A2-3. Momentum and Impulse

Linear Momentum

Momentum describes the "quantity of motion" an object possesses and indicates how difficult it is to bring that moving object to a halt.

Momentum ($\vec{p}$): The product of an object's mass and its velocity. Because velocity is a vector, momentum is also a vector. The direction of the momentum vector is exactly the same as the direction of the velocity vector.

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

It is measured in kilogram-metres per second ($\text{kg m/s}$).

Note on 2D Motion: Momentum can be broken down into perpendicular vector components: $p_x = mv_x$ and $p_y = mv_y$.

Impulse and the Impulse-Momentum Theorem

When an external force acts on an object for a specific duration of time, it changes the object's momentum. This effect is known as an impulse (e.g., kicking a soccer ball or a car crashing into a wall).

Impulse ($\vec{J}$): The product of the average force applied and the time interval over which it acts. It is a vector quantity, and its direction is the same as the direction of the applied force.

$$\vec{J} = \vec{F}_{avg}\Delta t$$

The Impulse-Momentum Theorem: The total impulse applied to an object is exactly equal to its change in momentum ($\Delta \vec{p}$).

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

Derivation from Newton's Second Law:
We know that average force equals mass times average acceleration. By expanding acceleration into the change in velocity over time, we get:
$$\vec{F}_{avg} = m\vec{a}_{avg} = m\dfrac{\Delta\vec{v}}{\Delta t} = \dfrac{m\vec{v}_f - m\vec{v}_i}{\Delta t} = \dfrac{\vec{p}_f - \vec{p}_i}{\Delta t}$$
Multiplying both sides by $\Delta t$ gives us the theorem: $\vec{F}_{avg}\Delta t = \vec{p}_f - \vec{p}_i$.

The unit for impulse is Newton-seconds ($\text{N s}$), which is dimensionally identical to $\text{kg m/s}$.

⚠ Force-Time Graphs & Real-world Applications

In most real-world collisions, force is not constant; it peaks during maximum compression and drops back to zero. On a Force-Time graph, the area under the curve represents the total Impulse (or total change in momentum $\Delta p$).

t (s) F (N) Favg Area = Impulse (J) ti tf Δt

Safety Applications: Devices like airbags, seatbelts, and crumple zones are designed to prolong the interaction time ($\Delta t$) of a crash. Because the total required change in momentum ($\Delta p$) is fixed to bring the passenger to a stop, increasing the collision time ($\Delta t$) significantly lowers the average impact force ($\vec{F}_{avg}$), thereby reducing severe injuries.

Example 1: Automotive Crash Test

Problem: To test vehicle safety, a 1000 kg car crashes into a solid test wall. Its initial velocity is 12 m/s moving to the right. After bouncing off the wall, its final velocity is 1.5 m/s moving to the left. The collision lasts for 0.010 s. Calculate the initial and final momentum, the impulse, and the average force exerted by the wall.

Solution:

  • Setup: Define the rightward direction as positive (+).
    $v_i = +12\text{ m/s}$ and $v_f = -1.5\text{ m/s}$.
  • a) Initial Momentum ($p_i$):
    $$p_i = mv_i = 1000 \times 12 = \mathbf{1.2 \times 10^4\text{ kg m/s}}$$ (Positive indicates rightward direction)
  • b) Final Momentum ($p_f$):
    $$p_f = mv_f = 1000 \times (-1.5) = \mathbf{-1.5 \times 10^3\text{ kg m/s}}$$ (Negative sign indicates leftward direction)
  • c) Impulse ($\vec{J}$):
    $$J = \Delta p = p_f - p_i = (-1.5 \times 10^3) - (1.2 \times 10^4) = \mathbf{-1.35 \times 10^4\text{ N s}}$$ (Negative sign indicates the impulse is directed to the left)
  • d) Average Force ($\vec{F}_{avg}$):
    $$F_{avg} = \dfrac{J}{\Delta t} = \dfrac{-1.35 \times 10^4}{0.010} = \mathbf{-1.35 \times 10^6\text{ N}}$$ (Negative sign indicates the wall exerted force to the left)