Newton's Second Law and Impulse

F = dp/dt is the most powerful equation in mechanics: it defines force, explains why F = ma is only a special case, and its time integral — impulse — is the tool for collision problems.

Newton's Second Law1 min · Free lecture

In this lesson

State the second law as F = dp/dt and derive F = ma for constant mass
Use the impulse–momentum theorem J = Δp
Solve JEE problems with time-varying forces via impulse

Newton's second law says the net force on a body equals the rate of change of its momentum: $\vec F = d\vec p / dt$ . When mass is constant, $d\vec p/dt = m\vec a$ , giving the familiar $\vec F = m\vec a$ — but for rockets, conveyors and variable-mass systems only the momentum form is correct.

\vec F = \frac{d\vec p}{dt}, \qquad \vec F = m\vec a \; (\text{constant } m)

Newton's second law — general and constant-mass forms

\vec F — net external force (N)
\vec p — linear momentum m\vec v (kg·m/s)
m — mass (kg)
\vec a — acceleration (m/s^2)

Integrate the second law over time and you get the impulse–momentum theorem: the impulse $\vec J$ delivered by a force over a time interval equals the change in momentum. It is the natural tool when forces act for short times — collisions, catches, rebounds.

\vec J = \int_{t_1}^{t_2} \vec F \, dt = \Delta \vec p

Impulse–momentum theorem

\vec J — impulse (N·s = kg·m/s)
\Delta \vec p — change in momentum

Impulse is the area under the F–t graph. JEE problems give you graphs or piecewise forces precisely so you integrate — never multiply F × t blindly when the force varies.

Worked example

A 150 g ball hits a wall at 20 m/s and rebounds at 15 m/s along the same line. Contact lasts 0.05 s. Find the impulse and the average force on the ball.

Take direction away from the wall as positive: u = −20 m/s, v = +15 m/s
J = Δp = m(v − u) = 0.15 × (15 − (−20)) = 0.15 × 35 = 5.25 N·s
F_avg = J / Δt = 5.25 / 0.05 = 105 N

Answer: J = 5.25 N·s (away from wall); F_avg = 105 N

Animated figure: free-body-diagram (rendered in video)

Free-body diagram: isolate the body, draw every force, then apply F = dp/dt

Transcript (1 min)

WEBVTT

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Newton's Second Law and Impulse
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F = dp/dt — the general second law
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J = ∫ F dt = Δp — impulse–momentum theorem
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Impulse = area under F–t graph
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Free-body diagram → one clean F = ma
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Example: m = 150 g, 20 → 15 m/s rebound → J = 5.25 N·s, F_avg = 105 N
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Recap: F = dp/dt · J = Δp · FBDs

Frequently asked

When is F = ma wrong and F = dp/dt necessary?

F = dp/dt is the real law; F = ma follows only when mass is constant. For variable-mass systems — rockets, conveyor belts, sand leaking from carts — you must use the momentum form.

Why is impulse useful for collision problems?

Collision forces act for milliseconds and vary wildly with time. Integrating them is hopeless — but the impulse–momentum theorem J = Δp bypasses the force entirely: you only need velocities before and after the contact.

Printable notes (free)

Practice

3 free · 1 premium

Q1 · Numerical · difficulty 2/5

A ball of mass 0.1 kg moving at 10 m/s hits a wall and rebounds along the same line with speed 0.8 × its original speed. Find the magnitude of the impulse delivered to the ball.

J = m(v - u), \quad v = fu

Hint available · solution with Premium.

Q2 · Numerical · difficulty 3/5

A force acts on a 4.5 kg ball initially at rest. The force–time graph is a triangle rising from 0 to a peak of 80 N in 2 s and back to 0. Find the final speed of the ball.

Hint available · solution with Premium.

Q3 · Numerical · difficulty 1/5

What net force (in N) is needed to accelerate a 2 kg body at 8 m/s²?

Hint available · solution with Premium.

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