Answer :
Answer:
a) [tex]a=(-2,-6,0)m/s^2[/tex], with a magnitude of [tex]6.3m/s^2[/tex]
b) [tex]\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2[/tex], with a magnitude of [tex]0.76Kgm/s^2[/tex]
c) [tex]F=(-0.24,-0.72,0)N[/tex], with a magnitude of [tex]0.76N[/tex]
Explanation:
We have:
[tex]v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg[/tex]
We can calculate each component of the acceleration using its definition [tex]a=\frac{\Delta v}{\Delta t}[/tex]
[tex]a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\[/tex]
The rate of change of momentum of the ball is [tex]\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma[/tex]
So for each coordinate:
[tex]\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2[/tex]
And these are equal to the components of the net force since F=ma.
If magnitudes is what is asked:
[tex]a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N[/tex]
(N and [tex]Kgm/s^2[/tex] are the same unit).