Question

A 0.25 N bullet traveling at 500 m/s hits a 44.75 N block suspended vertically by a 3 m long string. Find the maximum angle in which the string makes with the vertical if the bullet imbeds itself in the block

Answer 1

Answer:

88.34°

Explanation:
First we must do Conversion of Momentum:

$\[ m_{\text{bullet}} \times v_{\text{bullet, initial}} = (m_{\text{bullet}} + m_{\text{block}}) \times v_{\text{final}} \]\[ 0.25 \, \text{kg} \times 500 \, \text{m/s} = (0.25 \, \text{kg} + 4.57 \, \text{kg}) \times v_{\text{final}} \]\[ 125 = 4.82 \times v_{\text{final}} \]$
Next we do Energy:
$\[ \frac{1}{2} m_{\text{bullet}} v_{\text{bullet, initial}}^2 = m_{\text{block}} g h \]\[ \frac{1}{2} \times 0.25 \times 500^2 = 4.57 \times 9.8 \times 3 \]\[ 31250 = 134.136 \, h \]$
Finally we use trigonometry to figure out the max angle:
$\[ \sin(\theta) = \frac{h}{\text{length of the string}} \]\[ \sin(\theta) = \frac{232.954}{3} \]\[ \theta ≈ \sin^{-1}\left(\frac{232.954}{3}\right) \]\[ \theta ≈ \sin^{-1}(77.651) \]\[ \theta ≈ 88.34° \]$