8 - Momentum, Impulse, and Collisions

A progressive guide to momentum, impulse, conservation laws, collision types, and center-of-mass motion, with equations, examples, and problem-solving strategies.

Momentum as a Vector

Momentum describes how much motion an object has and depends on both mass and velocity. The defining relationship is

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

The arrow indicates that momentum is a vector. Its direction is the direction of the velocity. In a one-dimensional problem, select a positive direction first; motion opposite that direction has a negative velocity and therefore negative momentum.

For a collection of particles, add the individual momentum vectors:

\vec{p}_{\text{system}}=\sum_i m_i\vec{v}_i

For example, a $2.0\,\text{kg}$ cart moving at $3.0\,\text{m/s}$ and a $6.0\,\text{kg}$ cart moving at $1.0\,\text{m/s}$ have equal momentum magnitudes:

(2.0)(3.0)=(6.0)(1.0)=6.0\,\text{kg}\cdot\text{m/s}

Momentum is different from kinetic energy. Kinetic energy is

K=\frac{1}{2}mv^2

Because speed is squared in the kinetic-energy equation, equal momenta do not necessarily mean equal kinetic energies.

Takeaway: Momentum depends linearly on velocity and includes direction; kinetic energy depends on the square of speed and is scalar.

Impulse and Force-Time Interactions

The connects a force acting over a time interval with the resulting change in momentum:

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

For a constant net force,

\vec{J}=\vec{F}_{\text{net}}\Delta t

Impulse and momentum change have the same units:

1\,\text{N}\cdot\text{s}=1\,\text{kg}\cdot\text{m/s}

For a varying force, impulse is the area under the force-versus-time graph:

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

For a triangular force-time graph with maximum force $F_{\max}$ and duration $\Delta t$ , the magnitude is

J=\frac{1}{2}F_{\max}\Delta t

Consider a $0.15\,\text{kg}$ ball that changes velocity from $+40\,\text{m/s}$ to $-50\,\text{m/s}$ :

J=m(v_f-v_i)=(0.15)(-50-40)=-13.5\,\text{N}\cdot\text{s}

The negative sign indicates that the impulse points opposite the initial positive direction. If the change in momentum is fixed, increasing the collision time reduces the average force:

F_{\text{avg}}=\frac{\Delta p}{\Delta t}

This explains why airbags, padding, and crumple zones reduce average force: they lengthen the time over which momentum changes.

Takeaway: A force changes momentum through its impulse, and the same momentum change can occur with a smaller average force when the interaction lasts longer.

Systems and External Impulse

Before applying a conservation law, define the boundary. For a selected ,

\vec{F}_{\text{ext, net}}=\frac{\Delta\vec{p}_{\text{system}}}{\Delta t}

Over a finite interval, this becomes

\vec{J}_{\text{ext}}=\Delta\vec{p}_{\text{system}}

If the net external impulse is zero, total momentum is constant. Forces between objects inside the are internal: they can transfer momentum from one object to another but cannot change the total momentum of the by themselves.

During a brief collision, friction or gravity may produce an external impulse much smaller than the collision impulse. When that approximation is justified, treat the colliding objects as an isolated and conserve momentum. If an important external impulse acts, include it instead of assuming momentum is conserved.

A useful recoil example is a $5.0\,\text{kg}$ launcher initially at rest that fires a $0.020\,\text{kg}$ projectile at $300\,\text{m/s}$ . With negligible external horizontal impulse,

0=m_pv_p+m_Lv_L

v_L=-\frac{m_pv_p}{m_L}=-\frac{(0.020)(300)}{5.0}=-1.2\,\text{m/s}

The launcher recoils opposite the projectile, while the total momentum remains zero.

Takeaway: Conservation depends on the chosen and the external impulse, not merely on whether objects are interacting.

Momentum Conservation in Collisions

When the net external impulse is zero, write total initial momentum equal to total final momentum. For two objects moving along one axis,

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

The signs encode direction. In two dimensions, conserve each component separately:

\sum p_{x,i}=\sum p_{x,f}

\sum p_{y,i}=\sum p_{y,f}

For a collision in which objects stick together, both objects have the same final velocity. This is a perfectly , described by

m_1v_{1i}+m_2v_{2i}=(m_1+m_2)v_f

For example, a $0.50\,\text{kg}$ cart moving at $4.0\,\text{m/s}$ collides with a stationary $1.5\,\text{kg}$ cart. If they stick,

(0.50)(4.0)+(1.5)(0)=(0.50+1.5)v_f

which gives

v_f=1.0\,\text{m/s}

The carts move together to the right. Their total momentum is conserved, but some kinetic energy becomes thermal energy, sound, and deformation energy.

Takeaway: Use signed velocities in the momentum equation, and impose a shared final velocity only when the objects stick together.

Elasticity and Energy Transfer

Classify a collision by checking what is conserved. An conserves both momentum and total kinetic energy:

\vec{p}_i=\vec{p}_f,\qquad K_i=K_f

An conserves momentum but not total kinetic energy:

\vec{p}_i=\vec{p}_f,\qquad K_f<K_i

The decrease in kinetic energy is not destruction of energy. It represents transformation into internal energy, sound, deformation, or other forms. The kinetic-energy change can be written as

\Delta K=K_f-K_i

For an , $\Delta K<0$ .

Do not conserve kinetic energy merely because momentum is conserved. Momentum conservation follows from negligible net external impulse; kinetic-energy conservation requires the additional condition that the collision be elastic or that another energy relationship explicitly establishes it. In a perfectly , objects stick together, so the final velocity is found from momentum conservation rather than from kinetic-energy conservation.

Takeaway: All isolated collision types conserve momentum, but only elastic collisions conserve total kinetic energy.

Center-of-Mass Motion

The gives a single position and velocity that represent the overall motion of a . For particles along one axis,

x_{\text{CM}}=\frac{\sum_i m_ix_i}{\sum_i m_i}

For two objects,

x_{\text{CM}}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

A $2.0\,\text{kg}$ object at $x=0\,\text{m}$ and a $6.0\,\text{kg}$ object at $x=4.0\,\text{m}$ have

x_{\text{CM}}=\frac{(2.0)(0)+(6.0)(4.0)}{2.0+6.0}=3.0\,\text{m}

The result is closer to the more massive object. The center-of-mass velocity is related to total momentum by

\vec{v}_{\text{CM}}=\frac{\sum_i m_i\vec{v}_i}{M},\qquad M=\sum_i m_i

and therefore

\vec{p}_{\text{system}}=M\vec{v}_{\text{CM}}

For a of particles,

\vec{F}_{\text{ext, net}}=M\vec{a}_{\text{CM}}

If the net external force is zero, then $\vec{a}_{\text{CM}}=0$ , so the moves at constant velocity. Internal explosions, collisions, or spring forces may produce complicated relative motion without changing the center-of-mass motion.

Takeaway: Internal forces rearrange a , while the net external force determines how the moves.

A Reliable Problem-Solving Method

Use this sequence when solving a momentum or collision problem:

Define the . Include every object whose momentum transfer is part of the analysis.
Choose coordinates. State the positive direction before assigning velocity signs.
Assess external impulse. Decide whether it is negligible during the interaction.
Represent the initial and final states. Label each mass and signed velocity.
Write the momentum equation. Use components for two-dimensional motion.
Add energy information only when justified. Conserve kinetic energy for an or when another explicit energy relationship supports it.
Check the result. Verify signs, units, direction, and limiting behavior.

Common errors include conserving the momentum of one object instead of the selected , assuming every collision conserves kinetic energy, ignoring direction, and treating an external force as internal. For example, gravity is external if Earth is outside the , but gravitational forces between parts are internal if Earth is included.

Impulse should also not be confused with force. A large force acting briefly can produce the same impulse as a smaller force acting for a longer time because impulse depends on force multiplied by time, or more generally on the area under a force-time graph.

Final checklist: Identify the , set a sign convention, account for external impulse, distinguish momentum from kinetic energy, and use only the conservation laws supported by the physical situation.