The speed of the second rock after the collision is 4.15m/s
Given the direction of rock in space is x , the direction with speed is 6m/s
After the collision the speed of the first rock is at 4 m/s
The angle given is 60 degrees with respect to x - axis
We need to calculate the speed of the second rock after the collision is
We know that
[tex]m_1=m_2[/tex]
[tex]v_1 = 6m/s\\v_2 = 4m/s[/tex]
θ = 60°
We need to find [tex]v_x[/tex] and [tex]v_y[/tex]
Therefore ,
[tex]m_1v_1= m_2v_2cos60+m_2v_2\\[/tex]
So, 6 = 4 cos 60° + [tex]v_2[/tex]
[tex]v_x = 6-4cos60\\v_x = 2m/s[/tex]
The same way we will find the momentum for [tex]v_y[/tex]
The total momentum is 0
0+0 = [tex]m_1(4)sin60 + m_2v_y[/tex]
[tex]v_y = -3.64 m/s[/tex]
Now ,
V (resultant) = [tex]\sqrt{(v_x)^{2} + (v_y)^2} \\v = \sqrt{(2)^2+(-3.64)^2} \\v = 4.15 m/s[/tex]
Hence the speed of the second rock after the collision, in m/s is 4.15m/s
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