Linear Momentum and the Second Law: MDCAT Physics notes

Linear momentum

The linear momentum of a body is the product of its mass and velocity:

$$\vec p = m\vec v$$

It is a vector in the direction of velocity; SI unit $\text{kg m s}^{-1}$ (same as $\text{N s}$). Two equal masses moving at equal speeds have equal momentum only if they move in the same direction; otherwise their momenta differ (their kinetic energies, being scalars, are equal).

Newton's second law in terms of momentum

The time rate of change of linear momentum equals the net applied force:

$$\vec F = \frac{\Delta \vec p}{\Delta t} = \frac{m\vec v_f - m\vec v_i}{t}$$

For constant mass this reduces to $F = ma$. This form is more general (useful for rockets, jets of water, bullets).

Impulse

When a large force acts for a short time, we use impulse:

$$\vec J = \vec F\,\Delta t = \Delta \vec p$$

Impulse equals the change in momentum; its unit is the newton-second ($\text{N s}$), equivalent to $\text{kg m s}^{-1}$.

Example: a force of 50 dyne ($= 50\times10^{-5}\ \text{N} = 5\times10^{-4}\ \text{N}$) for 3 s gives $J = 1.5\times10^{-3}\ \text{N s}$.

Rebound from a wall

Take the initial direction as positive. A ball of momentum $mv$ bouncing back with the same speed has final momentum $-mv$:

$$\Delta p = (-mv) - (mv) = -2mv$$

The magnitude of the change is $2mv$ (e.g. $8\ \text{kg m s}^{-1}$ ball: change of magnitude $16\ \text{N s}$). If the ball sticks, the change is only $mv$.

Applications

• Seat belts, air bags, crumple zones, padded floors: the belt applies an opposite (retarding) force and increases the stopping time, so for the same $\Delta p$ the force $F = \Delta p/\Delta t$ is smaller.
• Stopping distance: a car and a truck at the same velocity, same retarding force: the car has less momentum, so it stops first ($t = \Delta p / F$).
• Machine gun: if $n$ bullets of mass $m$ leave per second at speed $v$, the force on the gun is $F = nmv$. With $m = 0.04\ \text{kg}$, $v = 1200\ \text{m s}^{-1}$, each bullet carries $48\ \text{N s}$; a man exerting 144 N can fire 3 bullets per second.

Worked example

A 150 kg car slows from 20 to $10\ \text{m s}^{-1}$ in 3.0 s. $F = m\Delta v/\Delta t = 150 \times 10/3 = 500\ \text{N}$ (retarding).

Key formulas

• $p = mv$
• $F = \Delta p/\Delta t$
• $J = F t = m v_f - m v_i$
• Rebound with same speed: $|\Delta p| = 2mv$
• Stream of bullets: $F = n m v$

Common MDCAT traps

• Force × time is change in momentum, not change in velocity or displacement.
• Rebound change is $2mv$ (sign negative if initial direction is positive), not zero.
• Momentum is a vector: same speed and mass does not guarantee the same momentum.
• Convert dyne to newton: $1\ \text{dyne} = 10^{-5}\ \text{N}$; grams to kg.

Quick revision

• Force = rate of change of momentum.
• Impulse unit: N s.
• Longer impact time means smaller force.
• $1\ \text{N s} = 1\ \text{kg m s}^{-1}$.
• Same force, same speed: smaller mass stops first.