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Answer & Explanation:
Crashing into the asteroid would cause the satellite to slow down, stop, or reverse direction, because it is a force in the opposite direction to the satellite's motion. Whichever crash was a stronger force would cause it to change motion more. It takes a stronger force to change the velocity of a more massive object.
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Answer:
The net force acting on the tennis ball while it is in contact with the racquet is 50.73 N
Explanation:
The impulse-momentum theorem said that the net impulse is equal to the change of the momentum, this is:
![\overrightarrow{I}=\varDelta\overrightarrow{P}\,\,(1)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BI%7D%3D%5CvarDelta%5Coverrightarrow%7BP%7D%5C%2C%5C%2C%281%29)
but the net impulse is too the net force times the change in time:
![\overrightarrow{I}=\overrightarrow{F}\varDelta t\,\,(2)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BI%7D%3D%5Coverrightarrow%7BF%7D%5CvarDelta%20t%5C%2C%5C%2C%282%29)
so using (2) on (1) we have:
![\overrightarrow{F}\varDelta t=\varDelta\overrightarrow{P}\,\,(3)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%5CvarDelta%20t%3D%5CvarDelta%5Coverrightarrow%7BP%7D%5C%2C%5C%2C%283%29)
Decomposing that on x and y components:
![F_{x}\varDelta t=\varDelta P_{x}=P_{fx}-P_{ox}\,\,(3)](https://tex.z-dn.net/?f=F_%7Bx%7D%5CvarDelta%20t%3D%5CvarDelta%20P_%7Bx%7D%3DP_%7Bfx%7D-P_%7Box%7D%5C%2C%5C%2C%283%29)
![F_{y}\varDelta t=\varDelta P_{y}=P_{fy}-P_{oy}\,\,(4)](https://tex.z-dn.net/?f=F_%7By%7D%5CvarDelta%20t%3D%5CvarDelta%20P_%7By%7D%3DP_%7Bfy%7D-P_%7Boy%7D%5C%2C%5C%2C%284%29)
(See figure below) with Pfx = m*vfx= m*vf*cos(15°)=(0.058kg)(40m/s)cos(15°),
Pox = -m*vox= m*vo*cos(15°)=-(0.058kg)(30m/s)cos(15°), the same analysis to Pfy and Poy gives
Pfy=(0.058kg)(40m/s)sin(15°), Poy=-(0.058kg)(30m/s)sin(15°), using those values on (3) and (4) and solving for Fy and Fx:
![F_{x}=\frac{(0.058)(40)cos(15\text{\textdegree})-(-(0.058)(30)cos(15\text{\textdegree}))}{0.08}\simeq49N\,\,(5)](https://tex.z-dn.net/?f=F_%7Bx%7D%3D%5Cfrac%7B%280.058%29%2840%29cos%2815%5Ctext%7B%5Ctextdegree%7D%29-%28-%280.058%29%2830%29cos%2815%5Ctext%7B%5Ctextdegree%7D%29%29%7D%7B0.08%7D%5Csimeq49N%5C%2C%5C%2C%285%29)
![F_{y}=\frac{(0.058)(40)sin(15\text{\textdegree})-(-(0.058)(30)sin(15\text{\textdegree}))}{0.08}\simeq13.13N\,\,(6)](https://tex.z-dn.net/?f=F_%7By%7D%3D%5Cfrac%7B%280.058%29%2840%29sin%2815%5Ctext%7B%5Ctextdegree%7D%29-%28-%280.058%29%2830%29sin%2815%5Ctext%7B%5Ctextdegree%7D%29%29%7D%7B0.08%7D%5Csimeq13.13N%5C%2C%5C%2C%286%29)
So the net force acting on the tennis ball while it is in contact with the racquet is:
![F=\sqrt{F_{x}^{2}+F_{y}^{2}}\simeq50.73N](https://tex.z-dn.net/?f=F%3D%5Csqrt%7BF_%7Bx%7D%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%5Csimeq50.73N)