Question

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?

Answer 1

Answer:

$-1.7\cdot 10^6 kg m/s$

Explanation:

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

$I=F \Delta t$

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

$I = \Delta p= m (v_f - v_i)$

where

$m=11.5 \cdot 10^4 kg$ is the mass of the spacecraft

$v_f = 1.2 m/s$ is the final velocity

$v_i = 16 m/s$ is the initial velocity

Substituting these numbers into the formula, we find

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

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).

Answer 2

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