Momentum & Impulse
Welcome to Momentum & Impulse. The Newton's Laws topic followed a force from instant to instant: at each moment the net force sets the acceleration. The Work & Energy topic followed a force through a distance and measured its effect as work. This topic follows a force through a time. Two quantities do the measuring. Momentum, mass times velocity, says how much motion a body carries and how hard it is to stop. Impulse, force times time, says how much push a force has delivered while it acted. This page defines both and practises them one force at a time. The next page joins them in the impulse–momentum theorem; the third shows when the total momentum of a group of bodies stays the same; the last applies that to collisions, recoil and explosions.
The topic has had 9 of the 192 questions so far, at least one in every paper, and five of them belong to this page. Two ask for the momentum of a moving body (December 2025, Q15), once paired with its kinetic energy (January 2026, Q19). Two are the same question with new numbers: the impulse of gravity on a block sliding down a 30° slope (December 2025, Q16; April 2026, Q18). The fifth asks for the impulse of static friction on a block resting on a slope (March 2026, Q9). All five are worked through below. None needs more than two lines of working. The marks are lost by multiplying the wrong force, and the wrong options are built from the forces that were not asked about. This page shows how to pick the right one.
Momentum: mass times velocity
Which is harder to stop, a tennis ball or a lorry, both moving at 20 m/s? The lorry, because it has far more mass. Which is harder to stop, a lorry at walking pace or the same lorry at 20 m/s? The fast one. How hard a body is to stop depends on its mass and its velocity together, and the second law says exactly how. A steady braking force F gives a body of mass m a deceleration a = F/m, and losing a velocity v at that rate takes a time t = v/a = mv/F. So the same force needs twice as long to stop a body with twice the mass, or with twice the velocity. What the force has to undo is the product mv. Newton called it the quantity of motion; today it is called momentum.
Because velocity has a direction, so does momentum, and the direction matters as much as the size. Two equal trolleys rolling apart at the same speed have momenta of the same size pointing opposite ways, so their momenta are different. Along a straight line, choose one direction as positive, as you did for velocities in the Describing Motion topic. A momentum the other way is then negative, and adding momenta along the line means adding the signed numbers. December 2025, Q1 asked which of four quantities is not a vector; momentum was one of the vectors, and the odd one out was potential energy.
An object with mass m = 0.2 kg moves in a straight line at a constant speed v = 20 m/s on a smooth horizontal surface. Its momentum is ( ).
A. 0.4 kg·m/s B. 4 kg·m/s C. 0.8 kg·m/s D. 8 kg·m/s
p = mv = 0.2 × 20 = 4 kg·m/s, pointing the way the object moves. The answer is B.
Check the decimal point, because the options are built around it: 0.2 kg is a fifth of a kilogram, and a fifth of 20 is 4. The smooth surface and the constant speed play no part in the sum. They only tell you that the momentum stays at 4 kg·m/s while the object moves.
A, 0.4 kg·m/s, drops a zero from the speed: 0.2 × 2. C, 0.8 kg·m/s, and D, 8 kg·m/s, do not follow from any common slip.
Momentum and kinetic energy
Kinetic energy, from the Work & Energy topic, is also built from a body's mass and speed: Ek = ½mv². The two quantities are easy to confuse, and January 2026 asked for both at once. They differ in three ways.
First, how they grow with speed. Momentum is proportional to v, kinetic energy to v². Double the speed of a car and its momentum doubles, but its kinetic energy grows four times.
Second, direction. Kinetic energy is a scalar: it has a size and no direction, and it is never negative. The two trolleys in the figure above have momenta that add up to zero, yet they carry 18 J of kinetic energy between them. A ball whirled round a circle at a steady speed keeps the same kinetic energy all the way round, while its momentum turns with its velocity and so changes all the time.
Third, units. Momentum is in kg·m/s; kinetic energy is in joules, and 1 J = 1 kg·m²/s². The unit is often the quickest way to tell which quantity an option is giving you.
Because both are made from m and v, one gives the other. Multiply the top and the bottom of ½mv² by m: ½mv² = m²v²/(2m) = (mv)²/(2m). The top is the momentum squared, so Ek = p²/(2m), and turned round, p = √(2mEk).
An object with a mass of 50 kg moves horizontally towards the right at a speed of 10 m/s. The momentum and kinetic energy are, respectively,
A. 500 kg·m/s, 2500 J B. 2500 kg·m/s, 500 J C. 500 kg·m/s, 5000 J D. 5 kg·m/s, 500 J
Step 1, the momentum: p = mv = 50 × 10 = 500 kg·m/s, to the right, the way the object moves.
Step 2, the kinetic energy: Ek = ½mv² = ½ × 50 × 10² = ½ × 50 × 100 = 2500 J. It has no direction.
Check: p²/(2m) = 500² ÷ 100 = 250 000 ÷ 100 = 2500 J, the same. The answer is A.
C, 5000 J, drops the ½: 50 × 100 = 5000. B has the right two numbers in the wrong places, 2500 as the momentum and 500 as the energy; the units decide which is which, kg·m/s for momentum and J for energy. D divides the mass by the speed, 50 ÷ 10 = 5, which is not any quantity, and gives 500 J, the momentum's number with the energy's unit.
Impulse: force multiplied by time
A force changes a body's momentum, and how much depends on how hard it pushes and for how long. Put a 2 kg trolley on a smooth, level track and push it from rest in two ways. A 20 N push for 1 s gives a = 20 ÷ 2 = 10 m/s², and after 1 s the trolley moves at 10 m/s. A 10 N push for 2 s gives a = 5 m/s², and after 2 s the trolley again moves at 10 m/s. Half the force for twice as long has the same effect, because both pushes have the same product of force and time, 20 N·s. That product is called the impulse of the force. The next page turns this example into a general rule, the impulse–momentum theorem; this page studies the impulse itself.
On a graph of force against time, a constant force is a horizontal line. The rectangle under it is F high and t wide, so its area is Ft, the impulse. Areas work for forces that change too, which the next page uses.
An impulse needs only a force and a time. Nothing has to move, and the force does not have to do any work. Compare work, W = Fs cos θ from the Work & Energy topic: it is zero when the body does not move, or when the force is at right angles to the motion. Impulse has neither escape. A book lying on a table for 10 s is pulled down by its weight and pushed up by the table. Neither force does any work, yet each has an impulse of mg × 10 s, one down and one up. The two impulses cancel, so the net impulse on the book is zero, but each force's own impulse is not.
Gravity's impulse on a slope
The papers' favourite impulse question puts a block on a 30° slope and asks for the impulse of gravity. Gravity on the block is its weight, mg, straight down, and it does not change while the block slides: the same size, the same direction, for the whole time. Its impulse is therefore I = mgt, whatever the angle of the slope, whether the slope is smooth or rough, and whether the block slides or rests. The details of the slope are there to tempt you into using a component.
The tempting number is mg sin θ × t. mg sin θ is the part of the weight along the slope, and on a smooth slope, as the Inclines & Connected Bodies page showed, it is the net force on the block. So mg sin θ × t is the net impulse, the combined impulse of gravity and the normal force. It is a real quantity, but it is not the one asked for.
An m = 5 kg object slides down a smooth incline plane with an angle of 30° to the horizontal. It starts at rest from a height of 20 m. The magnitude of its gravitational impulse over 2 s is ( ). (g = 10 m/s²)
A. 100 N·s B. 25 N·s C. 200 N·s D. 50 N·s
Step 1, name the force: the question asks for the impulse of gravity alone, not of the normal force and not of the two together.
Step 2, find it: gravity on the object is its weight, mg = 5 × 10 = 50 N, straight down. It has the same size and direction for the whole 2 s.
Step 3, multiply by the time: I = mgt = 50 × 2 = 100 N·s, straight down. The answer is A.
The angle, the smooth surface and the 20 m height were not used. The height does one quiet job: a slope 20 m high at 30° is 40 m long, and the object covers only 10 m in the first 2 s, so it is still sliding throughout. Gravity's impulse would be the same if it were not.
D, 50 N·s, is the trap: mg sin 30° × t = 25 × 2. On this smooth slope that is the net impulse of gravity and the normal force together, not gravity's own. B, 25 N·s, is mg sin 30° = 25 N, the force along the slope, with the time left out. C, 200 N·s, does not follow from any common slip.
December's paper used this object twice. Question 9 asked how far it slides in the same 2 s, and the Inclines & Connected Bodies page worked that out: a = g sin 30° = 5 m/s² and x = ½at² = 10 m. Question 16 asks for gravity's impulse, which needs none of that working. April 2026 paired its questions the same way, Q10 for the distance and Q18 for the impulse. When two questions share a set-up, read what the second one asks before reusing the first one's working.
The 50 N·s is not a meaningless number, either. After 2 s the object moves at v = at = 5 × 2 = 10 m/s, so its momentum is 5 × 10 = 50 kg·m/s down the slope, exactly the net impulse. That is no coincidence, and the next page shows why.
A force on a body at rest
Gravity is easy to find: it is always mg. The other forces on a body usually have to be worked out first, from the body's equilibrium or by resolving, and only then multiplied by the time. The routine is the same for any one force.
A block resting on a slope feels three forces: its weight mg, straight down; the normal force N, at right angles to the slope; and static friction f, up the slope, which stops it sliding down. The block does not move, so along the slope f = mg sin θ, and across it N = mg cos θ. A body at rest tempts students to answer 0 to every impulse question about it. Zero is the right answer to two other questions: the work done by each force, since nothing moves, and the net impulse, since the forces cancel. Each force on its own acts for the whole time and has an impulse.
A 0.4 kg object is at rest on a 30° inclined plane. The impulse of the frictional force acting on the object over 5 s is ( ). (g = 10 m/s²)
A. 10√3 N·s B. 20 N·s C. 0 D. 10 N·s
Step 1, name the force: friction. The object is at rest, so it is static friction, pointing up the slope.
Step 2, find its size: the object does not move, so along the slope friction balances the part of the weight that pulls it down: f = mg sin 30° = 0.4 × 10 × ½ = 2 N.
Step 3, multiply by the time: I = ft = 2 × 5 = 10 N·s, up the slope. The answer is D.
Each wrong option is the impulse of another force. A, 10√3 N·s, is the normal force's: N = mg cos 30° = 4 × (√3/2) = 2√3 N, and 2√3 × 5 = 10√3. B, 20 N·s, is gravity's: mgt = 4 × 5. C, 0, is the net impulse, because the three forces balance; it is also the work friction does, because the object does not move. None of them is the impulse of friction.
Impulse is not work
Work and impulse both multiply a force by something, and habits from one leak into the other. Work counts only the part of the force along the displacement, which is where the cos θ in W = Fs cos θ comes from. Impulse multiplies the whole force by a time. There is no displacement in it, so there is no direction to take a component along. A pull at an angle θ to the motion has impulse Ft along the rope, not Ft cos θ.
An angled pull changes a second impulse as well. Its upward part takes some of the body's weight, so the floor no longer has to push up with the whole of mg. Vertically the forces still balance, N + F sin θ = mg, so N = mg − F sin θ, and the normal force's impulse is (mg − F sin θ)t, not mgt.
A 10 kg sledge is pulled across smooth, level ice for 4 s by a rope that pulls with 50 N at 37° above the horizontal. Find the impulse of each force on the sledge, and the net impulse. (g = 10 m/s²; sin 37° = 0.6, cos 37° = 0.8)
Step 1, the forces: gravity, mg = 100 N, down; the pull, 50 N along the rope; the normal force N, up. The ice is smooth, so there is no friction.
Step 2, the normal force: resolve the pull. Its upward part is F sin 37° = 30 N and its forward part F cos 37° = 40 N. The sledge stays on the ice, so the vertical forces balance: N + 30 = 100, and N = 70 N, not 100 N.
Step 3, each force times 4 s: the pull, 50 × 4 = 200 N·s, along the rope; gravity, 100 × 4 = 400 N·s, down; the normal force, 70 × 4 = 280 N·s, up.
Step 4, the net impulse: vertically the forces balance, and horizontally only the 40 N forward part of the pull acts. The net force is 40 N forward, so the net impulse is 40 × 4 = 160 N·s, forward.
The pull's impulse is 200 N·s. The 160 N·s, Ft cos θ, is only the horizontal part of it; on smooth ice it also happens to be the net impulse, which is what makes it such a convincing wrong answer for the pull.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Momentum | p = mv, in kg·m/s, a vector pointing along v. December 2025 Q15: 0.2 × 20 = 4 kg·m/s. Convert grams to kilograms first. |
| Direction | Choose a positive direction; a momentum the other way is negative. The same speed in a different direction is a different momentum. |
| Momentum and kinetic energy | Ek = ½mv² = p²/(2m), a scalar in J; p = √(2mEk). Double v: p doubles, Ek grows four times. January 2026 Q19: 500 kg·m/s and 2500 J. |
| Impulse | I = Ft for a constant force, pointing along F. 1 N·s = 1 kg·m/s. On an F–t graph it is the area under the line. |
| Every force has its own impulse | Only a force and a time are needed: the body may be at rest, and the force may do no work or be cancelled by others. |
| Gravity's impulse | mgt, straight down, on any slope, smooth or rough, at any height. December 2025 Q16: 100 N·s; April 2026 Q18: 120 N·s. mg sin θ × t is the net impulse on a smooth slope, not gravity's. |
| A hidden force | Find it first. At rest on a slope: f = mg sin θ, N = mg cos θ. March 2026 Q9: 2 N × 5 s = 10 N·s. |
| Net impulse | The vector sum of all the impulses, Fnett. Zero for a body in equilibrium, though each force's impulse is not. |
| Impulse is not work | An angled pull has impulse Ft, not Ft cos θ. It also reduces the normal force: N = mg − F sin θ. |
| The common wrong options | The ½ dropped from Ek; p and Ek swapped; mg sin θ × t for gravity; 0 for a force on a body at rest; Ft cos θ for an angled pull; N = mg under an angled pull. |