Question

A car at rest on a horizontal surface (with coefficient of friction 0.1) has total mass $50\text{kg}$. Gases are ejected from this car backwards with relative velocity $20\text{m/s}$. The rate of ejection of gases is $2\text{kg/s}$. Total mass of gas is $20\text{kg}$. Find the maximum speed of the car in $\left(\text{m/s}\right)$
Take $g=10{\text{m/s}}^2,\text{ln}\left(\frac{4}{3}\right)=0.29$

• A. $0.2\text{m/s}$
• B. $0.8\text{m/s}$
• C. $0.6\text{m/s}$
• D. $0.4\text{m/s}$

Answer 1

The correct option is B $0.8\text{m/s}$

Given,
Total mass of the car $\left(m_0\right)=50\text{kg}$
Total mass of the gas $\left(m\right)=20\text{kg}$
Relative velocity of ejection of gases $v_r=20\text{m/s}$
& Rate of ejection $=2\text{kg/s}$

Car will start moving when the thrust force overcomes the limiting force of friction.
i.e $\mu mg=v_r\left(\frac{-dm}{dt}\right)$
$0.1\times (50-2t)\times 10=20\times 2$ $(\because m=m_0-\mu t)$
$\Rightarrow t=5\text{s}$
Therefore, car will start moving after $t=5\text{seconds}$

At $t=5\text{s}$, mass of the car will be
$50-2\times 5=40\text{kg}$
Total time taken to consume fuel $\left(t\right)=10\text{s}$
At $t=10\text{s}$, mass of the car will be $30\text{kg}$

Assuming $t=0$ when the car starts moving, this means max velocity will be achieved at $t=10-5=5\text{s}$
Thus, final velocity (max) of the car will be
$v=v_r\ln \left(\frac{m_0}{m}\right)-\mu gt$
$v_{\text{max}}=20\times \ln \left(\frac{40}{30}\right)-0.1\times 10\times (10-5)=0.80\text{m/s}$