Newton's Second Law of Motion — F = dp/dt, Impulse (NCERT 4.5)

1. What Does the Second Law Claim?

Newton's second law is the quantitative heart of mechanics: the rate of change of momentum of a body is proportional to the applied unbalanced force and happens in the force's direction. The first law told us that force is the only agent that changes a body's state of motion; the second law says how fast it changes — F = dp/dt. Everything computational in this chapter (and in most of mechanics) is this law plus bookkeeping: identify the net force along each axis, divide by mass, read off the acceleration. NTA's numerical-value questions — braking forces, average impact forces, recoil speeds, tension in strings — are all second-law calculations wearing different costumes.

The law is also a vector statement, which is the part students most often ignore: F = dp/dt means one independent equation per axis, F_x = dp_x/dt and F_y = dp_y/dt. A force applied at an angle changes only the momentum component along itself; the perpendicular component of momentum sails through untouched. That perpendicular-independence is the same axis discipline built in Chapter 3's two-dimensional kinematics — dynamics inherits it completely.

2. Complete Theory: From Momentum to Impulse

The exact form. Momentum p = mv, so F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt). For constant mass the second term dies and F = ma appears — the classroom form. But the momentum form is the true law: it survives when mass flows (a dropping sand-bag, a rocket burning fuel, a chain piling onto a table), where F = ma alone misleads. Unit consistency fixes the force unit: 1 newton is the force that gives a 1 kg mass an acceleration of 1 m s⁻², so 1 N = 1 kg m s⁻²; in CGS, 1 N = 10⁵ dyne. Dimensionally force is M L T⁻² — a check worth running on every derived expression, following Chapter 1's habit.

Component discipline with a quick example of the idea. A 2 kg body feels F = 10î + 20ĵ N. Then a_x = 10/2 = 5 m s⁻² and a_y = 20/2 = 10 m s⁻², computed separately — the x-force knows nothing of the y-motion. Any problem where forces act at an angle (a pull at 37°, a push down an incline) is this decomposition, and the normal-reaction scenarios on the common-forces page are precisely its consequence: pull upward at angle θ and N = mg − F sin θ because the vertical axis must obey its own second-law equation.

Impulse and the impulse–momentum theorem. Integrate the second law over a short collision: ∫F dt = Δp. The left side is the impulse J — for constant force, J = FΔt; for a spiking force, the area under the F–t graph. Hence J = Δp = mv − mu, the impulse–momentum theorem: for any short, violent interaction the momentum change is fixed by the event, and force trades against time. A hard stinger (large F, small Δt) and a soft long push (small F, large Δt) can carry identical impulse — the graph below makes the equality geometric. Because Δp is a vector, direction matters: a ball turned back by a bat has Δv = v − (−u) = v + u, and students who subtract magnitudes lose the mark.

Average force. Define F_avg = Δp/Δt — the constant force that would deliver the same impulse in the same time. Every "average force exerted by the wall/floor/glove" question is this definition. The practical corollary is the safety catalogue: cricketers pulling hands back, airbags, crumple zones, gymnasts bending knees on landing — all stretch Δt to shrink F_avg for an unavoidable Δp. And since Newton's third law (next page) guarantees the wall feels −F_avg, the same number answers "force on the wall" questions.

3. Visualising Impulse as Area

4. Solved Examples

5. Practice Questions

Q1. A force F = 6î − 8ĵ N acts on a 2 kg body. Find the magnitude of its acceleration and the direction of the net force.
ANSWER: a = F/m = 3î − 4ĵ m s⁻², magnitude √(9 + 16) = 5 m s⁻². The net force (6î − 8ĵ) has magnitude 2 × 5 = 10 N and points 53° below the +x axis (tan θ = 8/6 = 4/3). Components first, magnitude last — never add 6 and 8 as scalars.

Q2 (MCQ). A 0.5 kg ball strikes a wall perpendicularly at 10 m/s and rebounds at the same speed. If contact lasted 0.01 s, the average force on the wall is about: (a) zero (b) 500 N (c) 1000 N (d) 2000 N
ANSWER: (c) — Δp = 0.5 × (10 − (−10)) = 10 kg m/s; F = 10/0.01 = 1000 N on the ball, and by the third law 1000 N on the wall. Option (b) is the trap for students who used Δv = 10 instead of 20.

Q3. Sand falls at 2 kg/s onto a horizontal conveyor belt moving at 3 m/s. Find the force needed to keep the belt moving at constant speed, and the power spent.
ANSWER: Each second, 2 kg of sand must be accelerated from 0 to 3 m/s: F = dp/dt = (2 kg/s × 3 m/s) = 6 N. Power = Fv = 6 × 3 = 18 W. This is the v(dm/dt) term in action — the mass being accelerated changes with time, which is why the momentum form of the second law is the honest one.

6. Key Formulas & Takeaways

Where the momentum goes next: an isolated system of interacting bodies adds all the dp/dt's — internal (third-law) pairs cancel — and total momentum stands frozen. That is the conservation of momentum, the next page.

7. Frequently Asked Questions