Numerical Problems Based on Rocket Propulsion for Class 11 Physics

Mass of the rocket, $m = 3.5 \times 10^4\text{ kg}$

Upward acceleration, $a = 10\text{ ms}^{-2}$. Let’s take acceleration due to gravity, $g = 10\text{ ms}^{-2}$.

The total upward thrust ($F$) must overcome both the weight of the rocket and provide the required upward acceleration. Using Newton’s second law:

Rate of fuel consumption, $\frac{dm}{dt} = 50\text{ g s}^{-1} = 0.05\text{ kg s}^{-1}$

Relative velocity of exhaust gases, $v_r = 5 \times 10^5\text{ cms}^{-1} = 5000\text{ ms}^{-1}$

The thrust ($F$) exerted on the rocket by the exhaust gases is the rate of change of momentum of the ejected mass:

Final velocity to attain, $v = 11.2\text{ kms}^{-1}$

Exhaust velocity, $v_r = 1.6\text{ kms}^{-1}$

Initial velocity, $u = 0$. Neglecting gravity for the ultimate velocity calculation, we use the standard rocket equation:

Taking the exponential of both sides:

The required ratio of initial mass to final mass ($m_0/m$) is approximately $1096$.

Let initial mass be $m_0$. Rate of mass ejection, $\frac{dm}{dt} = \frac{m_0}{60}\text{ kg s}^{-1}$

Exhaust velocity, $v_r = 2073\text{ ms}^{-1}$

First, calculate the upward thrust ($F$):

The net force producing acceleration is the thrust minus the weight of the rocket ($m_0 g$). Taking $g = 9.8\text{ ms}^{-2}$:

Dividing by $m_0$:

(i) Thrust Exerted:

Rate of mass ejection, $\frac{dm}{dt} = 100\text{ kg s}^{-1}$

Exhaust velocity, $v_r = 6 \times 10^3\text{ ms}^{-1}$

(ii) Final Velocity:

Final mass, $m = \frac{1}{40}m_0 \implies \frac{m_0}{m} = 40$

Using the rocket equation (neglecting gravity):

Let initial mass be $m_0$. Rate of mass ejection, $\frac{dm}{dt} = \frac{1\% \text{ of } m_0}{1\text{ s}} = 0.01 m_0\text{ kg s}^{-1}$

Exhaust velocity, $v_r = 2000\text{ ms}^{-1}$

The thrust ($F$) provided by the engine is:

The acceleration imparted by this thrust (ignoring the opposing force of gravity as indicated by the final answer) is given by $F = m_0 a$:

(Note: If considering net acceleration from the Earth’s surface, one would subtract $g$, i.e., $a_{net} = 20 – 9.8 = 10.2\text{ ms}^{-2}$. The answer $20\text{ ms}^{-2}$ reflects the acceleration purely due to engine thrust.)

The phrasing “increased by 5 times” in the context of the answer implies the apparent weight ($R$) has become $5 \times$ his normal weight. So, $R = 5mg$.

Apparent weight in an upward accelerating frame is $R = m(g + a)$.

Now, calculate the total thrust force ($F$) required to lift the entire mass ($M = 1.0 \times 10^4\text{ kg}$) with this acceleration:

Let $U$ be the constant upthrust force of the air.

Case 1: Initial State
The net upward force on the mass $m$ is $U – mg = ma$.
From this, we get the upthrust: $U = m(g + a)$

Case 2: After mass is detached
Let the mass to be detached be $\Delta m$. The new mass is $(m – \Delta m)$.
The new upward acceleration is $2a$. The upthrust $U$ remains the same.

Substitute $U = m(g + a)$ into this equation:

Group the terms containing $\Delta m$ on one side:

This $\Delta m$ is the mass (and hence dictates the amount of weight) that must be detached. (Note: the original question text asks for the “fraction”, but gives the formula for the absolute mass to be detached. We have proven the required expression.)