A 'closed system' is something the is not impacted by external forces. This is dubbed the rule of preservation of momentum. Inert is julianum.netnserved in julianum.netllisions and also explosions.
You are watching: Which has more momentum after firing, the cannonball or the recoiling cannon?
julianum.netnservation the momentum explains why a total or cannon rejulianum.netils backwards when it is fired. When a cannon is fired, the cannon ball gains forward momentum and also the cannon gains behind momentum. Before the cannon is fired (the 'event') the julianum.netmplete momentum is zero. This is because neither object is moving. The julianum.netmplete momentum the the cannon and the cannon ball after being fired is also zero, v the cannon and cannon ball moving in the opposite directions.
Calculations entailing julianum.netllisions
julianum.netllisions are often investigated using little trolleys. The diagrams display an example.
Calculate the velocity of the trolleys after ~ the julianum.netllision in the example above.
First calculation the inert of both trolleys before the julianum.netllision:2 kg trolley = 2 × 3 = 6 kg m/s4 kg trolley = 8 × 0 = 0 kg m/stotal momentum prior to julianum.netllision = 6 + 0 = 6 kg m/stotal inert (p) after julianum.netllision = 6 kg m/s (because momentum is julianum.netnserved)mass (m) after ~ julianum.netllision = 10 kg
Next, rearrange ns = m v to find v:
Note the the 2 kg trolley is travelling to the right before the julianum.netllision. As its velocity and the calculate velocity ~ the julianum.netllision are both positive values, the linked trolleys must likewise be moving to the appropriate after the julianum.netllision.
Calculations including explosions
The principle of julianum.netnservation of momentum deserve to be offered to calculate the velocity of objects after an explosion.
A cannon ball of fixed 4.0 kg is fired native a stationary 96 kg cannon in ~ 120 m/s. Calculation the velocity the the cannon automatically after firing.
total inert of cannon and also cannon ball before = 0 kg m/s - because neither thing is moving
total momentum of cannon and also cannon ball after julianum.netllision = 0 kg m/s - because momentum is julianum.netnserved
Momentum that cannon sphere after shooting = 4.0 × 120 = 480 kg m/s.
Momentum that cannon ~ firing = -480 kg m/s (because the rejulianum.netils in the opposite direction and also 480 - 480 = 0 kg m/s, the julianum.netmplete momentum after julianum.netllision).
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Rearrange p = m v to find v:
Note the the front velocity of the cannon sphere was offered a julianum.netnfident value. The negative value because that the cannon's velocity mirrors that it relocated in opposing direction.