Changing acceleration of a rocket in terms of Law of Conservation of Momentum (1 Viewer)
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umbreller
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We can look at the rocket and it's fuel/exhaust as an isolated system. According to LCoM "the sum of momenta in an isolated system always remains constant", i.e. total momentum of the rocket and it's exhaust gases is always the same. So for any small amount of time, then any forward/upward increase in momentum of the rocket is equal and opposite to change in momentum of the exhaust gases being expelled.
Here's an algebraic way of thinking about it, but you won't ever need to use it in an answer, purely for your understanding
p(rocket) + p(gases) = k (k is the constant sum of momentum)
p(rocket) = k - p(gases) - (1)
Where 'p' here is really just initial momentum.
-> But as momentum changes, final momentum is given by p + Δp:
p(rocket) + Δp(rocket) = k - [p(gases) + Δp(gases)]
p(rocket) + Δp(rocket) = k - p(gases) - Δp(gases)
Sub in (1) in accordance with LCoM:
Δp(rocket) = - Δp(gases)
Which means
Δ(mv)rocket = -Δ(mv)gases
^ The last 3 lines you should mention straight up in your answers.
Because of the burning fuel, rocket mass is decreasing. Since the rocket is ofc under constant thrust, F=ma, (Newton's 2nd law) then acceleration must be increasing and hence velocity is increasing (at an increasing rate).
Here's an algebraic way of thinking about it, but you won't ever need to use it in an answer, purely for your understanding
p(rocket) + p(gases) = k (k is the constant sum of momentum)
p(rocket) = k - p(gases) - (1)
Where 'p' here is really just initial momentum.
-> But as momentum changes, final momentum is given by p + Δp:
p(rocket) + Δp(rocket) = k - [p(gases) + Δp(gases)]
p(rocket) + Δp(rocket) = k - p(gases) - Δp(gases)
Sub in (1) in accordance with LCoM:
Δp(rocket) = - Δp(gases)
Which means
Δ(mv)rocket = -Δ(mv)gases
^ The last 3 lines you should mention straight up in your answers.
Because of the burning fuel, rocket mass is decreasing. Since the rocket is ofc under constant thrust, F=ma, (Newton's 2nd law) then acceleration must be increasing and hence velocity is increasing (at an increasing rate).