Answer:
speed = 1.96 m/s
direction = 52.25° south of east
Explanation:
given that:
the velocity of the first ball [tex]v_1 = 2.0 \ m/s[/tex]
the velocity of the second ball [tex]v_2 = \ 0.8 \ m/s[/tex]
the final velocity of the second ball [tex]v_2f = 1.55 \ m/s[/tex]
The mass of the identical balls [tex]m_1 = m_2 = m[/tex]
However from law of conservation of momentum [tex]m_1v_1 + m_2v_2 = m_1v_1f + m_2v_2f[/tex]
along X- axis
[tex]m_1vx_1f = m_1v_1 + m_2v_2[/tex]
[tex]vx_1f =v_1 + v_2[/tex]
= ( 2.0 - 0.8 ) m/s
= 1.2 m/s
along Y-axis
[tex]0 = m_1vy_1f + m_2v_2f\\m_1vy_1f = -m_2v_2f\\then vyf_1 = - v_2f \\= -1.55 \ m/s[/tex]
then the final speed of the first ball
[tex]v_1f = \sqrt{(1.2)^2 + (1.55)^2[/tex]
[tex]v_1f = 1.96 \ m/s[/tex]
and direction
[tex]\theta = tan^{-1}( \frac{-1.55}{1.2} )[/tex]
[tex]\theta = 52.25^0[/tex] south of east.
