Momentum

Cambridge·Physics·1.6·Extended

Momentum

A moving object has momentum because it has mass and velocity. An object with greater mass or greater speed has a greater magnitude of momentum.

Momentum is the product of mass and velocity:

p=mvp = mv

Here, $$p$$ is momentum, $$m$$ is mass in kilograms and $$v$$ is velocity in metres per second. The SI unit of momentum is kilogram metre per second, written $$\mathrm{kg,m/s}$$.

Momentum is a vector quantity: it has both magnitude and direction. Its direction is the same as the object's velocity. For example, an object of mass $$2\ \mathrm{kg}$$ moving east at $$3\ \mathrm{m/s}$$ has momentum $$6\ \mathrm{kg,m/s}$$ east.

Rearranging the momentum equation

If momentum and velocity are known, mass can be found by dividing momentum by velocity. If momentum and mass are known, velocity can be found by dividing momentum by mass:

m=pvandv=pmm = \frac{p}{v} \qquad \text{and} \qquad v = \frac{p}{m}

When calculating along a straight line, a direction can be chosen as positive. Momentum in the opposite direction then has a negative value, as does velocity in that direction. Finding mass using $$m = p/v$$ requires a non-zero velocity.

Impulse and the Force–Momentum Relationship

A resultant force changes an object’s momentum. The longer the force acts, the greater the change in momentum, provided the force stays the same.

Impulse combines the force and the time interval for which it acts. It is equal to the change in momentum:

impulse=FΔt=Δp=Δ(mv)\text{impulse} = F\Delta t = \Delta p = \Delta(mv)

Here, $$F$$ is the resultant force, $$\Delta t$$ is the time interval, and $$\Delta p$$ is the change in momentum. Momentum is $$p = mv$$, where $$m$$ is mass and $$v$$ is velocity. The change in momentum is the final momentum minus the initial momentum.

For a constant resultant force, $$F$$ is that constant force. If the force varies during the interval, $$F$$ is the average resultant force over the interval.

Force as a rate of change of momentum

Rearranging the impulse relationship shows that the resultant force is the rate at which momentum changes:

F=ΔpΔtF = \frac{\Delta p}{\Delta t}

For the same change in momentum, a longer time interval means a smaller average resultant force. A shorter interval means a larger average resultant force.

Impulse and change in momentum are vectors: they have magnitude and direction. The impulse has the same direction as the change in momentum.

The unit of impulse is the newton second, $$\mathrm{N,s}$$. This is equivalent to the momentum unit $$\mathrm{kg,m/s}$$.

Conservation of Momentum

When objects interact, they can change each other’s momentum. The total momentum of the objects can nevertheless stay the same: momentum lost by one object is gained by another.

The conservation of momentum principle states that the total momentum of a system remains constant when there is no resultant external force on the system. A system is the set of objects being considered; an external force acts on the system from outside it.

∑pbefore=∑pafter\sum p_{\text{before}} = \sum p_{\text{after}}

The sums include the momentum of every object in the system, before and after the interaction. An individual object’s momentum can change even though the total remains constant.

Direction and momentum

Momentum is a vector, so its direction matters. For motion along one line, choose one direction as positive. Velocities in that direction are positive; velocities in the opposite direction are negative.

An object’s momentum is $$p = mv$$, where $$m$$ is its mass and $$v$$ is its velocity. Since mass is positive, the sign of the momentum is the same as the sign of the velocity. Add the signed momenta to find the total; do not simply add their magnitudes.

Finding an unknown velocity

A $$2,\mathrm{kg}$$ object moving at $$3,\mathrm{m/s}$$ collides with a $$1,\mathrm{kg}$$ object moving at $$1,\mathrm{m/s}$$ in the opposite direction. After the collision, the $$1,\mathrm{kg}$$ object moves at $$2,\mathrm{m/s}$$ in the original direction of the $$2,\mathrm{kg}$$ object. Assume there is no resultant external force on the two-object system.

Choose the original direction of the $$2,\mathrm{kg}$$ object as positive. The initial velocity of the other object is therefore $$-1,\mathrm{m/s}$$. Let $$v$$ be the final velocity of the $$2,\mathrm{kg}$$ object. Substituting $$p = mv$$ into the conservation equation gives:

(2)(3)+(1)(−1)=2v+(1)(2)(2)(3) + (1)(-1) = 2v + (1)(2)

Solving gives $$v = 1.5,\mathrm{m/s}$$. The positive result means the $$2,\mathrm{kg}$$ object continues in the chosen positive direction.

The same approach can be used to find an unknown mass: write the total momentum before and after using signed velocities, then solve for the missing quantity.