Respuesta :
Answer:
The magnitud of the velocity is
[tex]8.46m/s[/tex]
and the direccion:
[tex]-28.3[/tex] degrees from the horizontal.
Explanation:
Fist we define our variables:
[tex]m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?[/tex]
The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.
velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.
Using conservation of momentum:
[tex]m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}[/tex]
Clearing for the velocity of the stone after the crash:
[tex]v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}[/tex]
Substituting known values:
[tex]v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)[/tex]
The magnitud of the velocity is :
[tex]|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s[/tex]
and the direction:
[tex]tan^{-1}(-8.75/16.25)=-28.3[/tex]
this is -28.3 degrees from the +i direction or the horizontal direcction.
Note: i and j can also be seen as x and y axis.
