<span>The answer to this question depends upon Newton's third law of motion. For every action, there's an equal and opposite reaction. Because of this law, during the collision between two unequal masses, the impulse that each mass receives will be of equal magnitude and and opposite sign.</span>
<em><u>2</u></em><em><u>0</u></em><em><u>.</u></em><em><u>0</u></em><em><u>M</u></em><em><u>/</u></em><em><u>S</u></em><em><u>. </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>HORIZONTAL</u></em><em><u> </u></em><em><u>VELOCITY</u></em><em><u> </u></em><em><u>OF</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>BALL</u></em><em><u> </u></em><em><u>JUST</u></em><em><u> </u></em><em><u>BEFORE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>REA</u></em><em><u>CHES</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>GROUND</u></em>
<em><u>1</u></em><em><u>2</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>SECONDS</u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>APPROXIMATE</u></em><em><u> </u></em><em><u>TOTAL</u></em><em><u> </u></em><em><u>TIME</u></em><em><u> </u></em><em><u>REQUIRED</u></em><em><u> </u></em><em><u>FOR</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>BALL</u></em>
Answer:
The high of the ramp is 2.81[m]
Explanation:
This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.
If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.
We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.
![E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%5C%5Cwhere%3A%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5Cv%20%3D%202.8%5Bm%2Fs%5D%5C%5Cm%3D%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2Av%5E%7B2%7D%20%7D%20%5C%5Cm%3D%5Cfrac%7B3.8%7D%7B0.5%2A2.8%5E%7B2%7D%20%7D%20%5C%5Cm%3D0.969%5Bkg%5D)
Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.
![E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%2BW_%7Bf%7D%3DE_%7Bk%7D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5CW_%7Bf%7D%3D23%5BJ%5D%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5C)
And therefore
![m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%20%2B%20W_%7Bf%7D%3D3.8%5C%5C%200.969%2A9.81%2Ah%20-%2023%3D%203.8%5C%5Ch%20%3D%20%5Cfrac%7B23%2B3.8%7D%7B0.969%2A9.81%7D%5C%5C%20h%20%3D%202.81%5Bm%5D)