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 c [fwd] wrt the ground. How fast do the observers in the ship see the pulse travel wrt the ship?

Let's denote the speed of the pulse as \( v \) and the speed of the ship as \( v_{\text{ship}} \). 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 use the relativistic velocity addition formula to find the speed of the pulse with respect to the ship. The relativistic velocity addition formula is given by: \[ v_{\text{pulse}} = \frac{v + v_{\text{ship}}}{1 + \frac{v \cdot v_{\text{ship}}}{c^2}} \] Substitute the given values into the formula: \[ v_{\text{pulse}} = \frac{0.9c + 0.5c}{1 + \frac{0.9c \cdot 0.5c}{c^2}} \] Simplify the expression to find the speed 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: Add the terms: \(\frac{1.4c}{1+\frac{9}{20}}\) - step4: Add the numbers: \(\frac{1.4c}{\frac{29}{20}}\) - step5: Rewrite the expression: \(\frac{\frac{7}{5}c}{\frac{29}{20}}\) - step6: Rewrite the expression: \(\frac{\frac{7c}{5}}{\frac{29}{20}}\) - step7: Multiply by the reciprocal: \(\frac{7c}{5}\times \frac{20}{29}\) - step8: Reduce the fraction: \(7c\times \frac{4}{29}\) - step9: Multiply the terms: \(\frac{7c\times 4}{29}\) - step10: Multiply the terms: \(\frac{28c}{29}\) The speed of the pulse with respect to the ship is \( \frac{28c}{29} \) or approximately 0.9655c.