Respuesta :
Answer:
The speed of the combined honey drop after the collision is 9.05 m/s.
Explanation:
Given that,
The masses and velocities of the drops are
[tex]m_{1}=31.5\ g[/tex]
[tex]v_{1}=12.7\ m/s[/tex]
[tex]m_{2}=49.9\ g[/tex]
[tex]v_{1}=12.5\ m/s[/tex]
[tex]m_{1}=78.1\ g[/tex]
[tex]v_{1}=15.9\ m/s[/tex]
We need to calculate the total mass
Using formula of masses
[tex]m=m_{1}+m_{2}+m_{3}[/tex]
Put the value into the formula
[tex]m=31.5+49.9+78.1[/tex]
[tex]m=159.5\ g[/tex]
We need to calculate the velocity in x- direction
Using conservation of momentum
[tex]m_{1}v_{1}=(m_{1}+m_{2}+m_{3})v_{x}[/tex]
[tex]v_{x}=\dfrac{m_{1}v_{1}}{m_{1}+m_{2}+m_{3}}[/tex]
Put the value into the formula
[tex]v_{x}=\dfrac{31.5\times12.7}{159.5}[/tex]
[tex]v_{x}=2.50\ m/s[/tex]
We need to calculate the velocity in y- direction
Using conservation of momentum
[tex]m_{2}v_{2}=(m_{1}+m_{2}+m_{3})v_{y}[/tex]
[tex]v_{y}=\dfrac{m_{2}v_{2}}{m_{1}+m_{2}+m_{3}}[/tex]
Put the value into the formula
[tex]v_{y}=\dfrac{49.9\times12.5}{159.5}[/tex]
[tex]v_{y}=3.91\ m/s[/tex]
We need to calculate the velocity in z- direction
Using conservation of momentum
[tex]m_{3}v_{3}=(m_{1}+m_{2}+m_{3})v_{z}[/tex]
[tex]v_{z}=\dfrac{m_{3}v_{3}}{m_{1}+m_{2}+m_{3}}[/tex]
Put the value into the formula
[tex]v_{z}=\dfrac{78.1\times15.9}{159.5}[/tex]
[tex]v_{z}=7.78\ m/s[/tex]
We need to calculate the combined honey drop after the collision
Using formula of velocity
[tex]v=v_{x}+v_{y}+v_{z}[/tex]
Put the value into the formula
[tex]v=2.50i+3.91j+7.78k[/tex]
The magnitude of velocity
[tex]v=\sqrt{(2.50)^2+(3.9)^2+(7.78)^2}[/tex]
[tex]v=9.05\ m/s[/tex]
Hence, The speed of the combined honey drop after the collision is 9.05 m/s.
