Difference between revisions of "Momentum"

Revision as of 17:49, 8 November 2019

Key Stage 4 Higher

Meaning

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

About Momentum

Momentum is a vector so it has magnitude and direction.

The SI Units of momentum are kilogram metres per second (kgm/s).

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.

Equation

Momentum = Mass x Velocity

\(p = mv\)

Where

\(p\) = Momentum of the object.

\(m\) = Inertial Mass of the object.

\(v\) = Velocity of the object.

Example Calculations

Finding Momentum from Mass and Velocity

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. \(p = mv\) \(p = 7600 \times 32\) \(p = 243200kgm/s\) \(p \approx 240000kgm/s\)

Finding Velocity from Mass and Momentum

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. \(v = \frac{3.2 \times 10^9}{77 \times 10^6}\) \(v = 41.558441m/s\) \(v \approx 42m/s\)

Finding Mass from Velocity and Momentum

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. \(m = \frac{9.2 \times 10^4}{122}\) \(m = 754.0984kg\) \(m \approx 750kg\)

References

AQA

Momentum, pages 141, 164-5, GCSE Physics; Student Book, Collins, AQA

Momentum, pages 150-155, 157, GCSE Physics; Third Edition, Oxford University Press, AQA

Momentum, pages 166-7, GCSE Physics, Hodder, AQA

Momentum, pages 183, 184, GCSE Combined Science Trilogy; Physics, CGP, AQA

Momentum, pages 216-220, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA

Momentum, pages 244-5, GCSE Combined Science Trilogy 2, Hodder, AQA

Momentum, pages 70, 71, GCSE Physics; The Revision Guide, CGP, AQA

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 <math>m \approx 750kg</math>
 |}
+===References===
+====AQA====
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