Q2) The position of an artillery, with a speed of projectile 650, which can fire in any direction above the horizontal plane and
whose direction of fire can be adjusted necessarily and sufficiently fast, is (t) = 30 x 10m +20 x 10'my. Besides that, the position of a target is determined as Fr(t) =(-12 x 10 m + 100). Thus at which instant of time must the artillery be fired in order that the target is terminated as earlier as possible? (Ignore the friction, the air resistance and assume that there is no Coriolis force)
1 answer:
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Answer:
Approximately .
Explanation:
Let the increase in the rocket's velocity be . Let
represent the initial velocity of the rocket. Note that for this question, the exact value of
doesn't really matter.
The momentum of an object is equal to its mass times its velocity.
- Mass of the rocket with the 5 kg of fuel:
.
- Initial velocity of the rocket and the fuel:
.
- Hence the initial momentum of the rocket:
.
- Mass of the rocket without that 5 kg of fuel:
.
- Final velocity of the rocket:
.
- Hence the final momentum of the rocket:
.
- Mass of the 5 kg of fuel:
.
- Final velocity of the fuel:
(assuming that the the 500 m/s in the question takes the rocket as its reference.)
- Hence the final momentum of the fuel:
.
Momentum is conserved in an isolated system like the rocket and its fuel. That is:
Sum of initial momentum = Sum of final momentum.
.
Note that appears on both sides of the equation. These two terms could hence be eliminated.
.
.
Hence, the velocity of the rocket increased by around 2.5 m/s.