Question

A body slides on a frictionless surfaces $A$ to $B$ of the velocity of ball at $A$ is $5\text{m/s}$. Its velocity at $B$ is $(g=10{\text{m/s}}^2)$

• A. $8\text{m/s}$
• B. $\sqrt{65}\text{m/s}$
• C. $\sqrt{25}\text{m/s}$
• D. $\sqrt{40}\text{m/s}$

Answer 1

The correct option is B $\sqrt{65}\text{m/s}$

Let $v_B$ is the velocity at point B of a body mass $m$ .
Since total mechanical energy is conserved
$P{E}_i+K{E}_i=P{E}_f+K{E}_f$
(${}^{\prime }{i}^{\prime }$ is for initial and ${}^{\prime }{f}^{\prime }$ is for final)
So, the sum of kinetic energy $\left(K{E}_A\right)$ and potential energy $\left(P{E}_A\right)$ at point $A$ will be equal to the sum of kinetic energy $\left(K{E}_B\right)$ and potential energy $\left(P{E}_B\right)$ at point $B$
$P{E}_A+K{E}_A=P{E}_B+K{E}_B$
$\Rightarrow m\times g\times 2+\frac{1}{2}\times m\times v_A^2=0+\frac{1}{2}m\times v_B^2$
$\Rightarrow m\times g\times 2+\frac{1}{2}\times m\times 5^2=\frac{1}{2}m\times v_B^2$
$\Rightarrow v_B^2-25=4g$
$v_B=\sqrt{65}\text{m/s}$,
(Velocity at point $B=\sqrt{65}\text{m/s}$)