Question

2. A 14000 kg air jet accelerates from rest to 70 m/s before it takes off. What is the changein momentum of the jet?

Answer 1

980,000 kg m/s

Step-by-step explanation:

The change in momentum can be calculated as

`\begin{gathered} \Delta p=m\Delta v \\ \Delta p=m(v_f-v_i) \end{gathered}`

Where m is the mass, vf is the final velocity and vi is the initial velocity. Replacing m = 14000 kg, vf = 70 m/s and vi = 0 m/s, we get

`\begin{gathered} \Delta p=14000\text{ kg \lparen}70\text{ m/s - 0 m/s\rparen} \\ \Delta p=14000\text{ kg \lparen70 m/s\rparen} \\ \Delta p=980000\text{ kg m/s} \end{gathered}`

Therefore, the change in momentum of the jet is 980,000 kg m/s