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aleksklad [387]
3 years ago
7

A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of

the impulse.A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of the impulse.
Physics
1 answer:
Nonamiya [84]3 years ago
6 0

Answer: Impulse = 4 kgm/s

Explanation:

From the question, you're given the following parameters:

Momentum P1 = 12 kgm/s

Momentum P2 = 16 kgm/s

Time t = 0.2 s

According to second law of motion,

Force F = change in momentum ÷ time

That is

F = (P2 - P1)/t

Cross multiply

Ft = P2 - P1

Where Ft = impulse

Substitute P1 and P2 into the formula

Impulse = 16 - 12 = 4 kgm/s

The magnitude of the impulse is therefore 4 kgm/s.

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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

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