The final velocity of the cue ball is 0.85 m/s at E 25° S and the final velocity of the #8-ball is 1.27 m/s at N 77° E.
The final velocity of the balls after the collision is determined by applying the principle of conservation of linear momentum as shown below;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
Initial velocity of the balls in x-direction with respect to the horizontal
[tex]u_1_x = u_1 cos\theta = 1.2 \times cos(80) = 0.21 \ m/s\\\\u_2_x = u_2 cos\theta = -0.89 \times cos(20) = -0.836 \ m/s[/tex]
Initial velocity of the balls in y-direction with respect to the horizontal
[tex]u_1_y = u_2 sin\theta = 1.2 \times sin(80) = 1.18 \ m/s\\\\u_2_y = u_2 sin\theta = 0.89 \times sin(20) = 0.304 \ m/s[/tex]
[tex]m_1u_1_x + m_2u_2_x = m_1v_1_x + m_2v_2_x\\\\0.17(0.21) + 0.15(-0.836) = 0.17(v_1_x) + 0.15(v_2_x)\\\\-0.0897 = 0.17v_1_x + 0.15v_2_x \ ---(1)[/tex]
Apply one direction linear velocity formula;
[tex]u_1 + v_1 = u_2 + v_2\\\\u_1_x + v_1_x = u_2_x + v_2_x\\\\0.21 + v_1_x = -0.836 + v_2_x\\\\v_2_x = v_1_x + 1.046 \ ---(2)[/tex]
Solve (1) and (2) together;
[tex]-0.0897 = 0.17v_1_x + 0.15(v_1_x + 1.046)\\\\-0.0897 = 0.17v_1v_x + 0.15v_1_x + 0.157\\\\- 0.247 = 0.32v_1_x\\\\v_1_x = \frac{-0.247}{0.32} = -0.77 \ m/s\\\\v_2_x = -0.77 + 1.046 = 0.276 \ m/s[/tex]
[tex]m_1u_1_y + m_2u_2_y = m_1v_1_y + m_2v_2_y\\\\0.17(1.18) + 0.15(0.304)= 0.17v_1_y + 0.15v_2_y\\\\0.246 = 0.17v_1_y + 0.15v_2_y \ --(3)[/tex]
Apply one direction linear velocity formula;
[tex]u_1 + v_1 = u_2 + v_2\\\\u_1_y + v_1_y = u_2_y + v_2_y \\\\1.18 + v_1_y = 0.304 + v_2_y\\\\v_2_y = v_1_y+ 0.876 \ ---(4)[/tex]
Solve (3) and (4) together;
[tex]0.246 = 0.17v_1_y + 0.15(v_1_y + 0.876)\\\\0.246 = 0.17v_1_y + 0.15v_1_y + 0.131\\\\0.115 = 0.32v_1_y\\\\v_1_y = \frac{0.115}{0.32} = 0.36 \ m/s\\\\v_2_y = 0.36 + 0.876 = 1.236 \ m/s[/tex]
[tex]v(cue\ ball) = \sqrt{v_1_x^2 + v_1_y^2} = \sqrt{(-0.77)^2 + (0.36)^2} = 0.85 \ m/s\\\\tan \theta = \frac{v_1y}{v_1x} = \frac{0.36}{-0.77} \\\\\theta = E \ 25^0\ S \\\\v(8 \ ball) = \sqrt{v_2_x^2 + v_2_y^2} = \sqrt{(0.276)^2 + (1.236)^2} = 1.27 \ m/s\\\\tan \theta = \frac{v_2y}{v_2x} = \frac{1.236}{0.276} \\\\\theta = N\ 77\ ^0 \ E\\\\[/tex]
Thus, the final velocity of the cue ball is 0.85 m/s at E 25° S and the final velocity of the #8-ball is 1.27 m/s at N 77° E.
Learn more about conservation of momentum here: https://brainly.com/question/7538238
Answer:
cue ball final: 0.85 m/s at N65°W
8-ball final: 1.27 m/s at N12°E
Explanation:
In a perfectly elastic collision, both momentum and energy are conserved. The momentum is the sum of products of mass and velocity. The energy is half the product of mass and the square of velocity. The difference of "particle" velocities changes sign as a consequence of the collision.
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If we let the masses be m1 and m2, the corresponding initial velocities be v1 and v2, and the final velocities be u1 and u2, we have ...
m1·u1 +m2·u2 = m1·v1 +m2·v2 . . . . . . . conservation of momentum
u1 -u2 = v2 -v1 . . . . . . . . . . . . . . . relative velocity sign changes
This pair of linear equations can be solved for u1 and u2 in any of the usual ways to give ...
u1 = v1(m1 -m2)/(m1 +m2) + v2(2·m2)/(m1 +m2)
u2 = v2(m2 -m1)/(m1 +m2) +v1(2·m1)/(m1 +m2)
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Using angle measures as clockwise from north, we are given ...
cue ball: m1 = 0.17 kg, v1 = 1.2∠10° m/s
eight ball: m2 = 0.15 kg, v2 = 0.89∠-70° m/s
A suitable calculator can find the final velocities using the above equations.
The first attachment shows the calculations using a TI-84 work-alike calculator.
cue ball final: 0.85 m/s at N65°W
8-ball final: 1.27 m/s at N12°E
The second attachment shows a vector diagram of the velocities.
