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Momentum and Impulse
The Concept of Momentum
Momentum is a Vector quantity representing the 'quantity of motion' an object possesses, determined by its mass and velocity. The SI unit is kg·m/s, which is equivalent to N·s.
$$p = mv$$
Momentum is the product of how much stuff is moving and how fast it is going.
$p$=Linear Momentum(kg·m/s)
$m$=Mass(kg)
$v$=Velocity(m/s)
$v = 0$
→The object has zero momentum regardless of its mass.
$m$ is changing (e.g., rocket)
→Use $F = \frac{dp}{dt}$ instead of $F = ma$, since mass is not constant.
Directionality: Because velocity is a vector, momentum always points in the same direction as the motion. Direction matters in all momentum problems — assign +/− signs carefully.
Mass vs Velocity: A slow truck and a fast bullet can have equal momentum if their $m \times v$ products are the same.
Newton's Second Law Form: Force equals the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$. This is actually the more general form of $F = ma$.
Impulse: Changing Momentum
Impulse is the product of the Average Force applied and the time interval of contact, resulting in a Change in Momentum. When force varies during impact (like a bat hitting a ball), impulse captures the total effect.
$$J = F_{avg} \Delta t = \Delta p$$
Impulse links the force applied over time to the resulting change in an object's motion.
$J$=Impulse(N·s)
$F_{avg}$=Average impact force(N)
$\Delta t$=Contact time(s)
$\Delta p$=Change in momentum(kg·m/s)
Object bounces back at the same speed
→$\Delta v = 2v$ (not zero!), so impulse is $2mv$ — double that of simply stopping.
$\Delta t \to 0$ (instantaneous hit)
→Force becomes infinitely large — physically impossible, so real collisions always have some contact time.
Unit Equivalence: $1 \text{ N·s}$ is mathematically identical to $1 \text{ kg·m/s}$. Both measure impulse/momentum.
Graph Area: On a Force-Time graph, the total impulse is the area under the curve — even if the force varies.
Bouncing vs Stopping: A rubber bullet that bounces back delivers more impulse than a lead bullet that embeds — the reversal adds extra momentum change ($\Delta p = 2mv$ vs $mv$).
The Impulse-Momentum Theorem
This theorem states that the impulse applied to an object is exactly equal to its change in momentum, highlighting the trade-off between force and time.
$$F \Delta t = m(v_f - v_i)$$
Relates the external force acting on a mass to its velocity change over time.
$v_i$=Initial Velocity(m/s)
$v_f$=Final Velocity(m/s)
$\Delta t$ is very large
→Force becomes negligible — like a feather slowly drifting to rest.
Object comes to rest ($v_f = 0$)
→Simplifies to $F \Delta t = mv_i$.
Safety Design: Airbags and Crumple Zones increase Contact Time to reduce the average impact force for the same change in momentum.
The Soft Landing: Catching a ball by 'pulling back' your hands increases $\Delta t$, making the required $F$ smaller.
Limiting Case: As $\Delta t \to 0$ (rigid collision), the force approaches infinity — this is why hitting concrete hurts more than hitting a pillow.
Conservation of Linear Momentum
In a closed system where no external forces act, the total momentum before an interaction equals the total momentum after the interaction. This applies to all collisions and explosions.
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$
The sum of momenta in a system remains constant during collisions or explosions.
$u_1, u_2$=Initial velocities of objects 1 and 2(m/s)
$v_1, v_2$=Final velocities of objects 1 and 2(m/s)
One object is initially at rest ($u_2 = 0$)
→Simplifies to $m_1 u_1 = m_1 v_1 + m_2 v_2$
Both initially at rest (explosion: $u_1 = u_2 = 0$)
→Total momentum is zero, so $m_1 v_1 = -m_2 v_2$ — objects move in opposite directions
Internal Forces: Forces between colliding objects are internal and cancel out (Newton's Third Law), leaving the total system momentum unchanged.
Sign Convention: Always define a positive direction first. Objects moving in opposite directions get opposite signs.
Proportionality Shortcut: In explosions from rest, $\frac{v_1}{v_2} = -\frac{m_2}{m_1}$ — the lighter piece always moves faster.
2D Explosions: If a bomb splits into three fragments, conserve momentum separately in $x$ and $y$ directions, then use Pythagoras for the resultant.
Types of Collisions
In an elastic collision, both momentum AND kinetic energy are conserved. The objects bounce off each other with no permanent deformation or heat generation. The Coefficient of Restitution equals 1.
$$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$
Total kinetic energy before equals total kinetic energy after in elastic collisions.
$e$=Coefficient of restitution
Equal masses, one at rest ($m_1 = m_2$, $u_2 = 0$)
→The first object stops completely and the second moves with the initial velocity of the first — like a Newton's cradle.
A light object hits a very heavy stationary object ($m_1 \ll m_2$)
→The light object bounces back at nearly the same speed; the heavy object barely moves.
Coefficient of Restitution: $e = \frac{\text{relative speed after}}{\text{relative speed before}} = \frac{v_2 - v_1}{u_1 - u_2}$. For elastic: $e = 1$.
Heavy Hits Light ($m_1 \gg m_2$, $u_2 = 0$): The heavy object barely slows down; the light object flies off at nearly $2u_1$.
Real-world: Perfectly elastic collisions only occur between atoms/molecules. Billiard balls are approximately elastic.
In an inelastic collision, momentum is conserved but kinetic energy is NOT. Some kinetic energy converts to heat, sound, or deformation. In a perfectly inelastic collision, the objects stick together ($e = 0$) and move with a common final velocity.
$$m_1u_1 + m_2u_2 = (m_1 + m_2)v_f$$
When objects stick together after collision, use combined mass for the final state.
$v_f$=Common final velocity(m/s)
Maximum KE Loss: Perfectly inelastic collisions lose the maximum possible kinetic energy while still conserving momentum.
Final Velocity Shortcut: For target at rest: $v_f = \frac{m_1}{m_1 + m_2} u_1$ — always less than $u_1$.
KE Lost: $\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$ — depends on relative velocity and reduced mass.
Explosive Forces and Recoil
When an object at rest breaks apart or ejects mass (explosions, gunfire, rocket propulsion), the total momentum remains zero. The fragments must move in opposite directions to conserve momentum. This produces recoil.
$$m_1 v_1 + m_2 v_2 = 0$$
Starting from rest, the momenta of the fragments are equal in magnitude but opposite in direction.
$m_1 v_1$=Momentum of fragment 1 (e.g., bullet)(kg·m/s)
$m_2 v_2$=Momentum of fragment 2 (e.g., gun)(kg·m/s)
Recoil Velocity: $v_{\text{gun}} = -\frac{m_{\text{bullet}}}{m_{\text{gun}}} v_{\text{bullet}}$ — the gun recoils slowly because it is much heavier.
Rocket Propulsion: A rocket ejects gas backward at high velocity. Thrust = (mass ejected per second) × (exhaust velocity): $F_{\text{thrust}} = \frac{\Delta m}{\Delta t} v_{\text{exhaust}}$
Increasing Acceleration: As a rocket burns fuel, its mass $M$ decreases while thrust stays roughly constant, so acceleration $a = F/M$ increases over time.
Force from Continuous Flow
When a continuous stream of matter (like water from a hose) strikes a surface and stops, it delivers a steady force. The force equals the rate of momentum delivery.
$$F = \frac{\Delta m}{\Delta t} \times v$$
Force from a stream equals mass flow rate times velocity change.
$\frac{\Delta m}{\Delta t}$=Mass striking per second (mass flow rate)(kg/s)
$v$=Velocity of the stream(m/s)
Application: This formula applies to water jets, sand falling on conveyor belts, and any scenario where mass arrives continuously.
If it bounces back: If the stream bounces instead of stopping, the velocity change is $2v$ (stopping + reversing), so the force doubles.
Dimensional Check: $[\frac{\Delta m}{\Delta t}] \times [v] = \frac{\text{kg}}{\text{s}} \times \frac{\text{m}}{\text{s}} = \frac{\text{kg·m}}{\text{s}^2} = \text{N}$ ✓
Application Strategy
Worked Examples
Problem Statement
Calculate the momentum of a 1500 kg car traveling at 25 m/s north.
Linear Momentum$p = mv$
Strategic Toolkit
$$p = mv$$
Step 1
Identify Givens
We are given the mass and the velocity vector.
$m = 1500 \text{ kg}, v = 25 \text{ m/s north}$
Step 2
Apply Formula
Multiply mass by velocity to find the momentum vector.
$p = (1500 \text{ kg})(25 \text{ m/s}) = 37{,}500 \text{ kg·m/s north}$
Problem Statement
A 0.1 kg baseball is pitched at 40 m/s. The batter hits it straight back at 50 m/s. If the bat is in contact with the ball for 0.01 s, find the average force exerted by the bat.
Impulse-Momentum TheoremVector Change
Strategic Toolkit
$$F \Delta t = m(v_f - v_i)$$
Step 1
Define Directions
Velocity is a vector. Let incoming be negative and outgoing be positive.
$v_i = -40 \text{ m/s}, v_f = +50 \text{ m/s}$
Step 2
Calculate Velocity Change
The total change in velocity accounts for stopping the ball AND sending it back.
$\Delta v = v_f - v_i = 50 - (-40) = 90 \text{ m/s}$
Step 3
Solve for Force
Use the impulse theorem rearranged for force: $F = \frac{m \Delta v}{\Delta t}$.
$F = \frac{(0.1 \text{ kg})(90 \text{ m/s})}{0.01 \text{ s}} = 900 \text{ N}$
Problem Statement
A 4 kg rifle fires a 0.02 kg bullet at 500 m/s. What is the recoil velocity of the rifle?
Conservation of MomentumRecoil
Strategic Toolkit
$$m_1v_1 + m_2v_2 = 0$$
Step 1
Initial State
Both the rifle and bullet are at rest before firing. Total initial momentum is zero.
$p_i = 0$
Step 2
Apply Conservation
Since no external forces act, total momentum after must also be zero.
$m_{\text{bullet}} v_{\text{bullet}} + m_{\text{rifle}} v_{\text{rifle}} = 0$
Step 3
Solve for Recoil
Rearrange and substitute values.
$v_{\text{rifle}} = -\frac{m_{\text{bullet}}}{m_{\text{rifle}}} v_{\text{bullet}} = -\frac{0.02}{4} \times 500 = -2.5 \text{ m/s}$
Step 4
Interpret
The negative sign means the rifle moves opposite to the bullet. The rifle recoils at 2.5 m/s — much slower than the bullet because it's 200× heavier.
Problem Statement
A 3 kg ball moving at 8 m/s collides with a stationary 5 kg ball and they stick together. Find the final velocity and the kinetic energy lost.
Conservation of MomentumInelastic CollisionKE Loss
Strategic Toolkit
$$m_1u_1 = (m_1 + m_2)v_f$$
Step 1
Apply Conservation
Total momentum before = total momentum after. Objects stick, so combined mass.
$3(8) + 5(0) = (3 + 5)v_f$
Step 2
Solve for Final Velocity
Divide total momentum by combined mass.
$v_f = \frac{24}{8} = 3 \text{ m/s}$
Step 3
Calculate Initial KE
Only the 3 kg ball is moving before the collision.
$KE_i = \frac{1}{2}(3)(8^2) = 96 \text{ J}$
Step 4
Calculate Final KE
After collision, the combined 8 kg mass moves at 3 m/s.
$KE_f = \frac{1}{2}(8)(3^2) = 36 \text{ J}$
Step 5
Find KE Lost
The difference went into heat, sound, and deformation.
$\Delta KE = 96 - 36 = 60 \text{ J lost}$
Problem Statement
A 2 kg ball moving at 6 m/s collides elastically with an identical 2 kg ball at rest. Find the velocities of both balls after collision.
Elastic CollisionEqual MassVelocity Exchange
Strategic Toolkit
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \text{ and } e = 1$$
Step 1
Recognize the Special Case
Equal masses + one at rest + elastic collision = complete velocity transfer.
$m_1 = m_2 = 2 \text{ kg}, u_1 = 6 \text{ m/s}, u_2 = 0, e = 1$
Step 2
Apply the Equal-Mass Result
In this special case, the first ball stops and the second takes all the velocity.
$v_1 = 0 \text{ m/s}, v_2 = 6 \text{ m/s}$
Step 3
Verify Conservation
Check both momentum and KE are conserved.
$p_i = 2(6) = 12 = 2(0) + 2(6) = p_f$ ✓, $KE_i = \frac{1}{2}(2)(36) = 36 = \frac{1}{2}(2)(36) = KE_f$ ✓
Problem Statement
Water flows from a hose at 3 kg/s and strikes a wall at 5 m/s, coming to rest. What is the force on the wall?
Continuous Momentum DeliveryForce from Flow
Strategic Toolkit
$$F = \frac{\Delta m}{\Delta t} \times v$$
Step 1
Identify Given Values
Mass flow rate and velocity change.
$\frac{\Delta m}{\Delta t} = 3 \text{ kg/s}, \Delta v = 5 - 0 = 5 \text{ m/s}$
Step 2
Calculate Force
Force = rate of momentum change = mass per second × velocity change.
$F = 3 \times 5 = 15 \text{ N}$
Memorize formulas
Momentum Definition
Momentum is 'mass in motion'.
Formula
p = mv
Impulse
Impulse equals change in momentum.
Formula
$J = F_{avg} \Delta t = \Delta p$
Impulse-Momentum Theorem
Trade-off between Force and Time.
Formula
$F \Delta t = m(v_f - v_i)$
Conservation of Momentum
Total momentum is conserved in collisions.
Formula
$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
Perfectly Inelastic Collision
Objects stick together — use combined mass.
Formula
$m_1u_1 + m_2u_2 = (m_1 + m_2)v_f$
Recoil Velocity
Velocity of the heavier object after explosion from rest.
Formula
$v_{\text{gun}} = -\frac{m_{\text{bullet}}}{m_{\text{gun}}} v_{\text{bullet}}$
Force from Continuous Flow
Force from a stream of mass hitting a surface.
Formula
$F = \frac{\Delta m}{\Delta t} \times v$
Coefficient of Restitution
Measures bounciness of a collision (0 to 1).
Formula
$e = \frac{v_2 - v_1}{u_1 - u_2}$
Rocket Thrust
Thrust from ejecting mass at high velocity.
Formula
$F_{\text{thrust}} = \frac{\Delta m}{\Delta t} v_{\text{exhaust}}$
KE Lost in Perfectly Inelastic Collision
Energy lost when objects stick together.
Formula
$\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$
Review
1
A 0.3 kg ball hits a wall at 20 m/s and bounces back at 20 m/s. A second identical ball hits the wall at the same speed but stops on contact. The difference in impulse magnitudes between the two cases is:
Ans
6 N·s
Review
Tap to reveal
Compute the impulse in each case
Impulse magnitude is the change in momentum, $|\Delta p| = m|\Delta v|$. The rebound case changes velocity from $+20$ m/s to $-20$ m/s, while the stopping case changes it from $+20$ m/s to $0$.
$$|\Delta p_1| = 0.3(40) = 12\ \text{N·s},\qquad |\Delta p_2| = 0.3(20) = 6\ \text{N·s}$$
Take the difference
Subtract the smaller impulse from the larger one to get the requested difference.
$$12 - 6 = 6\ \text{N·s}$$
2
Two skaters, one 40 kg and one 80 kg, stand at rest on frictionless ice. They push off each other. If the 80 kg skater moves at 2 m/s, what is the velocity of the 40 kg skater?
Ans
4 m/s in the opposite direction
Review
Tap to reveal
Use conservation of momentum
The system starts at rest, so the total momentum after they push off must still be zero.
$$80(2) + 40v = 0$$
Solve for the lighter skater's velocity
Rearrange the equation and keep the sign, which shows the lighter skater moves opposite to the 80 kg skater.
$$v = -\frac{160}{40} = -4\ \text{m/s}$$
3
A 4 kg object moving at 6 m/s collides with a 2 kg object moving at 3 m/s in the same direction. If the collision is perfectly inelastic, what is the final velocity?
Ans
5 m/s
Review
Tap to reveal
Add the incoming momenta
Because the objects move in the same direction, their momenta add directly before the collision.
$$p_i = 4(6) + 2(3) = 30\ \text{kg·m/s}$$
Divide by the combined mass
After sticking together, the total mass is 6 kg, so the final velocity is total momentum divided by total mass.
$$v_f = \frac{30}{6} = 5\ \text{m/s}$$
4
A 0.15 kg cricket ball moving at 30 m/s is hit by a bat and returns along the same line at 40 m/s. If the contact time is 0.02 s, the average force exerted by the bat is:
Ans
525 N
Review
Tap to reveal
Find the change in velocity with direction
Take the incoming velocity as negative and the rebound velocity as positive, so the velocity change includes the reversal.
$$\Delta v = 40 - (-30) = 70\ \text{m/s}$$
Apply impulse over time
Average force equals change in momentum divided by contact time.
$$F = \frac{m\Delta v}{\Delta t} = \frac{0.15 \times 70}{0.02} = 525\ \text{N}$$
5
A stationary cannon of mass 200 kg fires a 2 kg ball horizontally. If the cannon recoils at 3 m/s on a frictionless surface, the speed of the ball is:
Ans
300 m/s
Review
Tap to reveal
Set total momentum to zero
The cannon-ball system starts at rest, so the cannon's backward momentum must balance the ball's forward momentum.
$$200(3) = 2v$$
Solve for the ball speed
Isolate $v$ and keep the magnitude, since the question asks for speed.
$$v = \frac{600}{2} = 300\ \text{m/s}$$
6
A 4 kg object moving at 5 m/s collides with a 1 kg object at rest. If they stick together, what fraction of the initial kinetic energy is lost?
Ans
$\frac{1}{5}$
Review
Tap to reveal
Find the initial and final kinetic energy
Compute the initial kinetic energy from the moving 4 kg object, then use momentum conservation to get the shared speed after they stick.
$$KE_i = \tfrac{1}{2}(4)(5^2) = 50\ \text{J},\qquad v_f = \frac{4\cdot 5}{4+1} = 4\ \text{m/s}$$
Compare the energies
Use the final common speed to find the remaining kinetic energy and then take the lost fraction.
$$KE_f = \tfrac{1}{2}(5)(4^2) = 40\ \text{J},\qquad \frac{KE_i-KE_f}{KE_i} = \frac{10}{50} = \frac{1}{5}$$
7
A 2 kg ball moving at 10 m/s collides head-on with a 2 kg ball at rest. If the collision is perfectly elastic, the first ball after collision will:
Ans
Stop completely
Review
Tap to reveal
Use the equal-mass elastic collision rule
For a head-on perfectly elastic collision with equal masses, the velocities are exchanged.
$$v_{1,\,\text{after}} = 0\ \text{m/s}$$
8
Sand falls vertically onto a conveyor belt at a rate of 5 kg/s. The belt moves horizontally at 4 m/s. The extra force required to keep the belt moving at constant speed is:
Ans
20 N
Review
Tap to reveal
Treat the sand as incoming mass
Each kilogram of sand arrives with zero horizontal speed and must be brought up to the belt speed.
$$F = \frac{dm}{dt}\,v$$
Substitute the flow rate and belt speed
Multiply the mass flow rate by the horizontal speed to get the extra force required.
$$F = 5 \times 4 = 20\ \text{N}$$
9
A ball of mass $m$ hits a wall at speed $v$ and bounces back at the same speed. A second identical ball hits the same wall at the same speed but sticks to it. The ratio of impulse in the first case to the second is:
Ans
2:1
Review
Tap to reveal
Write each impulse from momentum change
A bounce reverses the velocity, while sticking reduces it to zero, so the two momentum changes are not the same.
$$\Delta p_1 = 2mv,\qquad \Delta p_2 = mv$$
Form the ratio
Compare the two impulses directly.
$$\Delta p_1 : \Delta p_2 = 2:1$$
10
A bomb at rest explodes into three fragments of masses 1 kg, 2 kg, and 3 kg. The 1 kg and 2 kg fragments fly off perpendicular to each other with speeds 12 m/s and 8 m/s respectively. The speed of the 3 kg fragment is:
Ans
$\frac{20}{3}$ m/s
Review
Tap to reveal
Use vector momentum conservation
Because the bomb starts at rest, the third fragment must balance the perpendicular momentum components of the other two fragments.
$$p_3 = \sqrt{12^2 + 16^2} = 20\ \text{kg·m/s}$$
Convert momentum to speed
Divide the needed momentum by the 3 kg mass.
$$v_3 = \frac{20}{3}\ \text{m/s}$$
11
An airbag increases the time of collision from 0.005 s to 0.08 s during a car crash. The ratio of force WITH airbag to force WITHOUT airbag is:
Ans
1:16
Review
Tap to reveal
Use the inverse time dependence
The momentum change is the same in both crashes, so force is inversely proportional to collision time.
$$\frac{F_{\text{with}}}{F_{\text{without}}} = \frac{\Delta t_{\text{without}}}{\Delta t_{\text{with}}}$$
Insert the two contact times
The airbag case lasts much longer, so it produces a much smaller force.
$$\frac{0.005}{0.08} = \frac{1}{16}$$
12
A rocket has a thrust of 80,000 N and ejects exhaust gas at 2000 m/s relative to the rocket. The rate at which it burns fuel is:
Ans
40 kg/s
Review
Tap to reveal
Rearrange the thrust relation
Rocket thrust equals exhaust speed times mass burn rate, so solve the formula for $\frac{dm}{dt}$.
$$\frac{dm}{dt} = \frac{F}{v_e}$$
Substitute the given values
Divide the thrust by the exhaust speed to get the fuel burn rate.
$$\frac{dm}{dt} = \frac{80{,}000}{2000} = 40\ \text{kg/s}$$
13
A 2 kg ball moving at 6 m/s has a head-on elastic collision with a 6 kg ball at rest. The velocity of the 2 kg ball after collision is:
Ans
−3 m/s (bounces back)
Review
Tap to reveal
Apply the elastic collision formula
For a head-on elastic collision with the second ball initially at rest, use the standard one-dimensional result for the first ball's final velocity.
$$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1$$
Substitute the masses and initial speed
The heavier target makes the lighter ball reverse direction.
$$v_1 = \frac{2-6}{2+6}(6) = -3\ \text{m/s}$$
14
A rocket of mass 1000 kg (including fuel) ejects gas at 4000 m/s at a rate of 10 kg/s. After 50 seconds of burning, the acceleration of the rocket (ignoring gravity) compared to its initial acceleration is:
Ans
Doubled
Review
Tap to reveal
Keep thrust fixed and track the mass loss
The burn rate and exhaust speed are constant, so thrust stays constant while the rocket mass decreases with time.
$$F = 10 \times 4000 = 40{,}000\ \text{N}$$
Compare initial and final acceleration
Acceleration is thrust divided by mass, so halving the mass doubles the acceleration.
$$a_i = \frac{40{,}000}{1000} = 40\ \text{m/s}^2,\qquad a_f = \frac{40{,}000}{500} = 80\ \text{m/s}^2$$
Question 1 of 14
0:00
1
A 0.3 kg ball hits a wall at 20 m/s and bounces back at 20 m/s. A second identical ball hits the wall at the same speed but stops on contact. The difference in impulse magnitudes between the two cases is: