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)

16 m

Answer 2

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 E (i), $\frac{s_{1}}{s_{2}}=\frac{m_{2}^{2}}{m_{1}^{2}}=\left(\frac{300}{200}\right)^{2}=\frac{9}{4}$ $\frac{36}{s_{2}}=\frac{9}{4}$ $s_{2}=16\, m$