Respuesta :
Answer:
The initial velocity of the ball is 2.567 m/s
Solution:
Mass of the ball, m = 75 gm = 0.075 kg
Mass of the pendulum, m' = 200 gm = 0.2 kg
Height of rise of the ball, h = 2.5 cm = 0.025 m
Length, l = 30 cm = 0.3 m
Now,
Total mass, M = m + m' = 0.075 + 0.2 = 0.275 kg
Now,
Suppose the initial velocity of the ball be [tex]v_{i}[/tex]
The total mass is raised to a height of 0.025 m after the ball has been fired, thus the total potential energy of the system is given by:
[tex]PE = Mgh = 0.275\times 9.8\times 0.025 = 0.0674\ J[/tex]
Now, the initial kinetic energy of the system:
[tex]KE = \frac{1}{2}Mv^{2}[/tex]
Now, using law of conservation of energy:
PE = KE
[tex]\frac{1}{2}Mv^{2} = 0.0674[/tex]
[tex]v = \sqrt{\frac{0.13475}{0.275}} = 0.7\ m/s[/tex]
Now by using the principle of momentum conservation:
[tex]mv_{i} = Mv[/tex]
[tex]v_{i} = \frac{Mv}{m}[/tex]
[tex]v_{i} = \frac{0.275\times 0.7}{0.075} = 2.567\ m/s[/tex]
