Momentum

Key Stage 4 Higher

Momentum is a vector quantity and the product of mass and velocity of an object.

Momentum is a vector so it has magnitude and direction.

Momentum is conserved which means the total momentum before an interaction is the same as the total momentum after the interaction, see Conservation of Momentum.

Forces cause a change in momentum. The quicker the change in momentum the greater the force applied to an object.

Momentum = Mass x Velocity

\(p = mv\)

Where

\(p\) = Momentum of the object.

\(v\) = Velocity of the object.

A 21g bullet travels at a velocity of 600m/s. Calculate the momentum of the bullet correct to two significant figures.	A 7.6 tonne lorry travels at a velocity of 30m/s. Calculate the momentum of the bullet correct to two significant figures.
1. State the known quantities m = 21g = 0.021kg v = 610m/s	1. State the known quantities m = 7.6 tonnes = 7600kg v = 32m/s
2. Substitute the numbers into the equation and solve. \(p = mv\) \(p = 0.021 \times 610\) \(p = 12.81kgm/s\) \(p \approx 13kgm/s\)	2. Substitute the numbers into the equation and solve. \(p = mv\) \(p = 7600 \times 32\) \(p = 243200kgm/s\) \(p \approx 240000kgm/s\)

A 700kg car has a momentum of 2.4x10⁴kgm/s. Calculate the velocity of the car correct to two significant figures.	An asteroid with a mass of 77Mg has a momentum of 3.2x10⁹kgm/s. Calculate the velocity of the asteroid correct to two significant figures.
1. State the known quantities m = 700kg p = 2.4x10⁴kgm/s	1. State the known quantities m = 77Mg = 77x10⁶ p = 3.2x10⁹kgm/s
2. Substitute the numbers and evaluate. \(p = mv\) \(2.4 \times 10^4 = 700v\)	2. Substitute the numbers and evaluate. \(p = mv\) \(3.2 \times 10^9 = (77 \times 10^6 )v\)
3. Rearrange the equation and solve. \(v = \frac{2.4 \times 10^4}{700}\) \(v = 34.285714m/s\) \(v \approx 34m/s\)	3. Rearrange the equation and solve. \(v = \frac{3.2 \times 10^9}{77 \times 10^6}\) \(v = 41.558441m/s\) \(v \approx 42m/s\)

A skydiver has a velocity of 53m/s and a momentum of 4400kgm/s. Calculate the mass of the skydiver correct to two significant figures.	A plane travelling with a velocity of 122m/s has a momentum of 9.2x10⁴kgm/s. Calculate the mass of the [plane]] correct to two significant figures.
1. State the known quantities v = 53m/s p = 4400kgm/s	1. State the known quantities v = 122m/s p = 9.2x10⁴kgm/s
2. Substitute the numbers and evaluate. \(p = mv\) \(4400 = 53m\)	2. Substitute the numbers and evaluate. \(p = mv\) \(9.2 \times 10^4 = 122m\)
3. Rearrange the equation and solve. \(m = \frac{4400}{53}\) \(m = 83.01887kg\) \(m \approx 83kg\)	3. Rearrange the equation and solve. \(m = \frac{9.2 \times 10^4}{122}\) \(m = 754.0984kg\) \(m \approx 750kg\)