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A fielder who catches a hard ball with stiff hands feels it sting. One who lets the hands travel back with the ball barely notices. Both stopped the same ball. Impulse explains the difference, and the same idea sits behind force-time graph questions in JEE and NEET. This post gives the definitions, four worked examples and the mistakes to avoid.
What is impulse, and why does stopping time matter?
NCERT's Laws of Motion chapter defines impulse as force multiplied by the time it acts, and shows that this product equals the change in momentum:
J = F × Δt = Δp = m(v − u)
For a varying force, F is the average force over the contact time. NCERT uses the cricket example directly: a seasoned cricketer draws the hands back, allowing a longer time for the ball to stop, so a smaller force is needed. The same change in momentum in a shorter time needs a greater force.
Impulse is in the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus, under laws of motion. Check the current year's syllabus before your exam.
Example 1. A 0.16 kg ball arrives at 20 m/s and is caught.
Δp = 0.16 × 20 = 3.2 kg m/s, so J = 3.2 N s.
Stiff hands that stop it in 0.01 s: F = 3.2/0.01 = 320 N.
Hands drawn back so it stops in 0.10 s: F = 3.2/0.10 = 32 N.
Ten times the stopping time, one tenth of the average force. An airbag, a crash helmet's foam or a sand pit for long jumpers all work the same way: they cannot change the momentum you must lose, but they stretch the time over which you lose it.
How do you read impulse from a force-time graph?
Since J = ∫F dt, impulse is the area under a force-time graph. Real impacts rarely have a constant force. A typical question shows a triangle, a trapezium or a half-sine pulse.
Example 2. A 0.20 kg ball hits a wall at 10 m/s and rebounds at 8 m/s. The contact force rises linearly to a peak and falls back to zero in 5.0 ms. Find the impulse, the peak force and the average force.
Take the rebound direction as positive. Initial velocity −10 m/s, final +8 m/s.
J = m(v − u) = 0.20 × (8 − (−10)) = 0.20 × 18 = 3.6 N s.
The area of a triangle is ½ × base × height, so 3.6 = ½ × 0.005 × F_peak, giving F_peak = 1440 N.
Average force = J/Δt = 3.6/0.005 = 720 N, exactly half the peak for a triangular pulse.
Example 3. A 0.50 kg block at rest receives a horizontal force that rises from 0 to 40 N in 0.10 s, stays at 40 N for 0.20 s, then drops to 0 in 0.10 s. Find its final speed on a smooth floor.
Area of the trapezium = 2 triangles + 1 rectangle = ½ × 0.10 × 40 + 0.20 × 40 + ½ × 0.10 × 40 = 2 + 8 + 2 = 12 N s.
v = J/m = 12/0.50 = 24 m/s.
If part of a graph lies below the time axis, that area counts as negative impulse. Add the areas with their signs.
Why does weight matter when you land?
Example 4. A 60 kg student jumps off a wall and lands at 4.0 m/s. Find the average force from the ground if the student stops in 0.20 s by bending the knees, and in 0.020 s with stiff legs. Take g = 9.8 m/s².
The momentum to lose is 60 × 4.0 = 240 kg m/s.
The net upward force is N − mg, and (N − mg)Δt = 240.
Bent knees: N − 588 = 240/0.20 = 1200, so N = 1788 N, about three times body weight.
Stiff legs: N − 588 = 240/0.020 = 12000, so N ≈ 12600 N, over twenty times body weight.
For a ball bouncing off a wall in milliseconds, weight is tiny compared with the contact force and is usually ignored. For a landing lasting a fifth of a second, it is not.
What goes wrong in impulse problems?
Research suggests the hard part is not the formula. In Lawson and McDermott (1987), university students compared the momentum and kinetic energy changes of two pucks pushed by the same constant force. Most could not relate the algebra learned in class to the simple motion they watched. The fix is to connect each symbol to the physical event before calculating. Common slips:
No sign convention. Pick a positive direction and write it down before writing any velocity.
Peak force for average force. On a triangular graph the average is half the peak.
Ignoring weight in landings and slow catches.
Units. N s and kg m/s are the same unit; use whichever the question uses.
Thinking a cushion reduces momentum change. It reduces force by increasing time. Δp is fixed by the speeds.
Impulse then leads straight into collisions. Our post on collisions and the coefficient of restitution picks up from here, and reading physics graphs covers areas and slopes more generally.
A practice routine for this week
Day 1: five catch and landing problems. For each, sketch the motion and mark the positive direction first.
Day 2: five force-time graphs (triangle, trapezium, a graph with a negative part). Find impulse by area, then final velocity.
Day 3: three rebound problems where you also find average force, then two mixed with collisions.
When you get a graph question wrong, compare your set-up with a full solution rather than just the answer. The Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can see where your sign convention or area split went off.
Start with Example 2: change the rebound speed to 5 m/s and predict the new peak force before calculating.
