A $5000 ext{ kg}$ rocket is set for vertical firing. The exhaust speed is $800 ext{ ms}^{-1}$. To give an

Question

A $5000 	ext{ kg}$ rocket is set for vertical firing. The exhaust speed is $800 	ext{ ms}^{-1}$. To give an initial upward acceleration of $20 	ext{ ms}^{-2}$, the amount of gas ejected per second to supply the needed thrust will be ($g = 10 	ext{ ms}^{-2}$)

Accepted Answer

1. **Identify Forces:** The rocket experiences two main forces: the downward force of gravity ($mg$) and the upward thrust force ($F_t$) generated by the ejected gas. The net acceleration ($a$) is upwards.
2. **Apply Newton's Second Law:**
   $$ F_{net} = F_t - mg = ma $$
   $$ F_t = m(g + a) $$
3. **Relate Thrust to Mass Ejection:** The thrust is given by the product of the exhaust speed ($v$) and the rate of mass ejection ($dm/dt$).
   $$ F_t = v \frac{dm}{dt} $$
4. **Combine Equations:**
   $$ v \frac{dm}{dt} = m(g + a) $$
   $$ \frac{dm}{dt} = \frac{m(g + a)}{v} $$
5. **Substitute Values:**
   Given: $m = 5000 	ext{ kg}$, $v = 800 	ext{ m/s}$, $a = 20 	ext{ m/s}^2$, $g = 10 	ext{ m/s}^2$.
   $$ \frac{dm}{dt} = \frac{5000 (10 + 20)}{800} $$
   $$ \frac{dm}{dt} = \frac{5000 	imes 30}{800} = \frac{150000}{800} = \frac{1500}{8} $$
   $$ \frac{dm}{dt} = 187.5 	ext{ kg s}^{-1} $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Exercise 5.9).

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