Question

The blocks are of mass $2\text{kg}$ shown, is in equilibrium. At $t=0$ right spring in figure (i) and right string in figuree (ii) breaks. Find the ratio of instantaneous acceleration of blocks in fig (i) and fig (ii) (Take $g=10m/s^2$)

• A. $\frac{24}{25}$
• B. $\frac{24}{27}$
• C. $\frac{30}{27}$
• D. $\frac{25}{24}$

Answer 1

The correct option is D $\frac{25}{24}$

When the right string is cut, the body (of mass $M$) is constrained to move in the circular path. But when the right spring is cut, the body moves along and normal to the spring.

For string, let's say acceleration is $a_2$
$\Rightarrow Mg\cos {37}^{\circ }=M{a}_2\Rightarrow a_2=\frac{4g}{5}=8m/s^2$ ... (i)
For spring, let's say net acceleration is $a_1$ whose components are $a_1^{{}^{\prime }}$ and $a_1^{\prime \prime }$ along and normal to the direction of spring respectively.
$\Rightarrow kx-Mg\sin {37}^{\circ }=M{a}_1^{\prime }$ ... (ii)
$Mg\cos {37}^{\circ }=M{a}_1^{\prime \prime }\Rightarrow a_1^{\prime \prime }=8m/s^2$ ... (iii)


Also, initially $2kx\cos {53}^{\circ }=Mg$
$kx=\frac{5}{6}Mg$ ... (iv)
Putting this value in eq (ii) we get
$a_1^{{}^{\prime }}=\frac{7}{3}m/s^2$
Where: $a_1=\sqrt{a_1^{\prime 2}+a_1^{\prime \prime 2}}=\frac{25}{3}m/s^2$
$\therefore \frac{a_1}{a_2}=\frac{25}{3\times 8}=\frac{25}{24}$