Question

On an aircraft carrier a 20,608kg plane can be launched from 0m/s to 78m/s in 2seconds in order for the plane to take off in a relatively small space and time. What is the impulse experienced on this plane during launch?

Answer 1

To determine the impulse we use the fact that the impulse is equal to the change in momentum:

`I=p_2-p_1`

Where:

`\begin{gathered} I=\text{ impulse} \\ p_2=\text{ final momentum} \\ p_1=\text{ initial momentum} \end{gathered}`

The momentum is the product of the mass and the velocity, therefore, we have:

`\begin{gathered} I=m_v_2-m_v_1 \\ \end{gathered}`

Since the initial velocity is zero, we have:

`\begin{gathered} I=m_2v_2-m_(0) \\ I=m(v_2) \end{gathered}`

Now, we substitute the values:

`I=(20608kg)(78(m)/(s))`

Solving the operation:

`I=1607424kg(m)/(s)`

Therefore, the momentum is 1607424 kgm/s