Question

(Inelastic) Ballistic Pendulum. A bullet that has a mass of 120g is fired horizontally at a block of jelly of mass 15kg that is suspended from a thin wire. Unfortunately, the designer of the experiment didn't pick a large enough block of jelly, and the bullet passed through the jelly and kept moving. After the bullet passed through the jelly, the experimenters measured the bullet to have a speed of 20m/s and the jelly block swung to a maximum height of 45cm higher than its initial height. How much energy was lost to heating up the jelly and the bullet

Answer 1

Answer:

$\Delta E_{loss} = 9100.262\,J$

Explanation:

The bullet-Ballistic Pendulum system can be modelled after the Principles of Momentum Conservation and Energy Conservation:

Final speed of the ballistic pendulum:

$\frac{1}{2}\cdot (0.12\,kg)\cdot v^{2} = (0.12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.45\,m)$

$v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.45\,m)}$

$v \approx 2.971\,\frac{m}{s}$

Initial speed of the bullet:

$(15\,kg)\cdot \left(0\,\frac{m}{s}\right) + (0.12\,kg)\cdot v = (0.12\,kg)\cdot \left(20\,\frac{m}{s} \right) + (15\,kg)\cdot \left(2.971\,\frac{m}{s} \right)$

$v = 391.375\,\frac{m}{s}$

Energy loss:

$\Delta E_{loss} = \frac{1}{2}\cdot (15\,kg)\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(2.971\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (0.12\,kg)\cdot \left[\left(391.375\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right]$$\Delta E_{loss} = 9100.262\,J$