Question

A spaceprobe in outer space is flying with a constant speed of 1.795 km/s. The probe has a payload of 1635.0 kg and it carries 4092.0 kg of rocket fuel. The rocket engines of the probe are capable of expelling propellant at a speed of 4.161 km/s. Then the rocket engines are fired up. How fast will the spaceprobe travel when all the rocket fuel is used up?

Answer 1

The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

What is the law of conservation of linear momentum?

According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.

Unit conversion;

1 km/sec = 1000 m/sec

Given data;

Spaceprobe speed  = 1.795 km/s = 1795 m /sec

Probe mass = 635.0 kg

Fuel mass = 4092.0 kg

Expelled propellent velocity = 4.161 km/s = 41461 m/sec

From the momentum conservation principle;

$\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec$

Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

To learn more about the law of conservation of momentum refer to:

brainly.com/question/1113396

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