Given that, \[F=40\cos \left( \frac{\pi }{10} \right)t\]
A) 12.5 m/s
B) 8.5 m/s
C) 20 m/s
D) 9.5 m/s
Correct Answer: A
Solution :
[a] Change in liner momentum \[\Delta P=\int{Fdt}\] \[15({{v}_{f}}+u)=\int\limits_{0}^{15}{40\cos \left( \frac{\pi }{10} \right)t\,dt}\] \[{{v}_{f}}=-4+\frac{40}{15}{{\left[ \frac{\sin (\pi /10)t}{\pi /10} \right]}^{15}}_{0}=-4+\frac{400}{15\pi }(-1)\]\[=-12.5m/s\]
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