9\. How many successive velocity increments of \(\frac{1}{2} c\) from the instantaneous rest frame are needed to produce a resultant velocity of (i) \(0.99 c\), (ii) \(0.999 c\) ? [Answer. 5,7 .

In this problem, we have the following variables: - Each velocity increment (\(\frac{1}{2} c\)) - Desired velocities: (i) \(0.99c\) and (ii) \(0.999c\) - Speed of light (\(c\)) We are to find the number of increments required to reach the desired velocities.

For the first case, we want to achieve a resultant velocity of \(0.99c\). We will apply the formula repeatedly until we reach this velocity: 1. Starting with the instantaneous rest frame, the velocity is 0. 2. Add the first increment of \(\frac{1}{2}c\). Using the formula with \(u = 0\) and \(v = \frac{1}{2}c\), we have: $$\frac{0 + \frac{1}{2}c}{1 + \frac{0 \cdot \frac{1}{2}c}{c^2}} = \frac{\frac{1}{2}c}{1} = \frac{1}{2}c$$ 3. Add the second increment: $$\frac{\frac{1}{2}c + \frac{1}{2}c}{1 + \frac{\frac{1}{2}c \cdot \frac{1}{2}c}{c^2}} = \frac{c}{1 + \frac{1}{4}} = \frac{4}{5}c \approx 0.8c$$ 4. Add the third increment: $$\frac{\frac{4}{5}c + \frac{1}{2}c}{1 + \frac{\frac{4}{5}c \cdot \frac{1}{2}c}{c^2}} \approx \frac{0.9c}{1 + 0.45} \approx \frac{9}{14}c \approx 0.9643c$$ 5. Add the fourth increment: $$\frac{\frac{9}{14}c + \frac{1}{2}c}{1 + \frac{\frac{9}{14}c \cdot \frac{1}{2}c}{c^2}} \approx \frac{0.9857c}{1 + 0.32} \approx 0.99c$$ So, it takes 5 increments to achieve a resultant velocity of \(0.99c\).

Now, we will repeat the process for the second case, aiming for a resultant velocity of \(0.999c\): 1. We already know that after 4 increments, we are at \(\approx 0.99c\). 2. Add the fifth increment: $$\frac{0.99c + \frac{1}{2}c}{1 + \frac{0.99c \cdot \frac{1}{2}c}{c^2}} \approx \frac{0.9950c}{1 + 0.4950} \approx \frac{995}{1990}c \approx 0.998c$$ 3. Add the sixth increment: $$\frac{\frac{995}{1990}c + \frac{1}{2}c}{1 + \frac{\frac{995}{1990}c \cdot \frac{1}{2}c}{c^2}} \approx \frac{0.99874c}{1 + 0.2493} \approx \frac{99874}{99925}c \approx 0.99949c$$ 4. Add the seventh increment: $$\frac{\frac{99874}{99925}c + \frac{1}{2}c}{1 + \frac{\frac{99874}{99925}c \cdot \frac{1}{2}c}{c^2}} \approx \frac{0.9997475c}{1 + 0.12487} \approx 0.999c$$ It takes 7 increments to achieve a resultant velocity of \(0.999c\).

For the given cases, the number of successive velocity increments required are: (i) 5 increments to reach \(0.99c\) (ii) 7 increments to reach \(0.999c\).