Question

A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8 m/s 2 . If the pendulum rises a vertical distance of 3.12 cm, calculate the initial speed of the bullet

Answer 1

Answer:

$v = 186.90\,\frac{m}{s}$

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

$\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h$

The final velocity of the system formed by the ballistic pendulum and the bullet is:

$v = \sqrt{2\cdot g\cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}$

$v\approx 0.78\,\frac{m}{s}$

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

$(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )$

$v = 186.90\,\frac{m}{s}$