Question

Two masses $m$ and $M$ are connected by a light string passing over a smooth pulley. When set free, $m$ moves up by $1.4\text{m}$ in $2\text{s}$. The ratio $\frac{m}{M}$ is $(g=9.8{\text{ms}}^{-2})$

• A. $\frac{13}{15}$
• B. $\frac{15}{13}$
• C. $\frac{9}{7}$
• D. $\frac{7}{9}$

Answer 1

The correct option is A $\frac{13}{15}$



Given $s=1.4\text{m}$
$t=2s$
$s=\frac{1}{2}a{t}^2$
$1.4=\frac{1}{2}\times a\times 4$
$a=0.7{\text{m/s}}^2$
Applying the result of newton 2nd law $F=ma$
$Mg-T=M\left(0.7\right)$... $i$
$T-mg=m\left(0.7\right)$...$ii$ We get
$\left(\frac{M-m}{M+m}\right)g=0.7$
$\left(\frac{M-m}{M+m}\right)9.8=\frac{7}{10}$
$14M-14m=M+m$
$13M=14m$
$\frac{m}{M}=\frac{13}{15}$

Alternative;
For block having mass $m$ as it is going upward
$\begin{array}{}T=m(g+a)& \text{(a)}\end{array}$
For block having mass $M$ as it is going downward
$\begin{array}{}T=M(g-a)& \text{(b)}\end{array}$
As Tension is same equating both the equations
$m(g+a)=M(g-a)$
$\frac{m}{M}=\frac{(9.8-0.7)}{(9.8+0.7)}=\frac{9.1}{10.5}=\frac{1.3}{1.5}=\frac{13}{15}$