Answer:
the velocity (magnitude and direction) of the ball) just before the collision is [tex]v_i = 5.144 \ m/s[/tex]
The velocity (magnitude and direction) of the ball) just after the collision is [tex]v_f = - 1.129 \ m/s[/tex]
Explanation:
According to the law of conservation of energy;
[tex]m__{ball}} gh = \frac{1}{2}m__{ball}}v_i^2\\\\2* m__{ball}} gh = m__{ball}}v_i^2\\\\v_i^2 = \frac{2*m_{ball}gh}{m_{ball}}\\\\v_i^2 = 2gh\\\\v_i = \sqrt{2gh} \\\\v_i = \sqrt{2*9.8*1.35}\\\\[/tex]
[tex]v_i = 5.144 \ m/s[/tex]
Thus; the velocity (magnitude and direction) of the ball) just before the collision is [tex]v_i = 5.144 \ m/s[/tex]
Since, Air resistance is negligible, and the collision is elastic.
The equation for the conservation of momentum and energy can be expressed as:
[tex]v_f = [\frac{m_1 -m_2}{m_1+m_2}]v_i\\\\v_f = [\frac{m_{ball} -m_{block}}{m_{ball}+m_{block}}]v_i\\\\v_f = [\frac{1.6 -2.5}{1.6+2.5}]*5.144\\\\[/tex]
[tex]v_f = - 1.129 \ m/s[/tex]
The velocity (magnitude and direction) of the ball) just after the collision is [tex]v_f = - 1.129 \ m/s[/tex]
