resistance the net force on the pebble is not constant: going up, weight and drag both point downward (net force > mg); at the top the drag is zero (net force = mg); coming down, drag is upward (net force < mg). The pebble therefore takes longer to come down than to go up and returns with a speed smaller than the speed of projection. In the ideal case of this exercise time up = time down and speed on return = speed of projection.

Conclusion: The net force is 0.5 N (mg), vertically downward, in all three stages, and also for a 45° throw.

Concept. Use the second law with the forces actually acting on the stone at that instant. After a stone is released (a, b, c) the only force on it is gravity, mg = 0.1 × 10 = 1.0 N downward; the horizontal motion of the train gives the stone an initial horizontal velocity but no horizontal force (air resistance neglected). While lying on the floor of an accelerating train the stone is accelerated with the train by a real force, the friction of the floor.

1. (a), (b): the stone has the same velocity as the train at release; after release it follows a path under gravity alone (vertical downward force 1.0 N); in (b) the horizontal component of velocity (10 m s⁻¹) stays constant.
2. (c): the train is accelerating, but the force on the stone is not 0.1 N horizontal any more: once released the stone leaves contact with the train, so the horizontal force vanishes and the stone moves in a parabola relative to the ground; relative to the accelerating train it appears to fall backward.
3. (d): the stone accelerates along with the train: F = ma = 0.1 kg × 1 m s⁻² = 0.1 N. This horizontal force is provided by static friction between the stone and the floor (N and mg cancel vertically).

Key point (Fig. 4.5): acceleration at an instant is determined by the force at that instant (second law is a local relation), not by the earlier history of the motion. Therefore the acceleration of the train is irrelevant for a released stone but essential for the stone lying on the floor.

Extra depth. The speed of 36 km h⁻¹ = 10 m s⁻¹ in (b) is a distractor; in (c) students often think the force is still 0.1 N because the train accelerates. In the frame of the accelerating train (non-inertial) one would add a pseudo force −ma and the stone would appear to fall with a backward component; in the inertial ground frame there is no such force. The net force on the stone is determined in an inertial frame: only real forces count.

Conclusion: (a), (b), (c): 1.0 N vertically downward; (d): 0.1 N horizontally in the direction of the train’s acceleration.

Concept. The particle moves on a smooth horizontal table, so its weight is balanced by the normal reaction and the only horizontal force is the tension T in the string, directed along the string towards the peg (the centre). The net force towards the centre is therefore T. By Newton’s second law this net (centripetal) force equals mv²/l.

1. Vertical forces: weight mg (down) and normal reaction N (up) cancel since there is no vertical motion.
2. Horizontal force: only the tension T of the string towards the centre.
3. Net force towards the centre = T, which must equal mv²/l for circular motion at speed v: T = mv²/l.
4. Therefore the net force is T: alternative (i).

Important idea: the quantity mv²/l is not a force that can be added to T; it is the value that the net centripetal force must have. There is no separate ‘centripetal force’ in addition to the real forces: the resultant of the real forces pointing towards the centre plays that role.

Extra depth. Compare with a stone whirled in a vertical circle: there the weight and the tension both have components along the radius and the net force towards the centre is T − mg cosθ (or T + mg at the top). On a smooth horizontal table the net force is simply T. If the string is cut, T disappears and the particle moves along the tangent in a straight line with constant speed v (Ex 4.22).

Conclusion: The net force towards the centre is T (alternative (i)), and T = mv²/l.

Concept. A constant force produces a constant acceleration, so the equations of uniformly accelerated motion apply. The retarding force acts opposite to the velocity, so the acceleration (retardation) a = −F/m is negative if the initial direction of motion is positive.

1. Mass m = 20 kg, initial speed u = 15 m s⁻¹, final speed v = 0, retarding force F = 50 N.
2. By the second law the retardation is a = F/m = 50/20 = 2.5 m s⁻²; with the direction of u positive, a = −2.5 m s⁻².
3. Using v = u + at: 0 = 15 + (−2.5) t ⇒ t = 15/2.5 = 6 s.
4. Alternatively by impulse–momentum: F t = m u ⇒ t = mu/F = 20 × 15/50 = 6 s.

Check: the work done by the retarding force, 50 N × 45 m = 2250 J, equals the initial kinetic energy ½ × 20 × 15² = 2250 J (work–energy theorem, Chapter 5).

Extra depth. The impulse method (F Δt = Δp) avoids calculating a. It shows clearly that the time needed to stop depends on the momentum and the force: doubling the mass or the speed doubles the stopping time; doubling the force halves it. A similar idea is used in the design of crumple zones and air bags, where a longer stopping time reduces the average force on the passengers.

More to know — Newton’s second law in momentum form. F = Δp/Δt: the impulse of the retarding force removes the whole momentum of the body, so F × t = mu. This form is used whenever the force is large and acts for a short time (collisions) or when the mass changes (rockets). For a given initial momentum, a larger retarding force stops the body faster: with 100 N the stopping time would be 3 s and the stopping distance 22.5 m. The stopping distance can also be written s = mu²/2F = (20 × 225)/(2 × 50) = 45 m, showing s ∝ u².

Conclusion: The body stops in 6 s.

Concept. Since the direction of motion does not change, the problem is one-dimensional. The change in velocity gives the acceleration, and Newton’s second law gives the force. Because the speed increases in the same direction, the force acts in the direction of motion.

1. Initial speed u = 2.0 m s⁻¹, final speed v = 3.5 m s⁻¹, time t = 25 s, mass m = 3.0 kg.
2. Acceleration a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m s⁻² (positive: along the direction of motion).
3. Force F = ma = 3.0 × 0.06 = 0.18 N.
4. Direction: along the direction of motion of the body (the speed increases).

Check by impulse: change in momentum = m(v − u) = 3.0 × 1.5 = 4.5 kg m s⁻¹; F = Δp/Δt = 4.5/25 = 0.18 N.

Extra depth. If the speed had decreased from 3.5 m s⁻¹ to 2.0 m s⁻¹ the magnitude of the force would be the same 0.18 N but its direction would be opposite to the motion (a retarding force). The small value of the force reflects the slow change of speed (0.06 m s⁻², about 0.6% of g). This type of problem teaches that direction of the force follows the direction of the change in velocity, not the direction of the velocity.

More to know — the unit of force. The newton is the force that gives a 1 kg mass an acceleration of 1 m s⁻². Here F = 0.18 N is about the weight of an object of mass 18 g (a small coin). Such small forces are typical of slow changes of speed: the speed changes by only 1.5 m s⁻¹ in 25 s. For the same body, to change the speed by the same amount in 1 s the force would be 25 times larger (4.5 N). Newton’s second law says the force is proportional to the rate of change of velocity: F = m Δv/Δt.

Conclusion: The force is 0.18 N in the direction of motion.

Concept. The acceleration is along the resultant force, and its magnitude is a = F_net/m. For two perpendicular forces the resultant is found with the Pythagoras theorem (parallelogram law with θ = 90°), and its direction with a tangent.

1. Resultant force F = √(8² + 6²) = √(64 + 36) = √100 = 10 N.
2. Acceleration a = F/m = 10/5 = 2 m s⁻².
3. Direction: the angle θ made by the resultant (and hence the acceleration) with the 8 N force: tanθ = 6/8 = 0.75 ⇒ θ = 37° with the 8 N force (and 53° with the 6 N force).

Check using components: a_x = 8/5 = 1.6 m s⁻² and a_y = 6/5 = 1.2 m s⁻² give a = √(1.6² + 1.2²) = √4 = 2.0 m s⁻² and tanθ = 1.2/1.6 = 0.75 ✓.

Extra depth. This is an application of the vector nature of the second law: each component of acceleration depends only on the same component of force: a_x = F_x/m, a_y = F_y/m. The 6–8–10 triangle is a Pythagorean triple whose angles are 37° and 53°, which saves time in exams.

More to know — resultant of two forces. For forces F₁ and F₂ at angle θ: F² = F₁² + F₂² + 2F_1F2 cosθ and tanα = F₂ sinθ/(F₁ + F₂ cosθ) (Chapter 3). For θ = 90° this reduces to F = √(F₁² + F₂²) = 10 N and tanα = F₂/F₁ = 0.75. The acceleration is always in the direction of the resultant, and the magnitude of the resultant lies between |F₁ − F₂| = 2 N and F₁ + F₂ = 14 N depending on the angle: the corresponding accelerations are between 0.4 and 2.8 m s⁻².

Conclusion: a = 2 m s⁻² at 37° to the 8 N force (towards the 6 N force).

Concept. The vehicle and the driver move together, so the mass to be stopped is the total mass = 400 + 65 = 465 kg. The retarding force is found from the average retardation, a = (change in velocity)/time, with F = ma (or from impulse = change of momentum).

1. Speed u = 36 km h⁻¹ = 36 × 5/18 = 10 m s⁻¹; final speed v = 0; time t = 4.0 s.
2. Retardation a = (0 − 10)/4.0 = −2.5 m s⁻² (the magnitude of retardation = 2.5 m s⁻²).
3. Total mass m = 400 kg + 65 kg = 465 kg.
4. Average retarding force F = m|a| = 465 × 2.5 = 1162.5 N ≈ 1.2 × 10³ N, directed opposite to the motion.

Meaning. The stopping distance of 20 m shows how close the child was; the force of 1.16 kN is the average force exerted by the road (friction, via the brakes) on the vehicle. It is about 0.25 times the vehicle’s weight (465 × 10 = 4650 N).

Extra depth. The force is an average value; the actual braking force varies with time. We must include the mass of the driver because he decelerates with the vehicle (he is held by his seat and by friction). In more realistic situations the reaction time of the driver (about 0.2–0.7 s) adds a further distance v × t_r = 2–7 m during which the vehicle moves with the original speed.

More to know — road safety. A stopping force of 1.16 kN corresponds to a deceleration of 0.25 g; for ordinary dry-road braking a car can decelerate at 0.7–0.9 g. Doubling the speed to 20 m s⁻¹ would need four times the stopping distance (80 m) at the same retardation: the stopping distance is proportional to u². That is why the speed limit is lowered near schools. The force acts on the vehicle through the friction between the tyres and the road (or by skidding the wheels, kinetic friction) — the brakes themselves only convert kinetic energy into heat.

Conclusion: The average retarding force on the vehicle is about 1.2 × 10³ N (1162.5 N).

Concept. At lift-off two forces act on the rocket: the upward thrust F of the exhaust gases and the downward weight mg. The net upward force produces the upward acceleration: F − mg = ma. (The text for this exercise takes g = 10 m s⁻².)

1. Mass m = 20,000 kg = 2.0 × 10⁴ kg; acceleration a = 5.0 m s⁻² (upward); g = 10 m s⁻².
2. Newton’s second law (upward positive): F − mg = ma.
3. F = m(g + a) = 2.0 × 10⁴ × (10 + 5.0) = 2.0 × 10⁴ × 15 = 3.0 × 10⁵ N.
4. Thrust = 3.0 × 10⁵ N, directed upward (of this, 2.0 × 10⁵ N supports the weight and 1.0 × 10⁵ N produces the acceleration).

Why must thrust exceed the weight? If F = mg the net force would be zero and the rocket would hover; only if F > mg is there a net upward force and a positive upward acceleration. The thrust comes from the third law: the rocket pushes hot gases downward and the gases push the rocket upward with an equal force.

Extra depth. A real rocket loses mass quickly as fuel is burnt, so the acceleration increases with time even if the thrust remains constant (a = F/m − g). The ‘initial’ in the question refers to the instant of lift-off when the mass is the full 20,000 kg. A thrust-to-weight ratio of 1.5 as here is typical for launchers.

More to know — how a rocket works. The rocket expels gas at high speed backward; by the third law the gas pushes the rocket forward (thrust F = (dm/dt) v_exhaust). The thrust must exceed the weight for lift-off. For the 20,000 kg rocket the minimum thrust to just hover is 2.0 × 10⁵ N; to accelerate at 5.0 m s⁻² it is 3.0 × 10⁵ N. A rocket works in empty space too (it pushes against its own exhaust, not against the air), which is why a horse cannot move in empty space but a rocket can (Ex 4.23a).

Conclusion: The initial thrust of the blast is 3.0 × 10⁵ N.

Concept. Take north as the positive x-direction. Before t = 0 the body moves with constant velocity +10 m s⁻¹ (no force). From t = 0 to 30 s the force 8.0 N southward gives a constant acceleration a = F/m = −8.0/0.40 = −20 m s⁻². After t = 30 s the force is removed and the body again moves with constant velocity (the velocity it had at t = 30 s).

1. At t = −5 s (before the force acts): uniform motion at 10 m s⁻¹: x = ut = 10 × (−5) = −50 m, i.e. the body was 50 m south of the origin 5 s earlier.
2. At t = 25 s (force acting): x = ut + ½at² = 10 × 25 + ½(−20)(25)² = 250 − 6250 = −6000 m = −6.0 km (6.0 km south of the origin).
3. At t = 100 s: first find the state at t = 30 s: x₃₀ = 10 × 30 − 10 × 30² = 300 − 9000 = −8700 m; v₃₀ = u + at = 10 − 20 × 30 = −590 m s⁻¹ (590 m s⁻¹ towards the south).
4. After t = 30 s no force: uniform motion for the remaining 70 s: x₁₀₀ = x₃₀ + v₃₀ × 70 = −8700 − 590 × 70 = −8700 − 41300 = −5.0 × 10⁴ m = −50 km (50 km south of the origin).

Note. The negative signs mean ‘south of the origin’. The body reverses its direction at t = 0.5 s (v = 10 − 20t = 0), at x = +2.5 m, then moves ever faster to the south.

Extra depth. The same constant-acceleration formula cannot be used for t = 100 s because the force acts only for 30 s: the motion must be split into stages (the ‘piece-wise’ method). The negative time t = −5 s simply asks for the position before the force was applied; since the force was absent then, the velocity was constant at 10 m s⁻¹. A very large speed of 590 m s⁻¹ is physically unrealistic for a 0.4 kg body acted on by a force for such time; it is just a numerical exercise (the force of 8 N on 0.4 kg is huge).

Conclusion: x(−5 s) = −50 m; x(25 s) = −6.0 km; x(100 s) = −50 km (all measured with north positive, i.e. south of the starting point).

Concept. At the instant of release (t = 10 s) the stone has the same velocity as the truck: v_x = at = 2.0 × 10 = 20 m s⁻¹ in the direction of motion. After release the only force on it is gravity (no horizontal force since air resistance is neglected, and the stone is no longer in contact with the truck): so the horizontal velocity stays 20 m s⁻¹ while the vertical velocity grows under g. The truck’s acceleration is irrelevant after release (the second law is local in time).

1. Horizontal velocity at release: v_x = 0 + 2.0 × 10 = 20 m s⁻¹; it remains 20 m s⁻¹ at t = 11 s.
2. Vertical velocity 1 s after release (initial vertical velocity 0): v_y = 0 − g t = −10 × 1 = −10 m s⁻¹ (downward).
3. (a) Velocity at t = 11 s: v = √(20² + 10²) = √500 = 22.4 m s⁻¹, at angle θ = tan⁻¹(10/20) = 26.6° below the horizontal, in the forward direction of the truck’s motion.
4. (b) Acceleration: horizontal acceleration = 0 (no horizontal force); vertical acceleration = g = 10 m s⁻² downward. The stone’s acceleration at t = 11 s is 10 m s⁻² vertically down.

Check of the height: in 1 s the stone falls ½ × 10 × 1² = 5 m, which is less than 6 m, so it has not yet reached the ground at t = 11 s (it would hit the ground at t = 10 + √(2 × 6/10) = 11.1 s). Hence the answer for t = 11 s is valid.

Extra depth. Many students answer (b) as ‘2 m s⁻² horizontally plus 10 m s⁻² vertically’ because the truck accelerates; but acceleration is determined by the force acting now, and now the only force is gravity (Fig. 4.5 and the discussion of the second law as a local relation). A man on the truck would see the stone fall behind the vertical line with a backward acceleration of 2 m s⁻² relative to the truck; relative to the ground the path is a parabola.

Conclusion: (a) 22.4 m s⁻¹ at 26.6° below the horizontal (in the forward direction); (b) 10 m s⁻² vertically downward.

Concept. When the string is cut the only force on the bob is gravity (tension disappears), so the bob moves as a projectile with the velocity it had at the instant of cutting (first law): its path depends on that velocity. At an extreme position the bob is momentarily at rest; at the mean position it moves horizontally with speed 1 m s⁻¹.

1. (a) At the extreme position the speed is zero, so initial velocity = 0 and the only force is the weight: the bob falls vertically downward along a straight line (free fall).
2. (b) At the mean position the velocity is horizontal (1 m s⁻¹) and the weight acts vertically: the horizontal motion is uniform and the vertical motion is free fall: x = 1 × t and y = ½ g t², so y = 5x² (with g = 10 m s⁻²): a parabola.

Numbers for (b). If the height of the bob above the floor is h, it hits the floor after t = √(2h/g) with horizontal range 1 × t. For h = 1 m, t = 0.45 s and range 0.45 m.

Extra depth. The answer shows that the trajectory is decided by the initial conditions and not by the force alone (Point 7 of the chapter’s ‘Points to ponder’): under the same gravitational force the path can be a straight line or a parabola. For a cut at an intermediate position the bob would have a tangential velocity directed along the tangent to the arc and would follow a parabola projected at that angle. Speed at the mean position (1 m s⁻¹) means the bob was lifted initially through h = v²/2g = 0.05 m.

More to know — how high was the bob lifted? At the mean position the speed is 1 m s⁻¹, so by energy conservation (Chapter 5) the bob was released from a height h = v²/2g = 1/20 = 0.05 m above the lowest point (an angle of about 13° with the vertical for l = 2 m: cosθ = 1 − 0.05/2 = 0.975). The tension at the mean position is T = mg + mv²/l = 1.0 + 0.1 × 1/2 = 1.05 N, a little more than the weight. When the string is cut at the mean position the bob is also at the lowest point of its oscillation, so it falls from the height of the point of suspension minus 2 m.

Conclusion: (a) vertically downward straight line; (b) parabola (horizontal projectile).

Concept. The weighing scale reads the normal reaction R of the platform on the man (his apparent weight). Newton’s second law for the man with upward positive: R − mg = ma. The reading R = m(g + a) where a is the acceleration of the lift (upward positive). Take g = 10 m s⁻² and m = 70 kg (mg = 700 N).

1. (a) Uniform velocity means zero acceleration: the first law applies, so R = mg = 700 N: the same as at rest. The speed (10 m s⁻¹) is irrelevant.
2. (b) The lift accelerates downward: a = −5 m s⁻²: mg − R = 5m ⇒ R = m(g − 5) = 350 N. The man feels lighter.
3. (c) The lift accelerates upward: R − mg = 5m ⇒ R = m(g + 5) = 1050 N. He feels heavier.
4. (d) In free fall the lift and man have acceleration g downward: mg − R = mg ⇒ R = 0. The scale reads zero: the man is weightless (his true weight mg still acts, but he exerts no force on the platform).

Scale reading in kg: readings are quoted in newtons, but a bathroom scale calibrated in kg would show R/g: 70 kg, 35 kg, 105 kg and 0 respectively. The real weight of the man is mg = 700 N in all four cases.

Extra depth. The ‘apparent weight’ changes with the acceleration of the frame: this is why astronauts in orbit (free fall around the Earth) experience weightlessness, why one feels pressed down at the start of an upward lift and light at the start of a downward one. Equivalent idea in the lift frame (non-inertial): add a pseudo force −ma; the effective gravity is g_eff = g + a. If the lift accelerated downward with a > g, the man would be pressed against the ceiling.

Conclusion: (a) 700 N; (b) 350 N; (c) 1050 N; (d) zero.

Concept. Read the graph: for t < 0 the position is constant (x = 0): the particle is at rest. From t = 0 to t = 4 s the graph is a straight line rising from x = 0 to x = 3 m, i.e. a constant velocity v = 3/4 = 0.75 m s⁻¹. For t > 4 s the position is again constant (x = 3 m): the particle is at rest. A force changes only the velocity, so find where the velocity changes.

1. (a) In all three intervals the velocity is constant, hence the acceleration is zero and the force is zero.
2. (b) At t = 0 the velocity changes suddenly from 0 to 0.75 m s⁻¹; at t = 4 s from 0.75 m s⁻¹ back to 0. These sudden changes need large forces acting for a very short time (impulses). Impulse = change in momentum.
3. At t = 0: impulse = m(v_after − v_before) = 4 kg (0.75 − 0) = 3 kg m s⁻¹ (= 3 N s), in the +x direction.
4. At t = 4 s: impulse = 4 kg (0 − 0.75) = −3 kg m s⁻¹ (3 N s in the −x direction).

Check: the total impulse over the motion is 3 + (−3) = 0, consistent with the particle having the same velocity (zero) at the start and the end.

Extra depth. A sharp kink in an x-t graph corresponds to an instantaneous change of velocity, which would require an infinite force acting for zero time (an idealisation); in reality the change takes a short time Δt, during which an average force F = Δp/Δt (impulsive force) acts. Between the kinks the force is zero (first law). This is the same idea as the shape of velocity-time graphs in Chapter 2: slope of x-t is velocity, slope of v-t is acceleration, and F = m × slope of the v-t graph.

Conclusion: (a) The force is zero in all three intervals; (b) impulse is +3 kg m s⁻¹ at t = 0 and −3 kg m s⁻¹ at t = 4 s.

Concept. The two bodies are joined by a light, inextensible string, so they have the same acceleration a. The surface is smooth (no friction). The tension is the force with which the string pulls the body that is not being pulled directly by F. Treat the pair as one system to find a, then apply the second law to the body at the back.

1. Acceleration of the system (both cases): a = F/(m_A + m_B) = 600/(10 + 20) = 20 m s⁻².
2. (i) F applied to A: the tension T pulls B (20 kg) alone: T = m_B a = 20 × 20 = 400 N. Check on A: F − T = m_A a ⇒ 600 − 400 = 10 × 20 = 200 ✓.
3. (ii) F applied to B: the tension pulls A (10 kg) alone: T = m_A a = 10 × 20 = 200 N. Check on B: F − T = m_B a ⇒ 600 − 200 = 20 × 20 = 400 ✓.

Interpretation. The tension is not the same although the applied force is the same: in each case it equals the force needed to accelerate the ‘trailing’ body. The heavier the trailing body, the larger the tension. If the string is replaced by a rope with mass the tension would vary along its length.

Extra depth. The result T = F m_trailing/(m_A + m_B) shows that the string tension is a fraction of F: 2/3 of F in (i) and 1/3 of F in (ii). This is why a locomotive’s coupling nearest to the engine bears the largest tension (it has to accelerate all the coaches behind it) and the last coupling the smallest. The same method is used in the connected-bodies problems of Ex 4.16 and Example 4.9.

Conclusion: (i) T = 400 N; (ii) T = 200 N (acceleration 20 m s⁻² in both cases).

Concept. This is the Atwood machine. The string is light and inextensible and the pulley is frictionless, so the tension T is the same on both sides and both masses move with the same magnitude of acceleration a: the heavier mass (12 kg) goes down and the lighter (8 kg) goes up. Apply the second law to each mass separately (g = 10 m s⁻²).

1. 12 kg mass (moving down, take downward positive): 12g − T = 12a   (1)
2. 8 kg mass (moving up): T − 8g = 8a   (2)
3. Add (1) and (2): (12 − 8) g = (12 + 8) a ⇒ a = 4g/20 = g/5 = 2 m s⁻².
4. Substitute in (2): T = 8g + 8a = 8(10 + 2) = 96 N. (Check with (1): T = 12(10 − 2) = 96 N ✓.)

Comparison. T = 96 N lies between the weights of the two masses (80 N and 120 N), as it must: the tension must exceed 8g to accelerate the 8 kg mass upward, and be less than 12g to let the 12 kg mass accelerate downward. With g = 9.8 m s⁻²: a = 1.96 m s⁻², T = 94.1 N.

Extra depth. The tension is not equal to the weight of either mass: it equals 9.6 g = 96 N (between 8g = 80 N and 12g = 120 N). The pulley has to support a downward force of 2T = 192 N, which is less than the total weight 200 N because the heavier mass accelerates downward. Limiting cases: if m₁ = m₂, a = 0 and T = mg (equilibrium); if m₂ → 0, a → g and T → 0 (free fall).

Conclusion: The acceleration is 2 m s⁻² (12 kg down, 8 kg up) and the tension in the string is 96 N.

Concept. This is a direct application of the law of conservation of momentum. The nucleus is acted on by no external force during the disintegration (internal forces only), so the total momentum of the system is conserved. Initially the nucleus is at rest, so the initial total momentum is zero; hence the total momentum of the products must also be zero.

1. Let the nucleus (mass M) be at rest: initial momentum p_i = 0.
2. Suppose it disintegrates into two nuclei of masses m₁ and m₂ with velocities v₁ and v₂: final momentum p_f = m₁ v₁ + m₂ v₂.
3. By conservation of momentum: p_f = p_i = 0 ⇒ m₁ v₁ + m₂ v₂ = 0 ⇒ m₁ v₁ = −m₂ v₂.
4. The negative sign shows that v₁ and v₂ are in opposite directions (the two momentum vectors are equal in magnitude and opposite in direction). Hence the products move along the same line in opposite directions: v₁/v₂ = −m₂/m₁, the lighter product moving faster.

Why opposite directions only? If both fragments moved in the same direction, their momenta would add up to a non-zero vector, contradicting the zero initial momentum. In two or three dimensions the same argument shows that the momenta are equal and opposite vectors, so the two fragments move along the same straight line, back to back (a third fragment would be needed for non-collinear motion).

Extra depth. This is the principle of recoil: a gun and bullet, a rocket and its exhaust, and an alpha-decay of a nucleus at rest (the daughter nucleus recoils opposite to the α particle). The kinetic energies are inversely proportional to masses: K₁/K₂ = m₂/m₁, so in α decay the light α particle carries about 98% of the released energy. If the original nucleus were moving, the products need not move in opposite directions in the laboratory, only in the centre-of-mass frame.

Conclusion: By conservation of momentum m_1v1 + m_2v2 = 0, so the two nuclei must move in opposite directions.

Concept. Impulse = change in momentum of the ball (J = Δp = FΔt). Take the direction of motion of the first ball before the collision as positive: then ball 1 has initial velocity +6 m s⁻¹ and ball 2 has −6 m s⁻¹; after the collision they rebound with the same speed, so their velocities are reversed.

1. Ball 1: initial momentum p_i = 0.05 × (+6) = +0.30 kg m s⁻¹; final momentum p_f = 0.05 × (−6) = −0.30 kg m s⁻¹; impulse J₁ = p_f − p_i = −0.30 − 0.30 = −0.60 kg m s⁻¹ (magnitude 0.6 N s, directed opposite to its initial motion).
2. Ball 2: initial p_i = 0.05 × (−6) = −0.30; final p_f = 0.05 × (+6) = +0.30; J₂ = +0.30 − (−0.30) = +0.60 kg m s⁻¹ (magnitude 0.6 N s, in the direction opposite to the initial motion of ball 2).
3. The impulses on the two balls are equal in magnitude (0.6 N s) and opposite in direction: this is Newton’s third law in impulse form (F₁₂ Δt = −F₂₁ Δt).

Check: the total momentum of the system before and after the collision is zero, as required by conservation of momentum, and the kinetic energy is unchanged (elastic collision), ½(0.05)(6²) × 2 = 1.8 J before and after.

Extra depth. The magnitude 0.6 N s is the same as 2mv = 2 × 0.05 × 6: the impulse needed to reverse a momentum mv is 2mv. If the contact time were 0.01 s the average force would be 0.6/0.01 = 60 N, which is 120 times the weight of a ball (0.5 N). The average force cannot be determined without the time of contact, but the impulse can, as in Example 4.4.

Conclusion: Each ball receives an impulse of magnitude 0.6 N s (kg m s⁻¹), the two impulses being opposite in direction.

Concept. The gun and the shell form an isolated system during firing (the internal forces due to the explosion are huge compared with external forces, and the net external horizontal force is negligible). Before firing both are at rest, so the total momentum is zero; by conservation of momentum the total momentum after firing is also zero, and the gun recoils opposite to the shell.

1. m_s = 0.020 kg, v_s = 80 m s⁻¹ (forward), M = 100 kg, recoil velocity V (backward).
2. Conservation of momentum: 0 = m_s v_s + M V ⇒ V = −m_s v_s/M.
3. |V| = 0.020 × 80/100 = 1.6/100 = 0.016 m s⁻¹ (1.6 cm s⁻¹), directed opposite to the shell.

Meaning. The momenta of the shell and the gun are equal and opposite (1.6 kg m s⁻¹) but the speeds are inversely proportional to the masses; the gun’s recoil speed is only 1/5000 of the muzzle speed. The kinetic energies K = p²/2m are in the inverse ratio of the masses: the shell gets 5000 times the energy of the gun.

Extra depth. In practice the recoil of a heavy gun is checked with springs or a recoil mechanism; the recoil speed is small but the recoil force can still be large if the gun is stopped quickly (impulse 1.6 N s in 0.01 s gives 160 N). For a rifle (M = 4 kg, bullet 20 g at 400 m s⁻¹) the recoil speed is 2 m s⁻¹, which a shooter feels as a kick. The same principle propels rockets: exhaust gas goes backward, the rocket forward.

More to know — recoil in practice. For a rifle of mass 4 kg firing a 20 g bullet at 400 m s⁻¹ the recoil speed is 2 m s⁻¹; its kinetic energy is only 8 J compared with 1600 J for the bullet. The total momentum remains zero. Heavy guns use recoil cylinders and the sudden reaction on the shoulder can be reduced by pressing the rifle firmly (increasing the effective mass M). The same principle underlies the jet of a hose pushing the fire-fighter backwards and the backward kick of a bazooka.

Conclusion: The recoil speed of the gun is 0.016 m s⁻¹ (opposite to the shell).

Concept. Impulse is the change in momentum vector. The speed is unchanged but the direction of the ball’s velocity changes by 45°; so the momentum changes in direction and the impulse is the vector difference p_f − p_i of two vectors of equal magnitude mv at 45° to each other.

1. Speed v = 54 km h⁻¹ = 54 × 5/18 = 15 m s⁻¹; mass m = 0.15 kg; momentum magnitude p = mv = 0.15 × 15 = 2.25 kg m s⁻¹.
2. Take the initial velocity along the x-axis: p_i = (2.25, 0); after deflection through 45°: p_f = (2.25 cos45°, 2.25 sin45°) = (1.59, 1.59).
3. Δp = p_f − p_i = (1.59 − 2.25, 1.59) = (−0.66, 1.59) kg m s⁻¹.
4. Magnitude |Δp| = √(0.66² + 1.59²) = √(0.436 + 2.528) = √2.96 = 1.72 kg m s⁻¹ ≈ 1.7 N s.
5. Using the formula: |Δp| = 2 m v sin(θ/2) = 2 × 2.25 × sin22.5° = 4.5 × 0.3827 = 1.72 N s.
6. Direction: Δp makes an angle tan⁻¹(1.59/0.66) = 67.5° with the −x axis (i.e. 112.5° with the original direction): it is perpendicular to the bisector of the angle between initial and final directions, i.e. along the normal to the face of the bat.

Comparison: if the batsman had hit the ball straight back (deflection 180°), the impulse would be 2mv = 4.5 N s (Example 4.4 gave 3.6 N s for 0.15 kg at 12 m s⁻¹).

Extra depth. The speed is unchanged, so the kinetic energy is unchanged and the bat does no work on the ball, even though it exerts a large force: the force is perpendicular to the velocity change only in the sense of the normal to the bat. The result Δp = 2mv sin(θ/2) applies to every deflection by an angle θ at constant speed (also to reflection from a wall, Example 4.5). Using only magnitudes (2.25 − 2.25 = 0) is the common mistake.

Conclusion: The impulse imparted to the ball is about 1.7 N s (kg m s⁻¹), directed normal to the bat face.

Concept. The tension in the string supplies the centripetal force: T = mv²/R = mω²R. Convert 40 revolutions per minute to the angular speed in rad s⁻¹ first; then use the maximum tension to find the maximum speed v_max = √(T_max R/m).

1. m = 0.25 kg, R = 1.5 m, ν = 40 rev/min = 40/60 = 2/3 rev s⁻¹.
2. Angular speed ω = 2πν = 2 × 3.14 × (2/3) = 4.19 rad s⁻¹; speed v = ωR = 4.19 × 1.5 = 6.28 m s⁻¹.
3. Tension: T = mω²R = 0.25 × (4.19)² × 1.5 = 0.25 × 17.56 × 1.5 = 6.6 N (= mv²/R = 0.25 × 39.4/1.5 = 6.6 N).
4. Maximum speed: T_max = mv_max²/R = 200 N ⇒ v_max² = T_max R/m = 200 × 1.5/0.25 = 1200 m² s⁻² ⇒ v_max = √1200 = 34.6 m s⁻¹ (≈ 35 m s⁻¹).
5. Corresponding maximum frequency: ν_max = v_max/(2πR) = 34.6/9.42 = 3.7 rev s⁻¹ = 220 rev/min.

Remark: the string would break when the stone is whirled faster than 34.6 m s⁻¹ because T ∝ v²: the speed can be increased about 5.5 times before reaching T_max (tension rises 30-fold).

Extra depth. In the horizontal plane the stone’s weight is balanced by the vertical component of the string’s tension only if the string is slightly inclined; the exercise (and textbook) idealises the string as horizontal. The tension varies as the square of the speed (or the frequency), so a modest increase in speed can break the string; this is also what happens to a revolving rotor or a car’s tyre at excessive speed. After the string breaks the stone leaves along the tangent (Ex 4.22).

Conclusion: T = 6.6 N; the maximum speed is 34.6 m s⁻¹.

Concept. At the instant the string breaks, the stone has a velocity directed along the tangent to the circle at that point (velocity is always tangential in circular motion). After the break the only horizontal force that was acting, the tension, disappears; so by Newton’s first law the stone continues with the velocity it had, in a straight line along the tangent (while falling under gravity if the circle is not at the ground).

1. Velocity at any point of a circle is tangent to the circle in the direction of motion.
2. When T vanishes, the net horizontal force is zero: a = 0 in the horizontal plane (first law).
3. The stone therefore moves in a straight line along the tangent with the speed it had (more than 34.6 m s⁻¹).
4. So the correct answer is (b).

Example in daily life: mud flying off a wheel, sparks from a grinding wheel, a hammer released in the hammer throw and water drops leaving a spinning umbrella all move off tangentially.

Extra depth. In the rotating frame of the stone one may describe the stone as ‘flung outwards’ by a centrifugal pseudo-force, but in the inertial frame (the ground) there is no outward force and the stone just continues along the tangent. If the circle is vertical or the stone is above the ground, the path after breaking is a parabola whose initial velocity is along the tangent.

Conclusion: (b): the stone flies off tangentially from the instant the string breaks.

Concept. Each part is an application of one of Newton’s laws: (a) the third law, (b) the first law (inertia), (c) the second law together with friction (the normal reaction changes), and (d) the second law in the form F = Δp/Δt (impulse).

(a) Horse and cart in empty space

A horse pushes the ground backward with its hooves; by the third law the ground pushes the horse forward with an equal and opposite force (friction at the hooves); this forward reaction of the ground makes the horse and cart accelerate. In empty space there is no ground (no medium) to push against, so there is no external reaction force: the horse and cart together form an isolated system whose momentum cannot change; the horse cannot move forward and the cart cannot be pulled (the pull on the cart by the horse is an internal force of the system).

(b) Passengers thrown forward when a speeding bus stops suddenly

By the law of inertia a body continues in its state of uniform motion unless acted on by a force. The passengers move with the bus at its speed; when the bus is stopped suddenly the lower part of the body in contact with the seat/floor stops with the bus (friction), but the upper part continues to move forward with the original velocity: the passengers are thrown forward. Muscular forces then bring the body to rest. Seat belts supply the external force needed to stop the passenger together with the bus.

(c) Easier to pull a lawn mower than to push it

1. The handle is inclined at an angle θ to the horizontal. Pushing: the applied force F has a downward vertical component F sinθ; the normal reaction becomes N = mg + F sinθ, so the friction f = μN is larger.
2. Pulling: the vertical component of F is upward, so N = mg − F sinθ and friction f = μ(mg − F sinθ) is smaller.
3. The horizontal component F cosθ has to overcome friction; with less friction when pulling, a smaller force F is needed. Hence it is easier to pull the mower.

(d) Cricketer moves his hands backwards while catching

The ball has momentum p = mv that must be reduced to zero. The impulse = force × time = change of momentum: F = Δp/Δt. By drawing the hands back the cricketer increases the time Δt over which the ball is brought to rest; for the same Δp the average force on the hands is smaller, which reduces the chance of injury (Fig. 4.3). If the hands were held rigid, Δt would be very short and the force very large.

Extra depth. These four situations show that Newton’s laws govern everyday life: the third law needs a second body (ground, air, water) to push against — which is why rockets can accelerate in empty space (they push their own exhaust gas) but a horse cannot; inertia explains seat belts and head-rests; friction depends on the normal reaction, which can be changed by pulling or pushing; and the impulse formula underlies air bags, crumple zones, cushions and the use of nets in high-jump landing.

Conclusion: (a) Third law (no reaction); (b) first law (inertia); (c) pulling reduces the normal reaction and hence friction; (d) a longer time of stopping reduces the force (F = Δp/Δt).