Question

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.

Answer 1

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)

Step-by-step explanation:

(A) The force acting on the apple will be

`\begin{gathered} F=mg \\ =0.1*9.8\text{ } \\ =\text{ 0.98 N} \end{gathered}`

Impulse during the first 0.5 s will be

`\begin{gathered} Impulse\text{ = 0.98}*0.5 \\ =0.49\text{ N s} \end{gathered}`

Impulse during the second 0.5 s will be

`\begin{gathered} Impulse\text{ =0.98}*(0.5+0.5) \\ =0.98\text{ N s} \end{gathered}`

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

`\begin{gathered} d1=(1)/(2)g(t1)^2 \\ t1=\sqrt{(2d1)/(g)} \\ =\sqrt{(2*0.5)/(9.8)} \\ =0.319\text{ s} \end{gathered}`

The velocity will be

`\begin{gathered} v1=gt1 \\ =9.8*0.319 \\ =3.1262\text{ m/s} \end{gathered}`

The distance traveled by the apple in the second 0.5 m

`\begin{gathered} d2=(1)/(2)g(t2)^2 \\ t2=\sqrt{(2d2)/(g)} \\ =\sqrt{(2*0.5)/(9.8)} \\ =0.319\text{ s} \end{gathered}`

The velocity will be

'

`\begin{gathered} v2=v1+gt2 \\ =3.1262+(9.8*0.319) \\ =6.2524\text{ m/s} \end{gathered}`

The impulse will be

`\begin{gathered} Impulse=\text{ change in momentum} \\ =mv2-mv1 \\ =0.1*(6.2524)-0.1*(3.1262) \\ =0.62524-0.31262 \\ =0.31262\text{ N s} \end{gathered}`

(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.