1.25P-1.28P - Momentum and safety

Momentum as moving mass

An object has momentum when it has mass and velocity. A stationary object has no momentum because its velocity is zero.

The relationship is:

In this formula, $p$ is momentum, $m$ is mass in kilograms, and $v$ is velocity in metres per second. The unit of momentum is therefore:

Momentum has a direction because velocity has a direction. In one-dimensional calculations, choose one direction as positive. A velocity in the opposite direction is then negative. This is why a rebound involves a bigger change in momentum than simply stopping: the object loses all its momentum in the original direction and then gains momentum in the opposite direction.

Worked example: a $900\ \text{kg}$ car moving at $14\ \text{m/s}$ has momentum:

If the same car travels in the opposite direction at the same speed, its momentum would be $-12\,600\ \text{kg m/s}$ using that sign convention.

Force from momentum change

A resultant force changes an object's momentum. For the same momentum change, the average force is smaller if the change happens over a longer time.

The Edexcel relationship is:

Here $u$ is the initial velocity, $v$ is the final velocity, $m$ is the mass, and $t$ is the time taken for the change. In collision questions, $F$ is usually the average force during the contact time.

This is Newton's second law written in terms of momentum. Because $mv - mu = m(v - u)$, and $(v - u)/t$ is the acceleration, the equation is the same as $F = m \times a$.

Notice the order in the numerator: final momentum minus initial momentum. If you use signs consistently, the sign of $F$ tells you the direction of the force.

Worked example: a $0.20\ \text{kg}$ ball moving right at $18\ \text{m/s}$ is brought to rest by a wall in $0.030\ \text{s}$. Take right as positive.

Initial momentum:

Final momentum:

Average force:

The negative sign means the force on the ball acts left, opposite to its original motion. The magnitude of the force is $120\ \text{N}$.

Safety features and stopping time

If a person in a car crash goes from moving quickly to rest, their momentum must decrease to zero. For the same person and the same initial speed, the change in momentum is fixed. The safety question is: how quickly does that change happen?

[DIAGRAM: asset_name: Momentum and safety - diagram 01; asset_slug: p10_momentum_and_safety__diagram_01; recommended_method: matplotlib; description: NovaLearn pale-palette force-time comparison on identical scales: hard stop at 1800 N for 0.05 s and safety-feature stop at 450 N for 0.20 s, both with momentum change 90 kg m/s.]

The graph makes the comparison numerical: $1800\times0.050=90\,\mathrm{kg\,m/s}$ and $450\times0.20=90\,\mathrm{kg\,m/s}$. The momentum change is the same, but four times the stopping time gives one quarter of the average force. These rectangles represent average force; a real impact force varies during contact.

From

increasing the time taken reduces the average force. This is the momentum explanation behind many safety features.

A common mistake is to say that a crumple zone "absorbs the force". The better Physics answer is that it increases the collision time. For the same change in momentum, a longer time means a smaller average force.

In an exam explanation, link the steps in order: the person's momentum changes from $mv$ to zero whatever happens; the safety feature increases the time taken for that change; so, from $F = \text{change in momentum} / t$, the average force on the person is smaller and injury is less likely.

Conservation of momentum

In a collision or separation where external forces are negligible, total momentum is conserved. This means the total momentum before the event equals the total momentum after the event.

The "system" is the set of objects you are considering. In exam questions, this is usually two trolleys, two vehicles, or two people on ice. During a short collision, external forces such as friction are often treated as small enough to ignore.

Use a signed direction convention. For example, if right is positive, then left is negative:

Object A has lost $4\ \text{kg m/s}$ of momentum. Object B has gained $4\ \text{kg m/s}$. The total stays the same.

Why does this always happen? During the collision, A pushes on B and B pushes back on A with a force of the same size in the opposite direction, for the same contact time. From $F = \text{change in momentum} / t$, their changes in momentum are equal in size and opposite in direction, so they cancel in the total.

For one-dimensional calculations, the conservation equation is:

Then replace each momentum with $mv$.

Conservation calculations

The safest calculation method is to make a before-and-after momentum table.

Worked example: a $0.80\ \text{kg}$ trolley moving right at $3.0\ \text{m/s}$ collides with a stationary $1.20\ \text{kg}$ trolley. They stick together. Calculate their velocity after the collision.

Before the collision:

After the collision, the combined mass is:

Momentum is conserved, so the total momentum after is still $2.4\ \text{kg m/s}$.

The joined trolleys move right at $1.2\ \text{m/s}$.

The same idea works for objects pushing apart. If the total momentum before is zero, the total momentum after must also be zero. One object moving right must be balanced by momentum of the other object to the left.

Worked example: two skaters are initially at rest. After they push apart, skater A has momentum $+48\ \text{kg m/s}$. The total momentum before was $0$, so skater B must have momentum $-48\ \text{kg m/s}$. If skater B has mass $60\ \text{kg}$, then:

The negative sign means skater B moves in the opposite direction to skater A.

The same equation can be rearranged for whichever quantity is missing. If instead you were told that skater B moves at $-0.80\ \text{m/s}$, you could find B's mass:

Quick recap

• Momentum is $p=mv$, measured in $\mathrm{kg\,m/s}$; its direction matters in collisions and rebounds.
• Average resultant force is momentum change divided by time: $F=(mv-mu)/t$. Subtract initial momentum from final momentum.
• For the same momentum change, increasing stopping time reduces average force; this explains crumple zones, airbags and padding.
• When external forces are negligible, total momentum before equals total momentum after. Use consistent signs for all objects in the system.