Momentum, Collisions and Force

Momentum combines mass and velocity

A moving object has momentum, calculated by multiplying its mass by its velocity:

$$p=mv$$

Here, $p$ is momentum in kilograms metres per second (kg m/s), $m$ is mass in kilograms (kg), and $v$ is velocity in metres per second (m/s). An object at rest has zero momentum because its velocity is zero.

At the same velocity, an object with twice the mass has twice the momentum. For the same mass, doubling the velocity doubles the momentum. A light object moving quickly can therefore have the same momentum as a heavier object moving slowly.

Velocity describes both speed and direction, so momentum is also a vector. For motion along a straight line, choose one direction as positive. If right is positive, an object moving left has negative velocity and negative momentum. This lets you account for direction when adding momenta or calculating a change.

What is conserved in a collision?

A system is the object or group of objects you are considering. When two marbles collide, you can treat both marbles together as one system. The contact forces between them act within that system; friction from the surface is an external force because it comes from outside the system.

When the resultant external force on a system is zero, its total momentum is conserved:

$$\text{total momentum before}=\text{total momentum after}$$

In a short collision on a smooth surface, external forces may be small enough to neglect. One object can lose momentum while the other gains it: conservation applies to the total, not necessarily to each object separately.

For example, when one moving marble hits an identical stationary marble head-on, the first can stop while the second moves away with the first marble's original velocity. The momentum has been transferred between the marbles, but their total momentum is unchanged.

Objects can also stick together after a collision. They then move with a common velocity, so their final momentum is their combined mass multiplied by that velocity.

Following momentum through a collision

Suppose a 2 kg trolley moving right at 3 m/s collides with a stationary 1 kg trolley. They stick together, and external forces are negligible during the collision. Take right as positive.

The moving trolley initially has momentum $2\times3=6\text{ kg m/s}$. The stationary trolley has zero momentum, so the total before the collision is $6$.

Afterwards, the combined mass is $2+1=3\text{ kg}$. Conservation of momentum gives:

$$6=3v$$

Therefore, the joined trolleys move at $v=2\text{ m/s}$ to the right. They move more slowly than the original moving trolley because the same total momentum is now shared by a greater mass.

For any straight-line collision, calculate each object's momentum with its signed velocity, then add the momenta. Use the same positive direction before and after the collision.

A resultant force changes momentum

Newton's second law states that the resultant force is the rate of change of momentum. This means the change in momentum per second. In the form used to link force and time:

$$\Delta p=Ft$$

Here, $\Delta p$ is the change in momentum in kg m/s, $F$ is the resultant force in newtons (N), and $t$ is the time for which it acts in seconds (s). For a varying force, $F$ represents the average resultant force over that time.

A larger resultant force produces a larger momentum change in the same time. The same force acting for twice as long produces twice the momentum change. The change in momentum is in the direction of the resultant force.

To find the change, subtract initial momentum from final momentum. For an object whose mass stays constant:

$$\Delta p=mv-mu=m(v-u)$$

Here, $u$ is initial velocity and $v$ is final velocity. Keeping their directions matters, particularly when an object rebounds.

Calculating force or contact time

Rearrange the equation according to the quantity needed:

$$F=\frac{\Delta p}{t} \qquad t=\frac{\Delta p}{F}$$

Suppose a 0.20 kg ball initially travels right at 10 m/s and is brought to rest in 0.10 s. Taking right as positive, its initial momentum is $+2.0\text{ kg m/s}$ and its final momentum is zero. Its change in momentum is therefore $-2.0\text{ kg m/s}$.

The average resultant force is $-2.0/0.10=-20\text{ N}$. The negative sign means that the force acts left, opposing the ball's original motion.

If the same ball instead rebounds left at 10 m/s, its final momentum is $-2.0\text{ kg m/s}$. Its change is then $-2.0-(+2.0)=-4.0\text{ kg m/s}$: reversing direction produces a greater momentum change than merely stopping.

For a fixed change in momentum, increasing the contact time reduces the average force. Moving your hands backwards while catching a ball allows it to stop over a longer time, reducing the force needed to stop it.

Momentum

$$p=mv$$

$p$: kg m/s; $m$: kg; $v$: m/s. Momentum is a : use signed velocities to represent opposite directions.