{"id":707,"date":"2026-08-04T21:53:37","date_gmt":"2026-08-04T21:53:37","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=707"},"modified":"2026-08-24T13:03:52","modified_gmt":"2026-08-24T13:03:52","slug":"impulse-formula","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/impulse-formula\/","title":{"rendered":"Impulse Formula (J = F\u0394t): How to Calculate Impulse"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nThe impulse formula is J = F \u0394t: impulse equals the average force acting on an object multiplied by the time interval over which that force acts. Because force is the rate of change of momentum, the same impulse also equals the momentum change, J = \u0394p = mv &#8211; mu. Impulse is a vector measured in newton-seconds (N\u00b7s).\n<\/p><\/div>\n\n<p>Drop your phone on a tiled floor and it shatters. Drop it on a thick rug from the same height and it survives. The phone hits the ground with exactly the same momentum both times \u2014 so the change in momentum, and therefore the impulse, is identical.<\/p>\n\n<p>What changes is how long the stop takes. The rug stretches that stop over maybe ten times longer, and the force collapses by the same factor. That single trade-off is what the impulse formula lets you put a number on.<\/p>\n\n<h2>What Is the Impulse Formula?<\/h2>\n\n<p>The impulse formula is J = F \u0394t, where F is the average force in newtons and \u0394t is the contact time in seconds. Multiply the two and you get impulse in newton-seconds.<\/p>\n\n<p>There is a second version of the same equation, and you will use it just as often. Since a force acting for a time is exactly what changes an object&#8217;s momentum, impulse can also be worked out from mass and velocity alone: J = mv &#8211; mu.<\/p>\n\n<p>Which version you reach for depends entirely on what the question hands you. Force and a time? Use F \u0394t.<\/p>\n\n<p>Mass and two velocities? Use the momentum difference. Both routes give the same number in the same units.<\/p>\n\n<p>One detail that trips people up in the first line of their working: impulse is a <strong>vector<\/strong>. It points in the direction of the force, and in one-dimensional problems that means it carries a sign.<\/p>\n\n<h2>The Impulse Formula and Every Rearrangement You Need<\/h2>\n\n<p>Here are the two forms of the impulse equation, followed by every rearrangement an exam or a lab report will ask for.<\/p>\n\n<div class=\"pf-formula\">J = F \u0394t<\/div>\n\n<div class=\"pf-formula\">J = \u0394p = mv &#8211; mu<\/div>\n\n<p>Every symbol, with its SI unit:<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Symbol<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Quantity<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">SI unit<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Notes<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\"><strong>J<\/strong><\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Impulse<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">newton-second (N\u00b7s)<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Vector. Some textbooks write it as I or \u0394p.<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\"><strong>F<\/strong><\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Average resultant force<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">newton (N)<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">The time-average, not the peak value.<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\"><strong>\u0394t<\/strong><\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Time interval<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">second (s)<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Convert milliseconds first: 5 ms = 0.005 s.<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\"><strong>m<\/strong><\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Mass<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">kilogram (kg)<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Assumed constant during the interaction.<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\"><strong>u, v<\/strong><\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Initial and final velocity<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">metre per second (m\/s)<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Signed. A rebound flips the sign.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<h3>Rearranging for force or time<\/h3>\n\n<ul>\n<li><strong>Average force:<\/strong> F = J \/ \u0394t<\/li>\n<li><strong>Contact time:<\/strong> \u0394t = J \/ F<\/li>\n<li><strong>Final velocity:<\/strong> v = u + J \/ m<\/li>\n<li><strong>Mass:<\/strong> m = J \/ (v &#8211; u)<\/li>\n<\/ul>\n\n<p>Those four rearrangements cover essentially every impulse question you will meet. If you would rather check your algebra against a machine than trust it, our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/impulse\">Impulse Calculator<\/a> solves for impulse, force or time and shows the substitution line by line.<\/p>\n\n<h3>Why N\u00b7s and kg\u00b7m\/s are the same unit<\/h3>\n\n<p>A newton is a kg\u00b7m\/s<sup>2<\/sup>, so a newton-second is kg\u00b7m\/s<sup>2<\/sup> \u00d7 s = kg\u00b7m\/s. The two units are dimensionally identical, and both describe impulse.<\/p>\n\n<p>In practice, use N\u00b7s when you calculated from force and time, and kg\u00b7m\/s when you calculated from mass and velocity. Neither is wrong \u2014 examiners accept both.<\/p>\n\n<h2>How to Calculate Impulse in 4 Steps<\/h2>\n\n<p>To calculate impulse, pick the form of the formula that matches your data, convert everything to SI units, multiply while keeping track of sign, then quote the answer with a direction. Here is the routine in full.<\/p>\n\n<ol>\n<li><strong>Decide which pair you have.<\/strong> Force plus a time means J = F \u0394t. Mass plus two velocities means J = mv &#8211; mu. If you have all four, you can do it either way and cross-check.<\/li>\n<li><strong>Convert to SI and choose a positive direction.<\/strong> Grams to kilograms, milliseconds to seconds, km\/h to m\/s. Write down which way you are calling positive before any numbers go in.<\/li>\n<li><strong>Multiply or subtract, carrying the signs.<\/strong> A rebound means u and v have opposite signs, so the subtraction becomes an addition of magnitudes.<\/li>\n<li><strong>Quote units and direction, then sanity-check.<\/strong> An impulse of 4 N\u00b7s on a tennis ball is reasonable; 4000 N\u00b7s is not. Divide by the mass \u2014 if the implied velocity change is absurd, hunt for the slipped decimal point.<\/li>\n<\/ol>\n\n<p>Step 4 is the one students skip and the one that catches real errors. A quick division by mass turns an abstract answer back into a velocity you can judge by eye.<\/p>\n\n<p>Try it live below. Drag the change-in-momentum slider to fix the impulse, then stretch the contact time and watch the average force fall while the shaded area stays exactly the same.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Impulse and Momentum Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:560px}@media(max-width:760px){.pf-sim-frame{height:840px}}<\/style><iframe src=\"\/labs\/MomentumandImpuls.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>How to Find Impulse from a Force-Time Graph<\/h2>\n\n<p>Impulse is the area under a force-time graph. That one sentence replaces the whole formula, because it works whether the force is constant, spiky or wildly irregular.<\/p>\n\n<p>When the force is constant, the area is a rectangle and you are back to J = F \u0394t. When it is not constant, you break the shape into pieces you can measure \u2014 and the answer is still in N\u00b7s, because the axes are newtons and seconds.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/impulse-formula-four-force-time-shapes-area.webp\" width=\"1440\" height=\"540\" alt=\"Four force-time graph shapes and the impulse formula for the area under each\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:720px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Four common force-time shapes. Whatever the outline, the shaded area under it is the impulse in newton-seconds.<\/p>\n\n<h3>Working with an irregular curve<\/h3>\n\n<p>Real collisions produce a lopsided bump, not a neat triangle. Two methods handle it without calculus.<\/p>\n\n<ul>\n<li><strong>Count squares.<\/strong> Work out the impulse of one grid square (its width in seconds \u00d7 its height in newtons), count the squares under the curve, and multiply. Half-covered squares count as a half.<\/li>\n<li><strong>Split into strips.<\/strong> Chop the curve into trapezium slices, find each area, and add them. Narrower strips give a closer answer.<\/li>\n<\/ul>\n\n<p>If you have already met the area-under-a-graph trick in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/motion-graphs\/\">motion graphs<\/a>, this is the same skill on different axes \u2014 velocity-time gives displacement, force-time gives impulse.<\/p>\n\n<h2>Average Force vs Peak Force: The Number Students Get Wrong<\/h2>\n\n<p>The F in J = F \u0394t is the average force over the contact, never the peak force. Substituting the peak value is the single most common way an otherwise correct impulse calculation ends up wrong.<\/p>\n\n<p>Picture a symmetric triangular spike \u2014 force climbing from zero to a maximum and back down. Its area is \u00bd \u00d7 peak \u00d7 \u0394t, so the average force is exactly <strong>half<\/strong> the peak.<\/p>\n\n<p>Use the peak instead and your answer is out by a factor of two. In a crash-safety context, that is the difference between a survivable deceleration and a fatal one.<\/p>\n\n<p>This is also the reason impulse is the practical way to estimate collision forces at all. As Georgia State University&#8217;s <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/impulse.html\" target=\"_blank\" rel=\"noopener\">HyperPhysics notes on the impulse of force<\/a> point out, the force during an impact is almost never measurable directly \u2014 but the momentum change and the contact time usually are.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/impulse-formula-two-force-time-graphs-equal.webp\" width=\"1440\" height=\"994\" alt=\"Impulse formula - Two force-time graphs with equal shaded areas showing a rigid stop at 49 kilonewtons for 0.02 seconds and an airbag stop at 2.8 kilonewtons for 0.35 seconds\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:720px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Both shaded blocks have the same area, so both deliver the same 980 N\u00b7s impulse to a 70 kg driver stopping from 14 m\/s. Stretching the stop from 0.02 s to 0.35 s cuts the force by a factor of 17.5.<\/p>\n\n<p>That is the whole design brief for an airbag, a crash mat and a climbing rope. None of them reduces the impulse \u2014 the momentum change is fixed by the crash. They only stretch \u0394t.<\/p>\n\n<h2>Where the Impulse Formula Gets Used in Practice<\/h2>\n\n<p>The impulse formula turns up wherever a force acts briefly and you need a number for it. These are worked from the assumptions in each row, so you can follow the arithmetic rather than take it on trust.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Situation<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Mass and velocity change<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Contact time \u0394t<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Impulse J<\/th>\n<th style=\"border:1px solid #D9CFB8;padding:8px;text-align:left;\">Average force F = J \/ \u0394t<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Football kicked from rest<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.42 kg, 0 to 24 m\/s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.012 s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">10.1 N\u00b7s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">840 N<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Tennis return (ball reverses)<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.058 kg, 30 m\/s in to 40 m\/s out<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.005 s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">4.06 N\u00b7s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">810 N<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Driver stopped, rigid column<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">70 kg, 14 m\/s to 0<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.020 s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">980 N\u00b7s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">49,000 N<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Same driver, airbag deployed<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">70 kg, 14 m\/s to 0<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.35 s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">980 N\u00b7s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">2,800 N<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Model rocket motor burn<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Thrust profile, peak 90 N<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">1.70 s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">126 N\u00b7s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">74.1 N<\/td>\n<\/tr>\n<tr>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">Ball bouncing off the floor<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.20 kg, 6.0 m\/s down to 4.0 m\/s up<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">0.025 s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">2.0 N\u00b7s<\/td>\n<td style=\"border:1px solid #D9CFB8;padding:8px;\">80 N (net)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<h3>Rockets: impulse you can buy off a shelf<\/h3>\n\n<p>Rocket engineers care about impulse more than thrust, because thrust alone says nothing about how long it lasts. NASA defines the <a href=\"https:\/\/www.grc.nasa.gov\/WWW\/k-12\/airplane\/specimp.html\" target=\"_blank\" rel=\"noopener\">total impulse of a rocket<\/a> as the average thrust multiplied by the total firing time \u2014 J = F \u0394t with a rocket motor in place of a tennis racket.<\/p>\n\n<p>That is why motors are sold by total impulse in newton-seconds. It is the honest measure of how much momentum the motor can hand the vehicle.<\/p>\n\n<h3>Continuous jets: impulse per second<\/h3>\n\n<p>A hose firing 12 kg of water per second at 20 m\/s onto a wall delivers 240 kg\u00b7m\/s of momentum change every second. Divide impulse by time and you have a steady 240 N force.<\/p>\n\n<p>Same formula, rearranged. This rate form is how fire hoses kick back and how jet engines push aircraft forward.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/can-anyone-tell-me-what-causes-the-ball-to-warp-like-this-v0-00bcc09mbc6d1.webp\"\n       alt=\"Ball flattening against a racket, the short contact time used in the impulse formula J = F \u0394t\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" width=\"1920\" height=\"1280\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Contact lasts only a few milliseconds, which is why the average force reaches hundreds of newtons.<\/figcaption>\n<\/figure>\n\n<h2>4 Mistakes That Wreck Impulse Calculations<\/h2>\n\n<p>Most lost marks on impulse questions come from four specific slips, and none of them is about understanding the physics. They are all about the substitution line.<\/p>\n\n<h3>Mistake 1: using the peak force<\/h3>\n\n<p>J = F \u0394t needs the time-averaged force. Read the peak off a graph and you will typically double your answer, because a symmetric spike averages half its maximum.<\/p>\n\n<h3>Mistake 2: forgetting the sign flip on a rebound<\/h3>\n\n<p>If a ball arrives at 30 m\/s and leaves at 40 m\/s the other way, the velocity change is 70 m\/s, not 10 m\/s. Set your positive direction first, write u = -30 m\/s and v = +40 m\/s, and let the algebra handle it.<\/p>\n\n<p>Writing the sign convention down before you substitute costs one line and removes the error entirely.<\/p>\n\n<h3>Mistake 3: leaving time in milliseconds<\/h3>\n\n<p>A contact time of 5 ms is 0.005 s. Substitute 5 instead and any force from F = J \/ \u0394t lands a thousand times too small, while any impulse from J = F \u0394t lands a thousand times too big.<\/p>\n\n<p>Both wrong answers look tidy enough to survive an unchecked script. Convert the units on the line where you write them down, not later.<\/p>\n\n<h3>Mistake 4: confusing net impulse with the contact force<\/h3>\n\n<p>J = \u0394p gives the impulse of the <em>resultant<\/em> force. If a question asks for the force from the floor on a bouncing ball, weight acts during the contact too, so the floor&#8217;s force is the net impulse divided by \u0394t <em>plus<\/em> the weight.<\/p>\n\n<p>For a brief, violent impact the weight term is often tiny. Ignore it silently and you will still lose the mark for method.<\/p>\n\n<h2>How the Impulse Formula Connects to Momentum, Force and Energy<\/h2>\n\n<p>The impulse formula is Newton&#8217;s second law with the time multiplied through. Newton&#8217;s law says force is the rate of change of momentum, F = \u0394p \/ \u0394t, and rearranging that single line gives F \u0394t = \u0394p \u2014 the impulse-momentum theorem.<\/p>\n\n<p>If you want the concepts behind that step rather than the arithmetic, our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/momentum-and-impulse\/\">momentum and impulse<\/a> derives the theorem and explains why it matters, and the write-up on <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a> covers the F = ma form you probably met first.<\/p>\n\n<h3>Impulse and collisions<\/h3>\n\n<p>In a collision, the two objects exert equal and opposite forces on each other for exactly the same length of time. Equal force, equal time, so equal and opposite impulses \u2014 which is precisely why <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/conservation-of-momentum\/\">momentum is conserved<\/a>.<\/p>\n\n<p>That holds whether the collision is bouncy or sticky. The distinction between <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/elastic-inelastic-collisions\/\">elastic and inelastic collisions<\/a> is about kinetic energy, not impulse.<\/p>\n\n<h3>Impulse is not energy<\/h3>\n\n<p>Here is a case worth remembering. A ball bounces off a wall with the same speed it arrived at, so its <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a> is unchanged \u2014 zero energy transfer.<\/p>\n\n<p>Its impulse, though, is 2mv, the largest it could possibly be. Momentum depends on velocity linearly and carries direction; energy depends on speed squared and does not. They are different quantities and they behave differently.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A constant force of 25 N acts on a trolley for 4.0 s. Calculate the impulse delivered.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Force and time are given, so use the impulse formula J = F \u0394t.<\/p>\n<p>Step 2: Substitute with units: J = 25 N \u00d7 4.0 s.<\/p>\n<p>Step 3: Solve: J = 100 N\u00b7s, in the direction of the force.<\/p>\n<p><strong>Answer: J = 100 N\u00b7s (2 s.f.), along the direction of the applied force.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A 0.42 kg football is kicked from rest and leaves the boot at 24 m\/s. The boot is in contact with the ball for 0.012 s. Find the impulse and the average force on the ball.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Mass and two velocities are given, so use J = mv &#8211; mu with u = 0.<\/p>\n<p>Step 2: Substitute with units: J = 0.42 kg \u00d7 24 m\/s &#8211; 0.42 kg \u00d7 0 = 10.08 kg\u00b7m\/s.<\/p>\n<p>Step 3: Rearrange the impulse formula for force: F = J \/ \u0394t = 10.08 N\u00b7s \/ 0.012 s = 840 N.<\/p>\n<p><strong>Answer: J = 10.1 N\u00b7s (3 s.f.) and F = 840 N (2 s.f.), both forward.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A 0.058 kg tennis ball arrives at 30 m\/s and is returned at 40 m\/s in the opposite direction. Contact lasts 5.0 ms. Calculate the impulse and the average force.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Take the outgoing direction as positive, so u = -30 m\/s and v = +40 m\/s.<\/p>\n<p>Step 2: Velocity change: v &#8211; u = 40 &#8211; (-30) = 70 m\/s.<\/p>\n<p>Step 3: Impulse: J = m(v &#8211; u) = 0.058 kg \u00d7 70 m\/s = 4.06 kg\u00b7m\/s.<\/p>\n<p>Step 4: Convert the time, 5.0 ms = 0.0050 s, then F = J \/ \u0394t = 4.06 \/ 0.0050 = 812 N.<\/p>\n<p><strong>Answer: J = 4.1 N\u00b7s and F = 810 N (2 s.f.), both in the return direction.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A 70 kg driver travelling at 14 m\/s is brought to rest in a crash. Find the average force (a) against a rigid steering column stopping them in 0.020 s and (b) against an airbag stopping them in 0.35 s.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: The impulse is fixed by the momentum change: J = mv &#8211; mu = 0 &#8211; 70 kg \u00d7 14 m\/s = -980 kg\u00b7m\/s, so the magnitude is 980 N\u00b7s.<\/p>\n<p>Step 2: Rigid column: F = J \/ \u0394t = 980 \/ 0.020 = 49,000 N.<\/p>\n<p>Step 3: Airbag: F = J \/ \u0394t = 980 \/ 0.35 = 2,800 N.<\/p>\n<p>Step 4: Sanity check by dividing by mass: 49,000 \/ 70 = 700 m\/s<sup>2<\/sup>, about 71 g. The airbag gives 40 m\/s<sup>2<\/sup>, about 4 g.<\/p>\n<p><strong>Answer: (a) 49 kN, about 71 g. (b) 2.8 kN, about 4 g \u2014 a 17.5-fold reduction from stretching \u0394t alone.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">On a force-time graph the force rises linearly from 0 to 600 N over 0.030 s, then falls linearly back to 0 over the next 0.030 s. Find the impulse and the average force.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Impulse is the area under the graph, and this shape is a triangle of base 0.060 s and height 600 N.<\/p>\n<p>Step 2: Substitute: J = \u00bd \u00d7 base \u00d7 height = \u00bd \u00d7 0.060 s \u00d7 600 N.<\/p>\n<p>Step 3: Solve: J = 18 N\u00b7s.<\/p>\n<p>Step 4: Average force: F = J \/ \u0394t = 18 \/ 0.060 = 300 N, exactly half the 600 N peak.<\/p>\n<p><strong>Answer: J = 18 N\u00b7s and average F = 300 N (the peak is 600 N).<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A model rocket motor produces thrust that ramps from 0 to 90 N in 0.20 s, holds 90 N for 1.10 s, then falls to 0 over 0.40 s. Calculate the total impulse and the average thrust.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Split the area under the thrust-time graph into three pieces.<\/p>\n<p>Step 2: Ramp up (triangle): \u00bd \u00d7 0.20 s \u00d7 90 N = 9.0 N\u00b7s.<\/p>\n<p>Step 3: Steady burn (rectangle): 90 N \u00d7 1.10 s = 99.0 N\u00b7s.<\/p>\n<p>Step 4: Ramp down (triangle): \u00bd \u00d7 0.40 s \u00d7 90 N = 18.0 N\u00b7s.<\/p>\n<p>Step 5: Total impulse: J = 9.0 + 99.0 + 18.0 = 126.0 N\u00b7s over a burn of 0.20 + 1.10 + 0.40 = 1.70 s.<\/p>\n<p>Step 6: Average thrust: F = J \/ \u0394t = 126.0 \/ 1.70 = 74.1 N.<\/p>\n<p><strong>Answer: total impulse J = 126 N\u00b7s and average thrust F = 74.1 N (3 s.f.).<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 0.20 kg ball strikes the floor at 6.0 m\/s and rebounds at 4.0 m\/s. Contact lasts 0.025 s. Find (a) the net impulse on the ball and (b) the average normal force from the floor. Take g = 9.81 m\/s squared.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Take upwards as positive, so u = -6.0 m\/s and v = +4.0 m\/s.<\/p>\n<p>Step 2: Net impulse: J = m(v &#8211; u) = 0.20 kg \u00d7 (4.0 &#8211; (-6.0)) m\/s = 0.20 \u00d7 10.0 = 2.0 kg\u00b7m\/s upwards.<\/p>\n<p>Step 3: The net impulse comes from the normal force minus the weight, so J = (N &#8211; mg)\u0394t.<\/p>\n<p>Step 4: Rearrange for N: N = J \/ \u0394t + mg = 2.0 \/ 0.025 + 0.20 \u00d7 9.81.<\/p>\n<p>Step 5: Solve: N = 80.0 + 1.96 = 81.96 N.<\/p>\n<p><strong>Answer: (a) net impulse J = 2.0 N\u00b7s upwards. (b) average normal force N = 82 N (2 s.f.). The weight contributes under 3% here, but the method still needs it.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for impulse?<\/summary><div class=\"pf-faq-item-answer\">\nThe formula for impulse is J = F \u0394t, where F is the average resultant force in newtons and \u0394t is the time interval in seconds. Impulse also equals the change in momentum, so J = mv &#8211; mu is an equally valid form. Use whichever version matches the data in the question.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What are the units of impulse?<\/summary><div class=\"pf-faq-item-answer\">\nThe SI unit of impulse is the newton-second (N\u00b7s), which is identical to the momentum unit kg\u00b7m\/s. A newton is 1 kg\u00b7m\/s<sup>2<\/sup>, so multiplying by seconds gives kg\u00b7m\/s. Both notations are correct and interchangeable, and examiners accept either.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you calculate impulse from a force-time graph?<\/summary><div class=\"pf-faq-item-answer\">\nImpulse is the area under a force-time graph. For a constant force the area is a rectangle, F \u00d7 \u0394t. For a triangular spike it is \u00bd \u00d7 peak force \u00d7 time. For an irregular curve, count grid squares or split the shape into trapezium strips and add the areas.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Can impulse be negative?<\/summary><div class=\"pf-faq-item-answer\">\nYes. Impulse is a vector, so its sign shows direction relative to whichever way you chose as positive. A braking force acting against the motion gives a negative impulse and reduces momentum in that direction. The negative sign never means the impulse is somehow smaller.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you find impulse without knowing the force?<\/summary><div class=\"pf-faq-item-answer\">\nUse the momentum form, J = mv &#8211; mu. You only need the mass and the initial and final velocities, and the answer comes out in kg\u00b7m\/s. This is the standard route when a question gives a collision or a bounce but no force value.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is impulse the same as momentum?<\/summary><div class=\"pf-faq-item-answer\">\nNo. Momentum, p = mv, is a property an object has at an instant. Impulse is what a force delivers over a time interval, and it equals the change in momentum rather than the momentum itself. They share the same units, which is why the two are easily confused.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why is the average force used in the impulse formula and not the peak force?<\/summary><div class=\"pf-faq-item-answer\">\nThe impulse formula multiplies force by time, so it needs the value that, held constant for the whole interval, would give the same area under the force-time graph. That value is the time-averaged force. For a symmetric triangular pulse the peak force is exactly twice the average.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>The impulse formula is J = F\u0394t, and it also equals the change in momentum. Work through 7 solved examples, learn to read impulse off a force-time graph, and see the four mistakes that cost the most marks.<\/p>\n","protected":false},"author":1,"featured_media":708,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-707","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/707","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=707"}],"version-history":[{"count":7,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/707\/revisions"}],"predecessor-version":[{"id":1511,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/707\/revisions\/1511"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/708"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=707"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=707"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=707"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}