The landing of a spacecraft is cushioned with the help of airbags. During its landing on Mars, the velocity of downward fall is 16 meters/second. Immediately after the impact, the velocity is reduced to 1.2 meters/second. If the spacecraft has a mass of 11.5 × 104 kilograms, what is your estimate for the impulse if the time of impact is 0.8 seconds?

Respuesta :

You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same

Answer:

[tex]-1.7\cdot 10^6 kg m/s[/tex]

Explanation:

The impulse is equal to the product between the force (F) and the time of impact ([tex]\Delta t[/tex]):

[tex]I=F \Delta t[/tex]

However, the impulse is also equal to the change in momementum of the spacecraft:

[tex]I = \Delta p= m (v_f - v_i)[/tex]

where

[tex]m=11.5 \cdot 10^4 kg[/tex] is the mass of the spacecraft

[tex]v_f = 1.2 m/s[/tex] is the final velocity

[tex]v_i = 16 m/s[/tex] is the initial velocity

Substituting these numbers into the formula, we find

[tex]I=(11.5\cdot 10^4 kg)(1.2 m/s-16 m/s)=-1.7\cdot 10^6 kg m/s[/tex]

where the negative sign simply means that the impulse is in the opposite direction to the motion of the spacecraft (in fact, it makes it slowing down).

ACCESS MORE
EDU ACCESS