Physics 9702 · Paper 2

Momentum

The rest of this chapter is about collisions. To predict a collision you need one new quantity.
Two things decide how hard an object is to stop: its mass and its velocity. Multiply them and you get .

The definition

p = mv

Symbols

$p$ = linear momentum (kg m s⁻¹)
$m$ = mass (kg)
$v$ = velocity (m s⁻¹)

Linear momentum is the product of mass and velocity. That sentence, word for word, is a one-mark definition.
The unit kg m s⁻¹ has no special name. In base units it is the same as N s (lesson 3.12 shows why).
Momentum is a vector. Its direction is the direction of the velocity. In one dimension, choose a positive direction and give the opposite way a minus sign.

Worked example

Smallest case: one moving ball

A ball of mass 3.0 kg rolls at 4.0 m s⁻¹. Find its momentum.

$p = mv = 3.0 \times 4.0 = 12 \text{ kg m s}^{-1}$ in the direction of motion.

Answer

p = 12 \text{ kg m s}^{-1}

Very different objects can have the same momentum. A 20 000 kg lorry rolling at 1.5 m s⁻¹ and a 1500 kg car driving at 20 m s⁻¹ both have p = 30 000 kg m s⁻¹.

Your turn— tap to reveal the worked answer (9702/11/O/N/22 Q7)

Which is the correct definition of linear momentum? (9702/11/O/N/22 Q7)

Answer: the product of mass and velocity. It is a vector.

Momentum is not kinetic energy

Both are built from $m$ and $v$ , but they behave differently.
Momentum $p = mv$ is a vector: two equal objects moving opposite ways have total momentum zero.
Kinetic energy $E_k = \tfrac{1}{2}mv^2$ is a scalar: the same two objects have twice the kinetic energy of one, never zero.

Writing kinetic energy with momentum

Square the momentum: $p^2 = m^2v^2$ . Divide by $2m$ :

$\dfrac{p^2}{2m} = \tfrac{1}{2}mv^2 = E_k$

So $E_k = p^2/2m$ . Paper 1 uses this to jump straight from momentum to kinetic energy without finding $v$ first, and to ask how $E_k$ changes when $p$ changes (9702/11/M/J/25 Q19).

PreviousIdentifying Forces & Free-Body Diagrams NextConservation of Momentum