Momentum
A quantity relating to a moving object that is calculated by multiplying its mass by its velocity. is the product of The amount of matter an object contains. Mass is measured in kilograms (kg) or grams (g). and The speed of an object in a particular direction.. Momentum is also a A measurement having size (magnitude) and direction, eg a displacement of 4 m North. quantity – this means it has both a The size of a physical quantity. and an associated direction.
For example, an elephant has no momentum when it is standing still. When it begins to walk, it will have momentum in the same direction as it is travelling. The faster the elephant walks, the larger its momentum will be.
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Calculating momentum
Momentum can be calculated using the equation:
momentum = mass × velocity
\(p = m~v\)
This is when:
- momentum (p) is measured in kilogram metres per second (kg m/s)
- mass (m) is measured in kilograms (kg)
- velocity (v) is measured in metres per second (m/s)
Example
A lorry has a mass of 7,500 kg. It travels south at a speed of 25 m/s. Calculate the momentum of the lorry.
\(p = m~v\)
\(p = 7,500 \times 25\)
\(p = 187,500 kg~m/s~(south)\)
Question
An ice skater has a mass of 60 kg and travels at a speed of 15 m/s. Calculate the momentum of the skater.
\(p = m~v\)
\(p = 60 \times 15\)
\(p = 900~kg~m/s\)
Conservation of momentum
In a closed system:
total momentum before an event = total momentum after the event
A 'closed system' is something that is not affected by external forces. This is called the principle of The principle that the total momentum of a system remains the same. When bodies collide, whatever momentum is lost by one body, the other gains in equal amounts.. Momentum is conserved in When two objects meet and interact, eg two particles moving towards each other will collide. and When parts of a system separate and move apart. For example, a supernova is an exploding star - the outer layers are thrown out into space in all directions..
Conservation of momentum explains why a gun or cannon recoils backwards when it is fired. When a cannon is fired, the cannon ball gains forward momentum and the cannon gains backward momentum. Before the cannon is fired (the 'event'), the total momentum is zero. This is because neither object is moving. The total momentum of the cannon and the cannon ball after being fired is also zero, with the cannon and cannon ball moving in opposite directions.
Calculations involving collisions
Collisions are often investigated using small trolleys. For example:
Before collision:
After collision:
The principle of conservation of momentum can be used to calculate the velocity of the combined trolleys after the collision.
Example calculation
Calculate the velocity of the trolleys after the collision in the example above.
First calculate the momentum of both trolleys before the collision:
momentum of 2 kg trolley = 2 × 3 = 6 kg m/s
momentum of 8 kg trolley = 8 × 0 = 0 kg m/s
total momentum before collision = 6 + 0 = 6 kg m/s
total momentum (p) after collision = 6 kg m/s (because momentum is conserved)
mass (m) after collision = 10 kg
Next, rearrange \(p = m v\) to find v:
\(v = \frac{p}{m}\)
\(v = \frac{6}{10}\)
\(v = 0.6~m/s\)
Note that the 2 kg trolley is travelling to the right before the collision. As its velocity and the calculated velocity after the collision are both positive values, the combined trolleys must also be moving to the right after the collision.
Calculations involving explosions
The principle of conservation of momentum can be used to calculate the velocity of objects after an explosion.
Example calculation
A cannon ball of mass 4.0 kg is fired from a stationary 96 kg cannon at 120 m/s. Calculate the velocity of the cannon immediately after firing.
Total momentum of cannon and cannon ball before = 0 kg m/s (because neither object is moving)
Total momentum of cannon and cannon ball after collision = 0 kg m/s (because momentum is conserved)
Momentum of cannon ball after firing = 4.0 × 120 = 480 kg m/s
Momentum of cannon after firing = -480 kg m/s (because it recoils in the opposite direction)
Rearrange \(p = m v\) to find v:
\(v = \frac{p}{m}\)
\(v = \frac{-480}{96}\)
\(v = -5.9~m/s\)
Note that the forward velocity of the cannon ball was given a positive value. The negative value for the cannon's velocity shows that it moved in the opposite direction.
