Question

A ball of mass $2\text{kg}$ is suspended by a light string of length $10\text{m}$ as shown in the figure. It is imparted a horizontal velocity of $50\text{m/s}$ . Calculate the speed of the ball at point $\text{B}$. (Take $g=9.8{\text{m/s}}^2$ )

• A. $36\text{m/s}$
• B. $48\text{m/s}$
• C. $42\text{m/s}$
• D. $38\text{m/s}$

Answer 1

The correct option is B $48\text{m/s}$

Assume point $\text{A}$ to be at height $h_A$ from ground. Taking ground to be reference.


Potential energy $U_A$ at point A $=mg{h}_A$
Kinetic energy $K.E_A$ at point A $=\frac{1}{2}m{v}^2=\frac{1}{2}\left(2\right)\left(2500\right)\text{J}=2500\text{J}$
Hence, total mechanical energy at point $\text{A}$ is $U_A+K{E}_A=(mg{h}_A+2500)$

Potential energy $U_B$ at point $\text{B}$ $=mg{h}_B=mg(h_A+10)$
$=mg{h}_A+(2\times 9.8\times 10)$
$=mg{h}_A+196\text{J}$
Kinetic energy $K.E_B$ at point $\text{B}$ $=\frac{1}{2}m{v}_B^2$

Hence, total mechanical energy at point $\text{B}$ is $(U_B+K.E_B)$
By applying the law of conservation of energy,
$U_A+K.E_A=U_B+K.E_B$
$\Rightarrow U_A+2500=K.E_B+U_A+196$
Or, $K{E}_B=2500-196$
$\Rightarrow \frac{1}{2}\times 2\times v_B^2=2304$
$\Rightarrow v_B=\sqrt{2304}=48m/s$
Therefore, velocity of the mass at point $\text{B}$$=48\text{m/s}$