Answer:
The total energy of the system is 7.778 J
Solution:
As per the question:
Height of circular, H = 0.15 m
Radius of the circular arc, R = 0.27 m
Mass of the bullet, [tex]m_{1} = 30\ g = 0.03\ kg[/tex]
Mass of the block, [tex]m_{1} = 5.29\ kg[/tex]
Now,
By using the law of conservation of energy:
Kinetic Energy, KE = Potential Energy, U
Thus
[tex]\frac{1}{2}(m_{1} + m_{2})v^{2} = (m_{1} + m_{2})gH[/tex]
[tex]v = \sqrt{2gH}= \sqrt{2\times 9.8\times 0.15} = 1.71\ m/s[/tex]
Now, the total energy of the system after collision:
[tex]KE = \frac{1}{2}(m_{1} + m_{2})v^{2}[/tex]
[tex]KE = \frac{1}{2}(0.03 + 5.29)\times 1.71^{2} = 7.778\ J[/tex]
