Question

A string with one end fixed on a rigid wall passing over a fixed frictionless pulley at a distance of $2\text{m}$ from the wall, has a point mass $M=2\text{kg}$ attached to it at a distance of $1\text{m}$ from the wall. A mass $m=0.5\text{kg}$ attached at the free end is held at rest so that the string is horizontal between the wall and the pulley, and vertical beyond the pulley. What will be the speed which the mass $M$ will hit the wall when the mass $m$ is released? (Take $g=10{\text{m/s}}^2$)

• A. $3.4\text{m/s}$
• B. $4.1\text{m/s}$
• C. $5.5\text{m/s}$
• D. $6.6\text{m/s}$

Answer 1

The correct option is A $3.4\text{m/s}$

Let the speed of the mass $M$ with which it hits the wall be $V_1$, the mass $m$ is moving up with speed $V_2$.


From energy conservation principle,

Loss in gravitational potential energy = gain in kinetic energy of the two blocks,

$MgH-mgh=\frac{1}{2}M{V}_1^2+\frac{1}{2}m{V}_2^2$

From the geometry of figure,

$H=1\text{m}$

$h=$ Final length of string on the left side $-$ Initial length of string on the left side
$=(1+\sqrt{1^2+2^2})-2=(\sqrt{5}-1)\text{m}$

(since total length of string is constant)

Substituting the values, we get

$2\times 10\times 1-0.5\times 10\times (\sqrt{5}-1)=\frac{1}{2}\times 2\times V_1^2+\frac{1}{2}\times 0.5\times V_2^2$

$\Rightarrow 20-6.2=V_1^2+0.25V_2^2$

$\Rightarrow 13.8=V_1^2+0.25V_2^2...\left(1\right)$

By constraint relation of string, we have
$V_1\cos \theta =V_2...\left(2\right)$

From figure, $\cos \theta =\frac{2}{\sqrt{5}}$

Substituting the value of $\cos \theta$ in $\left(2\right)$, we get
$V_2=\frac{2V_1}{\sqrt{5}}...\left(3\right)$

From $\left(1\right)$ and $\left(3\right)$, we get

$13.8=V_1^2+0.25\times {\left(\frac{2V_1}{\sqrt{5}}\right)}^2$

$\Rightarrow 13.8=V_1^2+0.25\times 0.8V_1^2$

$\Rightarrow V_1^2=\frac{13.8}{1.2}$

$\therefore V_1\approx 3.4\text{m/s}$

So, the speed of the block $M$ will be $3.4\text{m/s}$ when it hits the wall.