Question

A body of certain mass is tied to one end of a light string of length $l$, whose other end is fixed at point $\text{O}$. The body is given a speed of $\sqrt{10gl}$ from the lowermost position $\text{A}$. Find the magnitude of change in velocity of body, while moving from position $\text{A}$ to $\text{B}$ as shown in figure.

• A. $\sqrt{gl}\text{m/s}$
• B. $\sqrt{7gl}\text{m/s}$
• C. $\sqrt{18gl}\text{m/s}$
• D. $\sqrt{9gl}\text{m/s}$

Answer 1

The correct option is C $\sqrt{18gl}\text{m/s}$

Reference axis is shown below


Given velocity at lowermost point $\text{A}$: $\overrightarrow{u_0}=\sqrt{10gl}\hat{i}$
Let velocity at lowermost point $\text{B}$: $\overrightarrow{u_1}=u_1\hat{j}$
Applying mechanical energy conservation between points $\text{A}$ and $\text{B}$, taking $P{E}_A=0$ (reference)
$K{E}_A+P{E}_A=K{E}_B+P{E}_B$
$\frac{1}{2}m{u_0}^2+0=\frac{1}{2}m{u_1}^2+mgl$
$\Rightarrow \frac{1}{2}m\times \left(10gl\right)=\frac{1}{2}m{u_1}^2+mgl$
$\Rightarrow u_1=\sqrt{10gl-2gl}=\sqrt{8gl}$
Hence, $\overrightarrow{u_1}=\sqrt{8gl}\hat{j}$

Magnitude of change in velocity from $A$ to $B$ is given by:
$|\Delta \overrightarrow{v}|=\overrightarrow{v_f}-\overrightarrow{v_f}=\overrightarrow{u_1}-\overrightarrow{u_0}....\left(i\right)$


From vector representation,
$|\Delta \overrightarrow{v}|=\sqrt{{|\overrightarrow{u_1}|}^2+{|\overrightarrow{u_0}|}^2}$
$\Rightarrow |\Delta \overrightarrow{v}|=\sqrt{(\sqrt{8gl}{)}^2+(\sqrt{10gl}{)}^2}$
$\therefore |\Delta \overrightarrow{v}|=\sqrt{18gl}\text{m/s}$