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 $B$
$\frac{1}{2}mv^{2} =$ mgh$ - mg2r = mg \left(h -2r\right)$ (i)
$v^{2} = 2g \left(h - 2r\right)$
Also, $\frac{mv^{2}}{r} = x$ 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 $2gh = \left(x + 1\right)gr + 4gr = \left(x + 5\right)gr$
$h = \left(\frac{x + 5}{2}\right)r$