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Two boys are sliding toward each other on a frictionless, ice-covered parking lot. jacob, mass 45 kg, is gliding to the right at 7.81 m/s, and ethan, mass 31.0 kg, is gliding to the left at 10.0 m/s along the same line. when they meet, they grab each other and hang on.

Respuesta :

AlpenGlow
If you are trying to find for the final velocity of the two boys, consider that the collision that has occurred is a perfect inelastic collision because they hung on to each other, meaning they moved together as one. This can be solved using the formula:

[tex]m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}[/tex]

m1 = mass of object 1
m2 = mass of object 2
v1i = velocity of object 1 before the collision
v2i = velocity of object 2 before the collision
vf = final velocity

Now based on your problem the given would be the following:
m1 = 45kg    (Mass of Jacob)
m2 = 31 kg   (Mass of Ethan)
v1i = 7.81 m/s  (Jacob)
v2i = -10 m/s    (Ethan)    (Now this is negative because he is coming from                                                            the opposite direction.) 

Now all you need to do is input the given by filling in what you know and solving for what you don't know.

[tex]m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}[/tex]
[tex](45kg)(7.81m/s)+(31kg)(-10m/s)=(45kg+31kg)v_{f}[/tex]
[tex](45kg)(7.81m/s)+(31kg)(-10m/s)=(45kg+31kg)v_{f}[/tex]
[tex](351.45kg.m/s)+(-310kg.m/s)=(76kg)v_{f}[/tex]
[tex](41.45kg.m/s)=(76kg)v_{f}[/tex]

Transpose the 76kg to the other side of the equation to isolate vf.
[tex](41.45kg.m/s)/(76kg)=v_{f}[/tex]
[tex]0.545m/s=v_{f}[/tex]

Because the value is positive, that means it is following the direction Jacob was going. So the final velocity is 0.545m/s to the right.

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