Momentum: definition, conservation and calculations

In a nutshell

All objects with mass that are moving have momentum, a larger momentum means that it's harder to bring the object to rest. The momentum of an object is the product of its mass and velocity. Momentum is a vector quantity, the direction that the object is travelling in is the direction of the momentum.

Equations

definiTION	SYMBOL EQUATION
Momentum	$p = mv$
Force	$F = \frac{m \Delta v}{t}$

Variable definitions

QUANTITY NAME	SYMBOL	dervied UNIT	si base UNITs
$momentum$	$p$	$kgms^{-1}$	$kg ms^{-1}$
$mass$	$m$	$kg$	$kg$
$velocity$	$v$	$ms^{-1}$	$ms^{-1}$
$force$	$F$	$N$	$kgms^{-2}$
$time$	$t$	$s$	$s$

Definition of momentum

The momentum of an object is the mass of the object multiplied by the velocity of the object:

$p = m v$

Example

If a car with a mass of $1200 \space kg$  is moving with a velocity of $5 \space ms^{-1}$ , what is the momentum of the car?

First state the variables from the question:

$m = 1200 \space kg$

$v = 5 \space ms^{-1}$

Next, write down the equation you need to use:

$p=mv$

Then, substitute these values into the equation:

$p = 1200 \times 5$

Don't forget to include your units:

$6000\, kg\, ms^{-1}$

The momentum of a $1200\,kg$ car travelling at $5\,ms^{-1}$ is $\underline{6000 \space kg \space ms^{-1}}$ .

Conservation of momentum

A closed system is one in which there are no external force acting. In a closed system, the momentum before an event is equal to the momentum after the event. This principle is called the conservation of momentum:

$mu = mv$

This is the reason that cannons move backwards after they fire a cannonball. The cannonball and cannon are not moving initially, so the total momentum before the cannonball is fired is zero.

Once the cannonball is fired, the ball moves forward with a very high speed. Momentum, which is a vector quantity, needs to be conserved so the cannon moves backwards - in the opposite direction to the ball.

Example

For the diagram below which shows two balls colliding, write a balanced equation for the conservation of momentum.

Physics; Newton's laws of motion and momentum; KS5 Year 12; Momentum: definition, conservation and calculations

Firstly, consider the momentum before the collision:

$momentum \space before = m_1u_1 + m_2u_2$

From the diagram, $u_2 = 0 \ ms^{-1}$ :

$momentum \space before = m_1u_1$

Now, consider the momentum after the collision:

$momentum \space after = m_1v_1 + m_2v_2$

Therefore as $momentum \space before = momentum \space after$ :

$m_1u_1 = m_1v_1 +m_2v_2$

Change in momentum

If a force is applied to a moving object, the momentum of the object must change. The force is proportional to the rate of change of the momentum:

$F = \dfrac{\Delta p}{t} = \dfrac{m \space \Delta v }{t}$

This is Newton's second law of motion. For a constant mass, the rate of change in velocity is the acceleration of the object and so the equation goes to:

$F=ma$

Example

A $15 \ g$ bullet accelerates to $350 \ ms^{-1}$ after being fired from a gun with mass $1.5 \ kg$ . What is the velocity of the recoiling gun?

First state the variables from the question:

$m_b = 1.5 \times 10^{-2} \ kg, \ \ \ v_b = 350 \ ms^{-1}, \ \ \ m _g = 1.5 \ kg$

Then state the equation needed:

$m_gu_g + m_bu_b = m_gv_g + m_bv_b$

Rearrange the equation:

$0 = m_gv_g + m_bv_b\newline \\[0.1in] m_gv_g =- m_bv_b\newline \\[0.1in] v_g =- \dfrac{m_bv_b}{m_g}$

Sub in and solve:

$v_g =- \dfrac{15 \times 10^{-3} \times 350}{1.5}\newline \\[0.1in]v_g = -3.5 \ ms^{-1}$

The velocity of the recoiling gun is $\underline{-3.5 \ ms^{-1}}$ .

Note: As the velocity is negative, it means that the gun is travelling in the opposite direction to the bullet.