Question

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

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

Answer 1

Answer: (C)

Given, $m_1=200kg$
$m_2=300kg$
$s_1=36m$
Using law of conservation of momentum,
$m_1{v}_1+m_2{v}_2=0$
or $m_1{v}_1=-m_2{v}_2$
$\frac{m_1}{m_2}=-\frac{v^2}{v_1}\dots (i)$
Kinetic energy of cart =work done against friction force.
For first cart, $\frac{1}{2}{m}_1{v}_1^2=f_s\times s_1$
$\frac{1}{2}{m}_1{v}_1^2=\mu m_{1}g\times s_1\dots (ii)$
For second car, $\frac{1}{2}{m}_2{v}_2^2=f_s\times s_2=\mu m_{2}g\times s_2\dots (iii)$
$\therefore \frac{\frac{1}{2}{m}_1{v}_1^2}{\frac{1}{2}{m}_2{v}_2^2}=\frac{\mu m_{1}g\times s_1}{\mu m_{2}f\times s_2}$
$\therefore \frac{s_1}{s_2}=\frac{v_1^2}{v_2^2}$
Using Eq. (i), $\frac{s_1}{s_2}=\frac{m_2^2}{m_1^2}={\left(\frac{300}{200}\right)}^2=\frac{9}{4}$
$\frac{36}{s_2}=\frac{9}{4}$
$s_2=16m$