Question

During the first 60 seconds of a rocket launch, the momentum of the rocket increased by 200,000 kgm/s. What was the impulse applied to the rocket during this time?

A. 120,000,000 kgm/s
B. 33,333 kgm/s
C. 500,000 kgm/s
D. 200,000 kgm/s

Answer 1

D. 200,000 kg*m/s


By the way I Hope this helps

Answer 2

impulse, J = 200,000 kg-m/s

Explanation:

It is given that,

The momentum of the rocket is increased by 200,000 kg-m/s during the first 60 seconds. The impulse applied to the rocket during this time is given by the change in momentum i.e.

`\Delta p=p_f-p_i`

Another definition of impulse is given by the force applied and the small interval of time i.e.

`J=\Delta p=F\Delta t`

So, in this case impulse applied to the rocket during 60 seconds is equivalent to the change in momentum i.e. 200,000 kg-m/s. Hence, the correct option is (d).