Answer:
[tex]v = 2.63 m/s[/tex]
[tex]\theta = -66.2 degree[/tex]
Explanation:
In the initial direction of motion we can use momentum conservation
so here we have
[tex]m_1v_1 = m_1v_{1x} + m_2v_{2x}[/tex]
[tex]m(6.00) = m(5.39 cos26) + mv_{2x}[/tex]
[tex]6 = 4.84 + v_{2x}[/tex]
[tex]v_{2x} = 1.16 m/s[/tex]
now by momentum conservation along perpendicular direction we can use as
[tex]0 = m_1v_{1y} + m_2v_{2y}[/tex]
[tex]0 = m(5.39 sin26) + mv_{2y}[/tex]
[tex]v_{2y} = -2.36 m/s[/tex]
So the magnitude of the final speed of the stuck ball is given as
[tex]v = \sqrt{v_{2x}^2 + v_{2y}^2}[/tex]
[tex]v = \sqrt{1.16^2 + 2.36^2}[/tex]
[tex]v = 2.63 m/s[/tex]
direction of motion given as
[tex]tan\theta = \frac{v_{2y}}{v_{2x}}[/tex]
[tex]tan\theta = \frac{-2.36}{1.16} = -2.27[/tex]
[tex]\theta = -66.2 degree[/tex]
The final velocity of the second billiard ball is 11.1 m/s.
The given parameters;
Apply the principle of conservation of linear momentum for elastic collision.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Apply one dimensional velocity in x-direction;
[tex]u_1 + v_1 = u_2 + v_2\\\\6 + (5.39\times cos \ 26) = 0 + v_2_x\\\\10.85 \ m/s = v_2_x[/tex]
Apply one dimensional velocity in y-direction
[tex]u_1 + v_1 = u_2 + v_2\\\\0 + (5.39\times sin \ 26) = 0 + v_2_y\\\\2.36 \ m/s = v_2_y[/tex]
The resultant of the second ball final velocity is calculated as follows;
[tex]v_2 = \sqrt{v_2_x^2 + v_2_y^2} \\\\v_2= \sqrt{(10.85)^2 +(2.36)^2} \\\\v_2 = 11.1 \ m/s[/tex]
Thus, the final velocity of the second billiard ball is 11.1 m/s.
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