In a 15 Joule impact test, the weight of the steel ball depends on the drop height required to generate 15 Joules of impact energy.
Formula for Impact Energy: E=mâ‹…gâ‹…hE = m \cdot g \cdot h Where: Weight Calculation: Rearranging the formula to find m: m=Egâ‹…hm = \frac{E}{g \cdot h} m=159.81â‹…1=159.81≈1.53 kgm = \frac{15}{9.81 \cdot 1} = \frac{15}{9.81} \approx 1.53 \, \text{kg} m=159.81â‹…0.5=154.905≈3.06 kgm = \frac{15}{9.81 \cdot 0.5} = \frac{15}{4.905} \approx 3.06 \, \text{kg} m=159.81â‹…0.2=151.962≈7.65 kgm = \frac{15}{9.81 \cdot 0.2} = \frac{15}{1.962} \approx …