Q.A rocket accelerates straight up by ejecting gas downwards. In a small time interval $\Delta t$, it ejects a gas of mass $\Delta m$ at a relative speed $u$. Calculate KE of the entire system at $t + \Delta t$ and $t$ and show that the device that ejects gas does work $= \left(\dfrac{1}{2}\right)\Delta m\, u^2$ in this time interval (neglect gravity).

Concept: Work-energy theorem applied to a variable-mass system — the work done by the internal ejection mechanism equals the increase in kinetic energy of the rocket–fuel system.

Let the rocket have mass $m$ and speed $v$ at time $t$. At $t+\Delta t$, the rocket mass is $m - \Delta m$ moving at $v + \Delta v$, and the ejected gas (mass $\Delta m$) moves downward at speed $u - v$ relative to the ground (since $u$ is relative to the rocket).

Step 1: KE at $t$

Step 2: KE at $t+\Delta t$

Rocket: $\frac12 (m - \Delta m)(v + \Delta v)^2$

Gas: $\frac12 \Delta m (v - u)^2$

Total:

Step 3: Change in KE

Expand to first order in $\Delta m$, $\Delta v$ (neglect $\Delta m \Delta v$ terms):

Simplify:

$\Delta K = m v \Delta v - \Delta m \, u v + \frac12 \Delta m \, u^2$ …

The key insight is that the ejection device does work only on the gas, not on the rocket directly. The rocket gains kinetic energy because it recoils, but that recoil energy comes from the same work done on the gas — the total system’s KE change is exactly the work input. Gravity is neglected, so no external forces act on the system.

Let’s set it up cleanly.

1. System definition and initial state At time $t$, the rocket of mass $M$ moves upward with speed $v$. Its kinetic energy is

No gas has been ejected yet, so the system’s total KE at $t$ is just that:

2. What happens in $\Delta t$ The rocket ejects a small mass $\Delta m$ of gas downward. Relative to the rocket, the gas leaves with speed $u$ downward. So relative to the ground, the gas has velocity

Taking upward as positive: the rocket’s velocity is $+v$, the gas is ejected downward relative to the rocket, so its ground velocity is

Since $u > 0$, this is less than $v$; it could even be negative if $u > v$.

3. Rocket’s new speed By momentum conservation (no external forces), the rocket’s mass becomes $M - \Delta m$ and its new speed $v + \Delta v$. The system’s total momentum is unchanged:

Expand:

The $M v$ cancels with $(M - \Delta m)v + \Delta m\, v$, leaving

So

For small $\Delta m$, this is approximately $\frac{\Delta m}{M} u$, but we keep the exact form.

4. Kinetic energy at $t + \Delta t$ The system now has two parts:
   • Rocket: mass $M - \Delta m$, speed $v + \Delta v$
   • Gas: mass $\Delta m$, speed $v - u$ So

5. Change in kinetic energy Subtract the initial KE:

Expand the first term:

The second term:

Now combine with $-\frac{1}{2}M v^2$. Notice $\frac{1}{2}(M - \Delta m)v^2 + \frac{1}{2}\Delta m\, v^2 = \frac{1}{2}M v^2$, so those cancel. We’re left with:

6. Substitute $\Delta v$ From step 3, $\Delta v = \frac{\Delta m}{M - \Delta m} u$. Then

This exactly cancels the $-\Delta m\, v u$ term. Good — the $v$-dependent terms vanish, as they must because the work done by the ejection device shouldn’t depend on the rocket’s speed.

7. Remaining terms Now

Substitute $(\Delta v)^2 = \left(\frac{\Delta m}{M - \Delta m}\right)^2 u^2$:

So …

Concept: Momentum Conservation + Work–Energy Theorem for a Variable-Mass System

Step 1: Write the KE of the system at time t

Step 2: Write the KE at $t+\Delta t$

Rocket (mass $m-\Delta m$, velocity $v+\Delta v$) plus ejected gas (mass $\Delta m$, ground velocity $v-u$, since it leaves at speed $u$ relative to the rocket, downward):

Step 3: Use momentum conservation to relate $\Delta v$ and $\Delta m$

No external force acts (gravity neglected), so total momentum is unchanged:

Expanding and dropping the second-order term $\Delta m\,\Delta v$:

Step 4: Expand $\Delta K = K_{t+\Delta t}-K_t$ to first order

Step 5: Substitute $m\Delta v = \Delta m\,u$ from Step 3 …