What Is The Magnitude Of The Change In Momentum Of The Cannonball During Firing
Forces, acceleration and Newton'due south laws of motion
Newton's laws of motion assistance explicate why an object needs a force to brand it move - inertia also explains this. Momentum, on the other paw, explains some of the almost important interactions in nature.
Conservation of momentum - Higher
In a closed system:
- total momentum before an consequence = total momentum after the consequence
A 'airtight system' is something that is not afflicted past external forces. This is called the principle of conservation of momentum . Momentum is conserved in collisions and explosions .
Conservation of momentum explains why a gun or cannon recoils backwards when it is fired. When a cannon is fired, the cannon ball gains forwards momentum and the cannon gains backward momentum. Earlier the cannon is fired (the 'consequence') the total momentum is zippo. This is considering neither object is moving. The total momentum of the cannon and the cannon ball later on existence fired is also zero, with the cannon and cannon ball moving in opposite directions.
Calculations involving collisions
Collisions are often investigated using small-scale trolleys. The diagrams show an example.
Earlier collision
After standoff
You tin can use the principle of conservation of momentum to summate the velocity of the combined (joined together) trolleys after the collision.
Case calculation
Calculate the velocity of the trolleys after the standoff in the example above.
Start calculate the momentum of both trolleys earlier the collision:
- 2 kg trolley = 2 × 3 = six kg m/s
- iv kg trolley = 8 × 0 = 0 kg grand/s
- total momentum before collision = 6 + 0 = 6 kg m/s
- total momentum (p) after collision = six kg m/due south (because momentum is conserved)
- mass (g) later collision = 10 kg
Next, rearrange p = one thousand 5 to detect v:
- \[v = \frac{p}{m}\]
- \[v = 6 \div10\]
- \[\underline{v=0.six\ thou/due south}\]
Note that the 2 kg trolley is travelling to the right before the collision. Every bit its velocity and the calculated velocity after the standoff 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 exist used to calculate the velocity of objects later an explosion.
Example calculation
A cannon ball of mass 4.0 kg is fired from a stationary 96 kg cannon at 120 m/s. Summate the velocity of the cannon immediately afterward firing.
full momentum of cannon and cannon ball before = 0 kg yard/s - because neither object is moving
full momentum of cannon and cannon ball later on collision = 0 kg k/s - because momentum is conserved
Momentum of cannon brawl afterward firing = 4.0 × 120 = 480 kg m/due south.
Momentum of cannon after firing = -480 kg m/s (because it recoils in the opposite direction and 480 - 480 = 0 kg yard/s, the total momentum after collision).
Rearrange p = m 5 to find v:
- \[five = \frac{p}{one thousand}\]
- \[v = 480 \div96\]
- \[\underline{v=v.0\ m/southward}\]
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 management.
Source: https://www.bbc.co.uk/bitesize/guides/zsvxdxs/revision/6
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