Question

A mass $m$ starting from $A$ reaches $B$ of a frictionless track. On reaching $B$, it pushes the track with a force equal to $x$ times its weight, then the applicable relation is

• A. $h=\frac{\left(x+5\right)}{2}r$
• B. $h=\frac{x}{2}r$
• C. $h=r$
• D. $h=\left(\frac{x+1}{2}\right)r$

Answer 1

Answer: (A)

$KE$ of blocks at $B=PE$ at $A-PE$ at $B$
$\frac{1}{2}m{v}^2=$ mgh$-mg2r=mg\left(h-2r\right)$ (i)
$v^2=2g\left(h-2r\right)$
Also, $\frac{m{v}^2}{r}=x$ mg + mg
or $v^2=\left(x+1\right)rg$ (ii)
Equating Eqs. (i) and (ii), we get $2g\left(h-2r\right)=\left(x+1\right)gr$
or $2gh=\left(x+1\right)gr+4gr=\left(x+5\right)gr$
$h=\left(\frac{x+5}{2}\right)r$