Linear Momentum and Impulse

In short

Linear momentum is \vec{p} = m\vec{v} — a vector that captures how much motion a body carries and in which direction. Newton's second law in momentum form is \vec{F} = d\vec{p}/dt. Impulse is the integral of force over time, \vec{J} = \int \vec{F}\,dt = \Delta\vec{p}. A large force for a short time and a small force for a long time can produce the same momentum change — it is the product that matters.

A fast bowler sends a cricket ball at 140 km/h toward the batsman. The batsman swings, and the ball flies back over the bowler's head at 140 km/h. The ball's speed did not change — 140 km/h before the bat, 140 km/h after. But something enormous happened in that one millisecond of contact. The ball was moving toward the batsman, and now it is moving away from the batsman. Whatever the bat did to the ball, it was not about changing the speed. It was about reversing the direction of motion.

That "motion with a direction" is what physicists call momentum. And the thing the bat did — exerting a massive force for the brief instant of contact — is called impulse. Momentum is what a moving body carries. Impulse is what changes it.

What momentum captures that speed alone cannot

Think about what it takes to stop a moving object.

A bicycle and a loaded truck are both moving at 10 m/s on the same road. Which is harder to bring to rest? The truck — it has more mass. Now consider two identical trucks: one crawling at 2 m/s, the other barrelling down the highway at 20 m/s. Which takes more effort to stop? The faster one.

So "how much effort it takes to stop something" depends on both mass and velocity. Newton called this the quantity of motion — the product of mass and velocity. The modern name is linear momentum, or simply momentum.

But there is a subtlety that speed alone misses entirely. A cricket ball flying toward the bowler at 5 m/s and the same ball flying toward the boundary at 5 m/s have the same speed. They do not have the same momentum. One carries its motion to the left, the other to the right. If you tried to stop each one, you would push in opposite directions. Momentum is not just a number — it is a vector. It has both magnitude and direction.

Same mass, same speed — but opposite momenta. Ball A carries momentum to the right (+0.8 kg·m/s). Ball B carries the same magnitude to the left (−0.8 kg·m/s). Speed is a number. Momentum is an arrow.

Linear momentum

The linear momentum of a body of mass m moving with velocity \vec{v} is

\vec{p} = m\vec{v}

Momentum is a vector. Its direction is the direction of velocity. Its SI unit is kg·m/s (kilogram-metre per second). There is no special name for this unit.

Reading the definition. m is the mass — a positive scalar. \vec{v} is the velocity — a vector with both magnitude and direction. Their product m\vec{v} scales the velocity vector by the mass. A 2 kg ball at 3 m/s to the right has momentum 2 \times 3 = 6 kg·m/s to the right. Double the mass and the momentum doubles. Reverse the velocity and the momentum reverses sign. The direction of \vec{p} is always the direction of \vec{v}.

A quick dimensional check: [p] = [m][v] = \text{kg} \cdot \text{m/s}. Mass tells you how much stuff is moving. Velocity tells you how fast and which way. Their product tells you how much motion the body is carrying, in which direction.

Newton's second law — the momentum version

You know Newton's second law as \vec{F} = m\vec{a}. But Newton himself did not write it that way. What he actually wrote in the Principia was closer to: the rate of change of the quantity of motion is proportional to the force applied. In modern symbols:

\vec{F} = \frac{d\vec{p}}{dt} \tag{1}

Here is how you get there. Start from the definition of momentum and differentiate with respect to time:

\frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}

Why: you want to know how momentum changes over time. Differentiate both sides of \vec{p} = m\vec{v}.

If the mass m is constant (the body is not gaining or losing matter), the mass comes out of the derivative:

\frac{d\vec{p}}{dt} = m\,\frac{d\vec{v}}{dt} = m\vec{a}

Why: the derivative of velocity with respect to time is acceleration by definition. Since m does not change with time, d(m\vec{v})/dt = m\,d\vec{v}/dt = m\vec{a}.

But m\vec{a} is exactly the net force \vec{F} by Newton's second law. Therefore:

\boxed{\vec{F} = \frac{d\vec{p}}{dt}} \tag{1}

Why: equating gives Newton's second law in its momentum form. The net external force on a body equals the rate at which its momentum changes. No net force means no change in momentum — that is Newton's first law as a special case.

This form is more general than \vec{F} = m\vec{a}. When mass changes — a rocket burning fuel, a chain piling up on a scale, monsoon rain falling into a moving cart — \vec{F} = m\vec{a} breaks down because m is not constant. But \vec{F} = d\vec{p}/dt still works, because it tracks the momentum of the whole system, not just the acceleration of a fixed mass.

Impulse — what a force does over time

Equation (1) says that force is the rate of change of momentum. Rearrange:

\vec{F}\,dt = d\vec{p} \tag{2}

Why: multiply both sides by the infinitesimal time interval dt. The left side is a tiny bit of "force times time." The right side is the tiny change in momentum that this bit of force-time produces.

Now integrate both sides from the initial time t_i to the final time t_f:

\int_{t_i}^{t_f} \vec{F}\,dt = \int_{t_i}^{t_f} d\vec{p} = \vec{p}_f - \vec{p}_i = \Delta\vec{p} \tag{3}

Why: the right side telescopes to the total change in momentum — final minus initial. The left side sums up all the tiny "force × time" contributions over the entire duration of the force.

The left side of equation (3) has a name: impulse, denoted \vec{J}.

Impulse and the impulse-momentum theorem

The impulse delivered by a force \vec{F} acting from time t_i to t_f is

\vec{J} = \int_{t_i}^{t_f} \vec{F}\,dt

The impulse-momentum theorem states:

\vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i

Impulse equals the change in momentum. Its SI unit is N·s (newton-second), which is dimensionally identical to kg·m/s.

Reading the theorem. The impulse-momentum theorem connects cause (a force acting over a span of time) to effect (a change in the body's momentum). It does not matter how the force varies during the interval — whether the force is steady, or spikes and drops, or oscillates wildly. All that matters is the integral. A constant force of 100 N for 0.1 s delivers the same impulse as a force that peaks at 1000 N but only lasts 0.01 s — both give \vec{J} = 10 N·s, and both produce the same change in momentum.

For a constant force, the integral simplifies:

\vec{J} = \vec{F}\,\Delta t \tag{4}

This is the version you will use in most introductory problems.

Why "how long" matters as much as "how hard"

The impulse-momentum theorem reveals something that \vec{F} = m\vec{a} alone does not make obvious: the total change in momentum depends on force multiplied by time, not force alone.

This is why a cricket fielder pulls their hands back while catching a ball. The ball arrives with momentum p = mv. To catch it, the fielder must reduce that momentum to zero — the change in momentum \Delta p is fixed by the ball's mass and speed. The impulse-momentum theorem says F_{\text{avg}} \cdot \Delta t = \Delta p. By pulling the hands back, the fielder increases the contact time \Delta t. Since the product F_{\text{avg}} \cdot \Delta t is fixed, a longer contact time means a smaller average force — and that means less pain and fewer broken fingers.

The same physics explains why cars have crumple zones, why you bend your knees when landing from a jump, why helmets have thick foam linings, and why a fall onto a muddy field hurts less than a fall onto a pakka road. In every case, the momentum change is the same — but spreading it over a longer time reduces the peak force on the body.

There is also a subtlety about direction. When a fielder catches a ball, the ball goes from +mv to zero — the momentum change is mv. But when the batsman hits the ball back at the same speed, the ball goes from +mv to -mv — the momentum change is 2mv, twice as large. Reversing direction requires twice the impulse that stopping requires.

Both balls arrive at the same speed (4 m/s). The caught ball (red) stops — its momentum change is $mv$. The batted ball (dark) reverses direction — its momentum change is $2mv$, twice as large. Batting the ball back requires twice the impulse of catching it. Click replay to watch again.

Impulse from a force-time graph

In real collisions, the force is not constant. When a cricket ball hits a bat, the force starts at zero the instant the ball touches the bat, climbs sharply to a peak as the ball compresses against the wood, and falls back to zero as the ball separates. The force-time profile looks like a narrow spike.

The impulse-momentum theorem says \vec{J} = \int \vec{F}\,dt — and you recognise an integral as the area under the curve. So impulse equals the area under the force-time graph.

The impulse is the area under the force-time curve (shaded). For a triangular pulse: $J = \frac{1}{2} F_{\max} \cdot \Delta t$. For a constant force, the area is a rectangle: $J = F \cdot \Delta t$. For any shape, integrate to find the area.

For a constant force, the graph is a rectangle and the area is F \cdot \Delta t. For a triangular pulse like the approximation above, the area is \frac{1}{2}F_{\max} \cdot \Delta t. For an irregular curve read from an experiment, you calculate the area by counting grid squares or using the trapezoidal rule.

The force-time graph makes the impulse-momentum theorem visible. Two force profiles with the same area deliver the same impulse and therefore the same momentum change, no matter how different the shapes look. A tall, narrow spike (bat hitting a ball — enormous force, tiny contact time) and a low, wide rectangle (hands pulling back while catching — gentle force, long contact time) can have exactly the same area.

Explore the trade-off yourself

The interactive figure below shows this trade-off in action. The impulse is fixed at J = 12.4 N·s — the momentum change when a 160 g cricket ball going at 38.9 m/s reverses completely. Drag the contact time and watch the average force adjust. Short contact time means enormous force. Long contact time means gentle force. The product is always the same.

Drag the red point to change the contact time. With $J = 12.4$ N·s fixed, shorter contact means a larger force. At 1 ms (a bat-ball collision), the force exceeds 12,000 N. At 50 ms (catching with soft hands), the force drops below 250 N. Same impulse, very different experience for your hands.

Worked examples

Example 1: The cricket ball reversal

A fast bowler delivers a cricket ball (mass 160 g) at 140 km/h. The batsman hits it straight back at 140 km/h. The bat-ball contact lasts about 1.0 ms. Find (a) the impulse delivered to the ball and (b) the average force on the ball during the hit.

The ball reverses direction. The speed is unchanged, but the momentum flips sign — the total change is $2mv$.

Step 1. Convert all quantities to SI units.

m = 160 \text{ g} = 0.160 \text{ kg}

u = 140 \text{ km/h} = \frac{140}{3.6} = 38.9 \text{ m/s}

\Delta t = 1.0 \text{ ms} = 0.001 \text{ s}

Why: SI units throughout — kilograms, metres per second, seconds. The conversion 1 km/h = 1/3.6 m/s comes from 1 km = 1000 m and 1 h = 3600 s.

Step 2. Define the sign convention and compute the momentum change.

Take the bowler-to-batsman direction as positive. The ball arrives at v_i = +38.9 m/s and leaves at v_f = -38.9 m/s (it reversed direction).

\Delta p = mv_f - mv_i = 0.160 \times (-38.9) - 0.160 \times (+38.9)

\Delta p = -6.22 - 6.22 = -12.44 \text{ kg·m/s}

Why: the ball reversed direction, so v_f and v_i have opposite signs. The momentum change is negative — directed back toward the bowler. The magnitude is |\Delta p| = 12.4 kg·m/s, which is 2mv because the ball bounced back at the same speed.

Step 3. The impulse equals the momentum change.

J = \Delta p = -12.44 \text{ N·s}

The magnitude is |J| = 12.4 N·s.

Why: the impulse-momentum theorem J = \Delta p directly. The unit N·s is dimensionally identical to kg·m/s.

Step 4. Compute the average force during the contact.

F_{\text{avg}} = \frac{J}{\Delta t} = \frac{-12.44}{0.001} = -12{,}440 \text{ N}

Why: for average force, J = F_{\text{avg}} \cdot \Delta t, so F_{\text{avg}} = J / \Delta t. The negative sign means the force on the ball is directed back toward the bowler — the bat pushes the ball backward.

Result: The impulse is 12.4 N·s. The average force on the ball during the hit is about 12{,}400 N — roughly the weight of 1,270 kg, or a small car pressing down. All concentrated into one millisecond.

The force is enormous because the contact time is tiny. The same momentum change spread over 0.5 seconds (the time it takes to catch the ball with soft hands) would need only about 25 N — the kind of force you apply when gently squeezing a stress ball.

Example 2: Bending your knees on landing

A person of mass 60 kg jumps from a wall 1.8 m high and lands on the ground. Compare the average force on their legs in two scenarios: (a) landing stiff-legged (contact time 10 ms) and (b) landing with bent knees (contact time 300 ms).

Both rectangles have the same area (impulse of 356 N·s). Stiff legs compress the time to 10 ms, driving the force to 35,600 N. Bending the knees stretches the time to 300 ms, dropping the force to 1,187 N.

Step 1. Find the landing speed using energy conservation.

The person falls freely through height h = 1.8 m. Using v^2 = 2gh:

v = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.28} = 5.94 \text{ m/s (downward)}

Why: all gravitational potential energy mgh converts to kinetic energy \frac{1}{2}mv^2 during free fall. Solving gives v = \sqrt{2gh}, which does not depend on the person's mass.

Step 2. Compute the impulse needed to bring the person to rest.

Take downward as positive. The person lands at v_i = +5.94 m/s and comes to rest: v_f = 0.

J = m(v_f - v_i) = 60 \times (0 - 5.94) = -356 \text{ N·s}

Why: the ground must push upward (negative in our convention) to stop the person. The impulse magnitude is 356 N·s.

Step 3. Compute the average force for stiff-legged landing.

F_{\text{stiff}} = \frac{|J|}{\Delta t} = \frac{356}{0.010} = 35{,}600 \text{ N}

Why: stiff legs allow almost no give — the contact time is about 10 ms. The impulse-momentum theorem gives the average force as |J|/\Delta t.

Step 4. Compute the average force for bent-knee landing.

F_{\text{bent}} = \frac{|J|}{\Delta t} = \frac{356}{0.300} = 1{,}187 \text{ N}

Why: bending the knees stretches the contact to 300 ms — thirty times longer. Since the impulse is the same, the force drops by the same factor of 30.

Result: Stiff legs produce an average force of 35,600 N (about 3,630 kg-weight — enough to fracture the tibia). Bent knees produce 1,187 N (about 121 kg-weight — well within the capacity of leg muscles). The force ratio is exactly the ratio of contact times: 300/10 = 30.

Every martial artist, gymnast, and paratrooper knows this instinctively: when you land, bend your knees. The impulse-momentum theorem tells you why — spreading the same impulse over a longer time reduces the force on your body.

Common confusions

"Momentum and kinetic energy are the same thing." They are not. Momentum is p = mv (linear in v), while kinetic energy is K = \frac{1}{2}mv^2 (quadratic in v). A car at 20 m/s has twice the momentum of the same car at 10 m/s, but four times the kinetic energy. Momentum is a vector; kinetic energy is a scalar. They are conserved under different conditions — momentum is conserved whenever external forces are absent, while kinetic energy is conserved only in elastic collisions.
"A heavier object always has more momentum." Not necessarily. A 0.05 kg bullet at 600 m/s has momentum 0.05 \times 600 = 30 kg·m/s. A 70 kg person walking at 0.4 m/s has momentum 70 \times 0.4 = 28 kg·m/s. The bullet carries more momentum despite being 1,400 times lighter — because it is moving 1,500 times faster.
"Impulse is just another name for force." Impulse is force multiplied by time. A 10 N force for 5 s and a 50 N force for 1 s deliver the same impulse (50 N·s) but are very different forces. Impulse has units of N·s; force has units of N. They are dimensionally different quantities.
"When a ball bounces off a wall, the momentum change is mv." That is the momentum change when a ball stops. When a ball bounces back at the same speed, the momentum changes from +mv to -mv, a total change of 2mv — twice as much. Reversing direction doubles the required impulse compared to merely stopping.
"The impulse-momentum theorem only works for constant forces." The theorem works for any force — constant, triangular, sinusoidal, or wildly irregular. The general form is \vec{J} = \int \vec{F}\,dt = \Delta\vec{p}. For constant forces, the integral simplifies to F\Delta t, but that is a special case.

If you came here to understand what momentum and impulse are, you have everything you need — you can stop here. What follows is for readers who want to see the deeper connections.

Why N·s and kg·m/s are the same unit

This is not a coincidence — it is a direct consequence of Newton's second law. One newton is defined as the force that gives a 1 kg mass an acceleration of 1 m/s²:

1 \text{ N} = 1 \text{ kg·m/s}^2

Multiply both sides by seconds:

1 \text{ N·s} = 1 \text{ kg·m/s}^2 \times \text{s} = 1 \text{ kg·m/s}

So the unit of impulse (N·s) and the unit of momentum (kg·m/s) are identically the same unit, expressed in different forms. This dimensional identity is the unit-level shadow of the impulse-momentum theorem — cause and effect share the same unit because the theorem says they are equal.

Variable mass: rockets and the momentum formulation

Consider a rocket. At time t, it has mass m and velocity v. In a small time interval dt, it expels exhaust of mass |dm| at velocity v_e relative to the rocket. The rocket's mass decreases to m + dm (where dm is negative because mass is lost) and its velocity increases to v + dv.

Apply conservation of momentum to the system (rocket + exhaust). Before the burn, the total momentum is mv. After the burn:

mv = (m + dm)(v + dv) + |dm|(v - v_e)

Expand and keep only first-order terms (dm \cdot dv \approx 0):

mv = mv + m\,dv + v\,dm + v\,|dm| - v_e\,|dm|

Since |dm| = -dm (mass leaving the rocket):

0 = m\,dv + v_e\,dm

m\,dv = -v_e\,dm

Why: the momentum lost by the rocket (through its exhaust) equals the momentum gained as thrust. The negative sign reflects that exhaust moves opposite to the rocket's acceleration.

This is the Tsiolkovsky rocket equation in differential form — and it only makes sense in the momentum formulation. Writing F = ma for a rocket is ambiguous because m is changing. The momentum form F = dp/dt handles variable-mass systems naturally.

ISRO's PSLV, carrying satellites to orbit, follows exactly this equation as it burns through its solid and liquid propellant stages. The payload reaches orbit not because any single force is large enough, but because the impulse — the integral of thrust over time across all four stages — is large enough to give the satellite the required momentum.

Momentum in two and three dimensions

Everything in this article generalises to two and three dimensions. Momentum is a vector:

\vec{p} = m\vec{v} = m(v_x\hat{i} + v_y\hat{j} + v_z\hat{k})

The impulse-momentum theorem becomes a vector equation:

\vec{J} = \Delta\vec{p}

In component form: J_x = \Delta p_x, J_y = \Delta p_y, J_z = \Delta p_z. Each component is independent. When a cricket ball hits the bat at an angle, you decompose the velocity into components along and perpendicular to the bat face, compute the momentum change in each direction separately, and combine the results.

This component-wise independence is what makes two-dimensional collision problems tractable — and it is the foundation of the next article on conservation of momentum.

Where this leads next

Conservation of Momentum — when the net external force on a system is zero, total momentum is constant. The most powerful tool in collision analysis.
Elastic Collisions — collisions where both momentum and kinetic energy are conserved. Derive the exact final velocities for any mass ratio.
Inelastic Collisions — collisions where kinetic energy is lost. Perfectly inelastic collisions, the coefficient of restitution, and real-world energy dissipation.
Centre of Mass — the point that moves as if all the system's mass and all external forces were concentrated there. The momentum of a system equals total mass times the velocity of its centre of mass.