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Topic 9 of 20

Linear Momentum & Impulse

Linear momentum is the "quantity of motion" of an object. Newton's Second Law is most generally stated as: net force equals the rate of change of momentum. This leads to the impulse–momentum theorem and, crucially, to conservation of momentum — one of the most broadly useful principles in physics.

Key Concepts

Linear Momentum

The product of mass and velocity:

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

. A vector quantity with units of kg·m/s. More massive objects and faster objects have more momentum.

Newton's Second Law (momentum form)

The most general statement:

\vec{F}_\text{net} = d\vec{p}/dt

. When mass is constant this reduces to

\vec{F}_\text{net} = m\vec{a}

Impulse

Impulse is the integral of force over time:

\vec{J} = \int\vec{F}\,dt

. For a constant force:

\vec{J} = \vec{F}\Delta t

. Impulse equals the change in momentum.

Impulse–Momentum Theorem

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

. A large force over a short time produces the same impulse as a smaller force over a longer time.

Conservation of Momentum

If the net external force on a system is zero, total momentum is conserved:

\vec{p}_\text{total} = \text{const}

. Apply separately in each direction. Internal forces do not change total momentum.

Center of Mass

The mass-weighted average position of a system:

\vec{r}_\text{cm} = \sum m_i\vec{r}_i / M

. The total external force equals

M\vec{a}_\text{cm}

— the CM moves as if all mass were concentrated there.

Key Equations

Linear momentum

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

Momentum is a vector; direction matches velocity.

Impulse–Momentum Theorem

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

Impulse (area under F–t graph) equals change in momentum.

Conservation of momentum

\vec{p}_{1i} + \vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f} \quad (\vec{F}_\text{ext}=0)

Total momentum is constant when there is no net external force.

Center of mass (two bodies)

x_\text{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

Position of the center of mass; generalizes to any number of bodies.

CM velocity

\vec{v}_\text{cm} = \frac{\sum m_i \vec{v}_i}{M}

Velocity of the center of mass; constant when net external force is zero.

Worked Example

Recoil of a Rifle

Problem

A 4 kg rifle fires a 10 g bullet at 600 m/s. Find the recoil speed of the rifle.

Solution

System: rifle + bullet. Initially both at rest, so $p_\text{total,i} = 0$ .

Conservation of momentum (no external horizontal force):

0 = m_\text{bullet}\,v_\text{bullet} + m_\text{rifle}\,v_\text{rifle}

v_\text{rifle} = -\frac{m_\text{bullet}}{m_\text{rifle}}\,v_\text{bullet} = -\frac{0.010}{4}\times 600

v_\text{rifle} = -1.5 \text{ m/s}

Answer The rifle recoils at 1.5 m/s in the direction opposite the bullet.

Practice

Exercises

7 problems

1 of 7

A $3 \text{ kg}$ object moves at $5 \text{ m/s}$ . What is the magnitude of its momentum?

Your answer

kg·m/s

2 of 7

A $0.5 \text{ kg}$ ball moving at $8 \text{ m/s}$ is brought to rest in $0.02 \text{ s}$ . What is the magnitude of the average force?

Your answer

3 of 7

A $1500 \text{ kg}$ car decelerates from $20 \text{ m/s}$ to rest in $5 \text{ s}$ . What is the magnitude of the average braking force?

Your answer

4 of 7

A $0.20 \text{ kg}$ ball moving at $10 \text{ m/s}$ hits a wall and bounces straight back at $10 \text{ m/s}$ . What is the magnitude of the change in momentum?

Your answer

kg·m/s

5 of 7

A $5 \text{ kg}$ rifle fires a $0.01 \text{ kg}$ bullet at $500 \text{ m/s}$ . What is the recoil speed of the rifle?

Your answer

m/s

6 of 7

Two carts collide and stick: $2 \text{ kg}$ at $3 \text{ m/s}$ and $3 \text{ kg}$ at rest (frictionless track). What is the final speed?

Your answer

m/s

7 of 7

A $10 \text{ kg}$ object at rest explodes into two pieces: $4 \text{ kg}$ flying at $6 \text{ m/s}$ to the right. What is the speed of the other piece?

Your answer

m/s

Key Takeaways

Momentum $\vec{p} = m\vec{v}$ is a vector — direction matters, and it must be conserved in each direction independently.
The impulse–momentum theorem: $\vec{J} = \Delta\vec{p}$ . Airbags and crumple zones extend $\Delta t$ to reduce peak force.
Conservation of momentum applies whenever net external force is zero — even when energy is not conserved.
Internal forces (e.g., between colliding objects) never change the total momentum of the system.
The center of mass of an isolated system moves at constant velocity regardless of internal interactions.