Question

Solve the question in the attachment​

Answer 1

Step-by-step explanation:

Given:

F = 10t² + 10t - 20

To find:

Find the change in momentum in the interval t=0 to t= 10sec ?

Solution:

We know that

$\begin{gathered}\bold{F = \frac{dp}{dt}} \\ \\ or \\ \\ \int dp = \int \: F \: dt \\ \\ dp=∆p = \int_0^{10} \: 10 {t}^{2} + 10t - 20 \: dt \\ \\ ∆p = \frac{10 {t}^{3} }{3} + \frac{10 {t}^{2} }{2} - 20t \Bigg]_0^{10} \\ \end{gathered}$

apply limit

$\begin{gathered}∆p = \frac{10 {(10)}^{3} }{3} + \frac{10 {(10)}^{2} }{2} - 20(10) \\ \\ ∆p = \frac{10000}{3} + 5 \times 100 - 200 \\ \\ ∆p = 3333.33 + 500 - 200 \\ \\∆p = 3633.33 \:Kg-m/sec\\ \\ \end{gathered}$

Final answer:

Change in momentum from t=0 to t=10 is 3633.33 Kg-m/sec

Hope it helps you.

Note*: Not applied lower limit because it is zero and all terms have t.