Theory

Systems of particles, momentum and collisions

How momentum, impulse and the center of mass describe systems of several bodies, why the total momentum is conserved in the absence of external forces, and how this law, together with energy, solves collisions, explosions and the motion of a rocket.

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Momentum of a particle
Impulse and the impulse-momentum theorem

Momentum of a particle

1. Momentum

Let $P$ be a particle of mass $m$ moving with velocity $v$ . The momentum of $P$ is the vector

p = m v .

It has the direction and the sense of the velocity and is measured in $kg \cdot m/s$ , a unit that coincides with $N \cdot s$ . Two bodies with the same velocity have momenta proportional to their masses; for example a car of $1200 kg$ at $15 m/s$ has $p = 1.8 \cdot 1 0^{4} kg \cdot m/s$ , a tennis ball of $0.058 kg$ at $30 m/s$ has $p \approx 1.7 kg \cdot m/s$ . Momentum is the quantity that bodies exchange when they interact, and in collision problems it is the first one to be written.

2. Newton's second law in terms of momentum

Newton’s second law, which is an experimental law and for which we refer to the topic on Newton’s laws, can be written in the form

F = \frac{d p}{d t},

where $F$ is the resultant of the forces acting on the particle. If the mass is constant, then $\frac{d p}{d t} = m \frac{d v}{d t} = m a$ , so this form coincides with $F = m a$ . The form with $p$ states that the force measures the rate at which momentum changes, and from this form follow the impulse-momentum theorem and the equation of variable-mass systems. For example, if the momentum of a body increases uniformly by $6.0 kg \cdot m/s$ every $2.0 s$ , the resultant force has magnitude $3.0 N$ .

Impulse and the impulse-momentum theorem

1. Impulse of a force

Let $F (t)$ be a force acting on a particle in the time interval $[t_{1}, t_{2}]$ . The impulse of the force over the interval is the vector

J = \int_{t_{1}}^{t_{2}} F (t) d t,

computed component by component. For a constant force the impulse is $J = F Δ t$ , with $Δ t = t_{2} - t_{1}$ . If the force has a fixed direction, the magnitude of the impulse is the area under the graph of $F (t)$ , and this makes it possible to compute it even when the force is known only at certain points or from a graph.

The impulse of the force

F (t) = F_{m a x} sin \frac{π t}{τ}

, with

F_{m a x} = 3 kN

and

τ = 4 ms

, is the shaded area

J = \frac{2 F _{m a x} τ}{π} \approx 7.6 N \cdot s

; the dashed rectangle with base

τ

and height

F_{m} \approx 1.9 kN

has the same area

2. Impulse-momentum theorem

Let $F (t)$ be the resultant of the forces acting on a particle in the interval $[t_{1}, t_{2}]$ . Then the impulse of the resultant is equal to the change in momentum:

J = p (t_{2}) - p (t_{1}) = Δ p .

The theorem relates a quantity that depends on the whole history of the force, the impulse, to a difference between two instants. It is used when the force is intense, brief and cannot be measured instant by instant, as in the stroke of a racket or in a collision with a wall; in these cases the impulse is obtained from the change in velocity, and from the impulse the average force.

Proof

By Newton’s second law in terms of momentum, at every instant $F (t) = \frac{d p}{d t}$ . Integrating both sides between $t_{1}$ and $t_{2}$ , component by component, and applying the fundamental theorem of calculus to the right-hand side, we obtain

\int_{t_{1}}^{t_{2}} F (t) d t = \int_{t_{1}}^{t_{2}} \frac{d p}{d t} d t = p (t_{2}) - p (t_{1}),

which is the claim.

3. Average force

The average force over the interval $[t_{1}, t_{2}]$ is the constant force that produces the same impulse over the same interval:

F_{m} = \frac{J}{Δ t} = \frac{Δ p}{Δ t} .

In the graph of $F (t)$ it is the height of the rectangle with base $Δ t$ and the same area. For a given change in momentum, the average force is smaller the longer the contact time; for this reason an airbag or a foam mat reduces the force on the body that is being stopped.

Tennis ball against a wall

A tennis ball of mass $m = 0.058 kg$ hits a wall perpendicularly with speed $30 m/s$ and bounces back at $25 m/s$ ; the contact lasts $Δ t = 4.0 ms$ . With the $x$ axis pointing from the wall outward, $v_{1} = - 30 m/s$ and $v_{2} = 25 m/s$ , therefore

J_{x} = m (v_{2} - v_{1}) = 0.058 \cdot 55 kg \cdot m/s \approx 3.2 N \cdot s, F_{m} = \frac{J _{x}}{Δ t} \approx 8.0 \cdot 1 0^{2} N .

The change in velocity is the sum of the two magnitudes, because the velocity reverses its sense. The average force of the wall is about $1400$ times the weight of the ball, which is $0.57 N$ .

Impulsive forces

Forces that are very intense and act for a very short time, such as the contact forces in a collision, are called impulsive. During a small interval $Δ t$ , a force of bounded magnitude, such as weight or friction, produces an impulse of magnitude at most $F_{m a x} Δ t$ , which tends to zero with $Δ t$ ; impulsive forces instead produce a finite impulse. Therefore, over the interval of the collision, all non-impulsive forces can be neglected. In the example of the tennis ball, the impulse of the weight during contact is $0.57 \cdot 0.004 N \cdot s \approx 2 \cdot 1 0^{- 3} N \cdot s$ , compared with $3.2 N \cdot s$ for the impulse of the wall.

The theory continues with 7 more sections (systems of particles and center of mass, motion of the center of mass, conservation of momentum…), plus all worked examples, the solutions and the PDF.

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