Observer on earth sees a ship travelling at \( 0.5 \mathrm{c}[\mathrm{fwd} \) ] wrt the ground. A plasma cannon fires a pulse out the front of the ship so that the ground observer sees the pulse travel at \( 0.9 \mathrm{c}[\mathrm{fwd} \) ] wrt the ground. How fast do the observers in the ship see the pulse travel wit the ship?

Let's denote the speed of the ship as \( v \) and the speed of the pulse as \( u \). Given that the observer on earth sees the ship traveling at \( 0.5c \) and the pulse traveling at \( 0.9c \) with respect to the ground, we can set up the following equations based on the relativistic velocity addition formula: 1. The velocity of the ship with respect to the ground is \( v = 0.5c \). 2. The velocity of the pulse with respect to the ground is \( u = 0.9c \). We need to find the velocity of the pulse with respect to the ship, denoted as \( u' \). Using the relativistic velocity addition formula, we have: \[ u' = \frac{u - v}{1 - \frac{uv}{c^2}} \] Substitute the given values into the formula to find the velocity of the pulse with respect to the ship. Simplify the expression by following steps: - step0: Solution: \(\frac{\left(0.9c-0.5c\right)}{\left(1-\frac{\left(0.9c\times 0.5c\right)}{c^{2}}\right)}\) - step1: Remove the parentheses: \(\frac{0.9c-0.5c}{1-\frac{0.9c\times 0.5c}{c^{2}}}\) - step2: Reduce the fraction: \(\frac{0.9c-0.5c}{1-\frac{9}{20}}\) - step3: Subtract the terms: \(\frac{0.4c}{1-\frac{9}{20}}\) - step4: Subtract the numbers: \(\frac{0.4c}{\frac{11}{20}}\) - step5: Rewrite the expression: \(\frac{\frac{2}{5}c}{\frac{11}{20}}\) - step6: Rewrite the expression: \(\frac{\frac{2c}{5}}{\frac{11}{20}}\) - step7: Multiply by the reciprocal: \(\frac{2c}{5}\times \frac{20}{11}\) - step8: Reduce the fraction: \(2c\times \frac{4}{11}\) - step9: Multiply the terms: \(\frac{2c\times 4}{11}\) - step10: Multiply the terms: \(\frac{8c}{11}\) The velocity of the pulse with respect to the ship is \( \frac{8c}{11} \).