Question #95ec3

Oct 9, 2016

K E of an object of mass m and velocity v is given by the formula

${E}_{k} = \frac{1}{2} m {v}^{2}$

For first bullet $m = 4.0 g = 4 \times {10}^{-} 3 k g \mathmr{and} \text{velocity"= 100" m/s}$

Its KE ${E}_{k 1} = \frac{1}{2} \times 4 \times {10}^{-} 3 \times {100}^{2} = 20 J = 20 \times {10}^{7} e r g$

In Calorie KE ${E}_{k 1} = \frac{20}{4.2} c a l \approx 4.762 c a l$

For 2nd bullet $m = 8.0 g = 8 \times {10}^{-} 3 k g \mathmr{and} \text{velocity"= 100" m/s}$

Its KE ${E}_{k 2} = \frac{1}{2} \times 8 \times {10}^{-} 3 \times {100}^{2} = 40 J = 40 \times {10}^{7} e r g$

In Calorie KE ${E}_{k 2} = \frac{40}{4.2} c a l \approx 9.524 c a l$