Question

Two carts of masses $200\, kg$ and $300 \,kg$ on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are same. If the $200\, kg$ cart travels a distance of $36\, m$ and stops, then the distance travelled by the cart weighing $300\, kg$ is

• A. 32 m
• B. 24 m
• C. 16 m
• D. 12 m

Answer 1

Answer: (C)

Given, $m_{1}=200\, kg$
$m_{2}=300\, kg$
$s_{1}=36 \,m$
Using law of conservation of momentum,
$m_{1} v_{1}+m_{2} v_{2}=0$
or $m_{1} v_{1}=-m_{2} v_{2}$
$\frac{m_{1}}{m_{2}}=-\frac{v^{2}}{v_{1}} \ldots (i)$
Kinetic energy of cart =work done against friction force.
For first cart, $\frac{1}{2} m_{1} v_{1}^{2}=f_{s} \times s_{1}$
$\frac{1}{2} m_{1} v_{1}^{2}=\mu m_{1} g \times s_{1} \ldots(ii)$
For second car, $\frac{1}{2} m_{2} v_{2}^{2}=f_{s} \times s_{2}=\mu m_{2} g \times s_{2} \ldots (iii)$
$\therefore \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}}=\frac{\mu m_{1} g \times s_{1}}{\mu m_{2} f \times s_{2}}$
$\therefore \frac{s_{1}}{s_{2}}=\frac{v_{1}^{2}}{v_{2}^{2}}$
Using Eq. (i), $\frac{s_{1}}{s_{2}}=\frac{m_{2}^{2}}{m_{1}^{2}}=\left(\frac{300}{200}\right)^{2}=\frac{9}{4}$
$\frac{36}{s_{2}}=\frac{9}{4}$
$s_{2}=16\, m$