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

Variable definitions

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Definition of momentum

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

​$$p = m v$$

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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:

$$p=mv$$

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Then, substitute these values into the equation:

​$$p = 1200 \times 5$$

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Don't forget to include your units:

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

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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.

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?

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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:

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

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

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

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