Laws of motion and momentum for NDA

Newton's laws are the backbone of NDA mechanics. The syllabus names motion, Newton's laws, force and momentum, the parallelogram of forces, and stability and equilibrium, and questions on them come in two flavours. One is the everyday example: why passengers lurch forward, why a gun recoils, why a fielder draws the hands back. The other is a one-step numerical on momentum, impulse, a lift or friction.

Both are easy once the ideas are exact. Most wrong answers in this chapter come from a half-remembered idea, such as thinking action and reaction cancel out, rather than from arithmetic.

Typical question types are:

• identifying which law explains an everyday situation;
• inertia of rest, motion and direction;
• recoil velocity and other conservation-of-momentum problems;
• impulse and change in momentum, including a ball bouncing back;
• apparent weight in a lift;
• the resultant of two forces;
• static, sliding and rolling friction;
• centripetal force, banking of roads, and stability.

Newton's three laws

First law and inertia

Inertia is a body's resistance to a change in its state of rest or motion. Mass is the measure of inertia: a loaded truck is harder to start or stop than a bicycle.

• Inertia of rest: a coin on a card falls into the glass when the card is flicked away.
• Inertia of motion: a person jumping off a moving bus falls forward.
• Inertia of direction: passengers are pushed outward when a car takes a sharp turn.

Second law, momentum and impulse

Momentum p = mv is a vector, measured in kg m/s. The second law says F = Δp/Δt, which becomes F = ma when mass is constant. One newton is the force that gives a 1 kg mass an acceleration of 1 m/s².

Impulse = force × time = change in momentum. The same change in momentum can come from a large force acting briefly or a small force acting longer. That is why:

• a fielder draws the hands back while catching, increasing the time and reducing the force;
• vehicles have crumple zones and airbags, and high jumpers land on soft mats;
• a fall on concrete hurts more than a fall on sand.

Third law

Action and reaction are equal in size and opposite in direction, but they act on different bodies, so they never cancel each other. When a horse pulls a cart, the cart pulls back on the horse; the pair moves forward because of the friction force the ground exerts on the horse's hooves.

Conservation of momentum

If no external force acts on a system, its total momentum stays constant. This follows from the second and third laws together.

• Recoil of a gun: before firing, the total momentum is zero. Afterwards, mv (bullet) = MV (gun), so the gun recoils with V = mv/M, much slower because it is much heavier.
• Rockets push exhaust gases backward and move forward. They need no air to push against, so they work in space.
• Jumping from a boat pushes the boat backwards.

Apparent weight in a lift

A weighing scale reads the normal reaction R, not the true weight mg.

Astronauts in an orbiting spacecraft feel weightless for the same reason: they and the craft are in free fall together.

Resultant of forces

Two forces P and Q acting at a point with an angle θ between them add by the parallelogram law: the resultant R satisfies R² = P² + Q² + 2PQ cos θ.

• θ = 0°: R = P + Q (the largest possible).
• θ = 180°: R = P − Q in size (the smallest).
• θ = 90°: R = √(P² + Q²).
• Two equal forces F at 120° give a resultant of exactly F.

A body is in equilibrium when the net force (and net turning effect) on it is zero. Three forces in equilibrium can be drawn as the sides of a closed triangle.

Stability improves with a lower centre of gravity and a wider base. This is why buses are loaded with heavy luggage below rather than on the roof, and why a cone standing on its base is stable, on its tip unstable, and on its side in neutral equilibrium.

Friction

Friction opposes relative motion between surfaces in contact.

• Static friction adjusts itself up to a maximum, the limiting friction, which equals μN (μ is the coefficient of friction, N the normal reaction).
• Kinetic (sliding) friction is a little less than limiting friction, and rolling friction is much smaller still. So it is easier to keep a body moving than to start it, and wheels and ball bearings save effort.
• The angle of repose θ, the steepest slope on which a body rests without sliding, satisfies tan θ = μ.
• Friction is useful (walking, brakes, writing, gripping) and wasteful (wear, heating). Lubricants, polishing, ball bearings and streamlining reduce it.

Circular motion

A body moving in a circle at constant speed is accelerating, because its direction keeps changing. The acceleration points towards the centre and equals v²/r, so a centripetal force F = mv²/r is needed.

• For a stone whirled on a string, tension supplies it; for a car on a flat road, friction; for the Moon, gravity.
• Banking of roads: tilting the road lets a component of the normal reaction supply the centripetal force, reducing reliance on friction. For a speed v on a curve of radius r, the ideal banking angle satisfies tan θ = v²/(rg).
• Centrifugal force is not a real force in the ground frame. It is the outward push felt inside the turning vehicle, which is really inertia of direction.

Worked NDA-style MCQs

Q1. A 4 kg gun fires a 20 g bullet at 400 m/s. The recoil speed of the gun is:
(a) 0.2 m/s (b) 2 m/s (c) 20 m/s (d) 4 m/s

V = mv/M = (0.02 × 400)/4 = 8/4 = 2 m/s. Answer: (b).

Q2. A force of 10 N acts for 0.5 s on a 2 kg body initially at rest. Its final speed is:
(a) 2.5 m/s (b) 5 m/s (c) 10 m/s (d) 1 m/s

Impulse = 10 × 0.5 = 5 N s = change in momentum, so v = 5/2 = 2.5 m/s. Answer: (a).

Q3. A 60 kg person stands on a scale in a lift that accelerates upward at 2 m/s² (g = 10 m/s²). The scale reads:
(a) 600 N (b) 480 N (c) 720 N (d) 120 N

R = m(g + a) = 60 × 12 = 720 N. Accelerating downward, it would read 480 N. Answer: (c).

Q4. Two forces of 3 N and 4 N act at right angles. Their resultant is:
(a) 7 N (b) 1 N (c) 5 N (d) 12 N

√(9 + 16) = 5 N. 7 N and 1 N are the maximum and minimum possible. Answer: (c).

Q5. A 1,000 kg car takes a curve of radius 50 m at 10 m/s. The centripetal force needed is:
(a) 200 N (b) 500 N (c) 2,000 N (d) 20,000 N

F = mv²/r = 1,000 × 100/50 = 2,000 N. Answer: (c).

Q6. A 5 kg block rests on a floor with μ = 0.4 (g = 10 m/s²). A horizontal push of 15 N is applied. The friction force on the block is:
(a) 20 N (b) 15 N (c) 0 (d) 5 N

Limiting friction is 0.4 × 50 = 20 N. The push is smaller, so the block stays put and static friction matches the push: 15 N. Answer: (b).

Common mistakes

• Saying action and reaction cancel. They act on different bodies.
• Using μN as the friction force every time. It is only the maximum; below it, static friction equals the applied force.
• Forgetting direction in momentum change. A ball that bounces back reverses its velocity, so the change is m(v + u), not m(v − u).
• Calling uniform circular motion unaccelerated. The speed is constant; the velocity is not.
• Thinking a scale shows true weight in a moving lift. It shows the normal reaction.

Practice set

1. It is harder to stop a loaded truck than a bicycle moving at the same speed because the truck has greater: (a) velocity (b) momentum (c) acceleration (d) speed
2. The momentum of a 0.5 kg ball moving at 10 m/s is: (a) 5 kg m/s (b) 20 kg m/s (c) 0.05 kg m/s (d) 25 kg m/s
3. The force needed to accelerate a 1,500 kg car at 2 m/s² is: (a) 750 N (b) 1,500 N (c) 3,000 N (d) 30,000 N
4. The apparent weight of a person in a freely falling lift is: (a) mg (b) 2mg (c) mg/2 (d) zero
5. A horse can pull a cart because: (a) action and reaction cancel (b) the ground's friction on the horse exceeds the cart's backward pull (c) the cart pulls the horse forward (d) the horse's weight pushes the cart
6. The same force acts on bodies of mass 2 kg and 3 kg. The ratio of their accelerations is: (a) 2 : 3 (b) 3 : 2 (c) 4 : 9 (d) 1 : 1
7. A 0.2 kg ball hits a wall at 20 m/s and rebounds at 20 m/s. The change in its momentum is: (a) 0 (b) 4 kg m/s (c) 8 kg m/s (d) 40 kg m/s
8. If the coefficient of friction is 1, the angle of repose is: (a) 30° (b) 45° (c) 60° (d) 90°

Answers:

1. (b). At equal speed, the heavier body has more momentum, so a given braking force needs longer to remove it.
2. (a). 0.5 × 10.
3. (c). F = ma = 1,500 × 2.
4. (d). R = m(g − g) = 0.
5. (b). Action and reaction act on different bodies; the forward push comes from the ground.
6. (b). a is inversely proportional to m.
7. (c). 0.2 × [20 − (−20)] = 8.
8. (b). tan θ = 1.

What to do next

• For each law, write three everyday examples in your own words; NDA often rephrases the standard ones.
• Solve 20 numericals mixing recoil, impulse, lifts and friction, drawing a quick force sketch for each.
• Revise units and motion basics if the equations of motion felt slow, then move to work, energy and power.