Answer:
![v1 = \sqrt{(2)}Vi](https://tex.z-dn.net/?f=v1%20%3D%20%5Csqrt%7B%282%29%7DVi)
![v2 = \sqrt{(2/3)}Vi](https://tex.z-dn.net/?f=v2%20%3D%20%5Csqrt%7B%282%2F3%29%7DVi)
ANGLE is 35.3 degree celcius
Explanation:
Given data:
mass m and 3m
initial speed Vi
particle with mass m is moving toward left while particle with mass 3m is moving toward right
By using conservation of momentum :
![mVi + 3m(-Vi) = mv1 +3mv2](https://tex.z-dn.net/?f=mVi%20%2B%203m%28-Vi%29%20%3D%20mv1%20%2B3mv2)
![-2mVi = m(v1 + 3v2)](https://tex.z-dn.net/?f=-2mVi%20%3D%20m%28v1%20%2B%203v2%29)
![-2Vi = v1 + 3v2](https://tex.z-dn.net/?f=-2Vi%20%3D%20v1%20%2B%203v2)
conservation of energy :
![m(Vi^2) + 3m(-Vi^2) = mv1^2 + 3mv2^2](https://tex.z-dn.net/?f=m%28Vi%5E2%29%20%2B%203m%28-Vi%5E2%29%20%3D%20mv1%5E2%20%2B%203mv2%5E2)
![4mVi^2 = m(v1^2+3v2^2)](https://tex.z-dn.net/?f=4mVi%5E2%20%3D%20m%28v1%5E2%2B3v2%5E2%29)
![4Vi^2 = v1^2+3v2^2](https://tex.z-dn.net/?f=4Vi%5E2%20%3D%20v1%5E2%2B3v2%5E2)
After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,
x direction : ![-2mVi = 3m.v2i](https://tex.z-dn.net/?f=-2mVi%20%3D%203m.v2i)
![-2Vi = 3v2i](https://tex.z-dn.net/?f=-2Vi%20%3D%203v2i)
y direction : ![0 = m(v1)j+3m(v2)j](https://tex.z-dn.net/?f=0%20%3D%20m%28v1%29j%2B3m%28v2%29j)
![-v1j = 3v2j](https://tex.z-dn.net/?f=-v1j%20%3D%203v2j)
subsitute these value in energy conservation
![v1 = \sqrt{(2)}Vi](https://tex.z-dn.net/?f=v1%20%3D%20%5Csqrt%7B%282%29%7DVi)
![v2 = \sqrt{(2/3)}Vi](https://tex.z-dn.net/?f=v2%20%3D%20%5Csqrt%7B%282%2F3%29%7DVi)
35.3 degree from x-axis
I believe it would be x because you are trying to find the width of the wave but i could be wrong
If the box is light enough, the box will be pulled to the right. If it’s too heavy, then it won’t move
When do you gotta turn it in?