Newton’s third law and momentum

An object is in equilibrium when its resultant force is zero. It has no acceleration, so it can remain at rest or move at constant velocity.

Consider a book resting on a table. Earth’s gravitational pull on the book acts downwards, and the table’s contact force on the book acts upwards. These forces balance because the book is in equilibrium. However, they are not a third-law pair: both act on the book, and they are different types of force.

There are two separate interactions here:

• The table pushes upwards on the book; the book pushes downwards on the table. These are a contact-force pair.
• Earth pulls the book downwards; the book pulls Earth upwards. These are a gravitational-force pair.

Momentum describes an object’s motion — Higher tier

Momentum is the product of an object’s mass and velocity:

$$p = m \times v$$

Here, $p$ is momentum in kilogram metres per second (kg m/s), $m$ is mass in kilograms (kg), and $v$ is velocity in metres per second (m/s).

At the same velocity, a more massive object has more momentum. For the same mass, doubling the velocity doubles the momentum. An object at rest has zero momentum because its velocity is zero.

Momentum is a vector: it has both magnitude and direction. If rightwards is chosen as positive, leftwards velocities and momenta are negative. A negative momentum describes direction, not a negative mass.

For example, a 2 kg trolley travelling rightwards at 3 m/s has momentum $p = 2 \times 3 = 6\text{ kg m/s}$ rightwards. Travelling leftwards at the same speed, its momentum would be $-6\text{ kg m/s}$ with this sign convention.

What happens during a collision?

Imagine trolley A moving rightwards towards stationary trolley B. During contact, A pushes B rightwards, while B pushes A leftwards with an equally large force. B accelerates rightwards, while A slows down.

The equal forces do not necessarily produce equal accelerations. From $a = F/m$, the less massive trolley has the greater acceleration magnitude for the same force. Equal masses have equal acceleration magnitudes in opposite directions, provided these interaction forces are their resultant forces.

Newton’s third law also explains why momentum is transferred rather than created during the collision. The interaction forces are equal and opposite and act for the same contact time. The trolleys therefore undergo equal and opposite changes in momentum: the momentum gained by one is balanced by the momentum lost by the other.

For the two trolleys considered together, total momentum before the collision equals total momentum afterwards, provided there is no resultant external force. Forces between the trolleys are internal to this system; an external force, such as friction from the track, could change its total momentum.

Following the momentum through a collision — Higher tier

For example, let the moving trolley have mass 2 kg and velocity 3 m/s rightwards, and let the stationary trolley also have mass 2 kg. Suppose they stick together after colliding, with external forces negligible.

Their total initial momentum is $2 \times 3 + 2 \times 0 = 6\text{ kg m/s}$. Afterwards, the combined mass is 4 kg, so their shared velocity is $6/4 = 1.5\text{ m/s}$ rightwards.

The first trolley’s momentum falls from 6 to 3 kg m/s, while the second’s rises from zero to 3 kg m/s. Their individual momenta change, but the total remains 6 kg m/s.

Force is the rate of change of momentum — Higher tier

A resultant force changes an object’s velocity and therefore its momentum. Newton’s second law can be written as:

$$F = \frac{\text{change in momentum}}{\text{time taken}} = \frac{mv - mu}{t}$$

Here, $F$ is force in newtons (N), $m$ is the object’s constant mass in kilograms, $u$ is its initial velocity, $v$ is its final velocity, and $t$ is the time taken in seconds. If the force varies during contact, this calculation gives the average resultant force.

For the same change in momentum, a shorter contact time means a greater average force. For example, when a tennis ball hits a racket, changing its momentum over a shorter time produces a larger average interaction force. By Newton’s third law, the ball and racket exert equally large, oppositely directed forces on each other.

Direction matters especially when an object rebounds. For example, a 0.06 kg ball approaches a racket at $+10\text{ m/s}$ and leaves at $-15\text{ m/s}$ after 0.05 s of contact. Its change in momentum is $0.06 \times (-15) - 0.06 \times 10 = -1.5\text{ kg m/s}$. The average force on the ball is therefore $-1.5/0.05 = -30\text{ N}$: a force of magnitude 30 N opposite to its original motion.

The interacting objects exert equal and opposite forces for the same time, giving equal and opposite momentum changes.

Total momentum before = total momentum after, if there is no resultant external force.

Momentum — Higher tier

$$p = mv$$

$p$: kg m/s; $m$: kg; $v$: m/s. Momentum has direction; an object at rest has zero momentum.

Force and momentum — Higher tier

$$F = \frac{mv - mu}{t}$$

$u$ is initial velocity; $v$ is final velocity; $t$ is time in seconds; $F$ is in newtons.

For the same momentum change, shorter time means greater average force.