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How do I do this problem? 9.A) A 100 g apple is falling from a tree. What is the impulse that Earth exerts on it during the first 0.5s ofits fall? What about the next 0.5 s?9.B) The same 100 g apple is falling from the tree. What is the impulse that Earth exerts on it in the first0.5 m of its fall? What about the second 0.5 m?9.c) Give a clear explanation for why the answers from 9.a and 9.b are different.

Sagot :

Given:

The mass of the apple is m = 100g = 0.1 kg

To find (A) the impulse during the first 0.5 s and in the next 0.5 s

(B) Impulse during the first 0.5 m of its fall and about the second 0.5 m.

(C)

Explanation:

(A) The force acting on the apple will be

[tex]\begin{gathered} F=mg \\ =0.1\times9.8\text{ } \\ =\text{ 0.98 N} \end{gathered}[/tex]

Impulse during the first 0.5 s will be

[tex]\begin{gathered} Impulse\text{ = 0.98}\times0.5 \\ =0.49\text{ N s} \end{gathered}[/tex]

Impulse during the second 0.5 s will be

[tex]\begin{gathered} Impulse\text{ =0.98}\times(0.5+0.5) \\ =0.98\text{ N s} \end{gathered}[/tex]

(B) The distance traveled by the apple is d = 0.5 m

[tex]\begin{gathered} d1=\frac{1}{2}g(t1)^2 \\ t1=\sqrt{\frac{2d1}{g}} \\ =\sqrt{\frac{2\times0.5}{9.8}} \\ =0.319\text{ s} \end{gathered}[/tex]

The velocity will be

[tex]\begin{gathered} v1=gt1 \\ =9.8\times0.319 \\ =3.1262\text{ m/s} \end{gathered}[/tex]

The distance traveled by the apple in the second 0.5 m

[tex]\begin{gathered} d2=\frac{1}{2}g(t2)^2 \\ t2=\sqrt{\frac{2d2}{g}} \\ =\sqrt{\frac{2\times0.5}{9.8}} \\ =0.319\text{ s} \end{gathered}[/tex]

The velocity will be

'

[tex]\begin{gathered} v2=v1+gt2 \\ =3.1262+(9.8\times0.319) \\ =6.2524\text{ m/s} \end{gathered}[/tex]

The impulse will be

[tex]\begin{gathered} Impulse=\text{ change in momentum} \\ =mv2-mv1 \\ =0.1\times(6.2524)-0.1\times(3.1262) \\ =0.62524-0.31262 \\ =0.31262\text{ N s} \end{gathered}[/tex]

(C) Although the numerical value is the same in both the cases but in part A it is the time and in part B it is the distance.

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