Respuesta :
Answer:
The final velocity is [tex]\bf{562.9}[/tex].
Explanation:
Given:
The maximum angle that the pendulum and the ball system can travel, [tex]\theta_{m} = 45^{0}[/tex].
The length of the pendulum, [tex]l = 30~cm[/tex].
The mass of the pendulum, [tex]m_{p} = 250~g[/tex].
The mass of the ball, [tex]m_{b} = 76~g[/tex].
Consider that the initial velocity of the ball-pendulum system is [tex]v_{i}[/tex]. So the initial total energy of the system is given by
[tex]E_{i} = K.E. + P.E.\\~~~~= \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2} + 0\\~~~~= \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2}[/tex]
Consider the final velocity of the ball is [tex]v_{0}[/tex] and the final height attended by the system is [tex]h[/tex]. So the final total final energy of the system is given by
[tex]E_{f} = K.E. + P.E.\\~~~~~= 0 + (m_{b} + m_{s})gh\\~~~~~= (m_{b} + m_{s})gh[/tex]
From the conservation of energy,
[tex]E_{i} = E_{f}\\&or,& \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2} = (m_{b} + m_{s})gh\\&or,& v_{i} = \sqrt{2gh}[/tex]
From the conservation of momentum,
[tex]&& m_{b}v_{0} = (m_{b} + m_{s})v_{i}\\&or,& v_{0} = (1 + \dfrac{m_{p}}{m_{b}})\sqrt{2gh}[/tex]
The final height attended by the system is given by
[tex]h = l(1 - \cos 45^{0}) = 8.787~cm[/tex]
The final velocity is given by
[tex]v_{0} = (1 + \dfrac{250}{76}})\sqrt{2(980)(8.787)}\\~~~~= 562.9[/tex]
