Question

A small disc $A$ slides down with initial velocity equal to zero from the top of a smooth hill of height $H$ having a horizontal portion. What must be the height of the horizontal portion $h$ to ensure the maximum distance $s$ covered by the disc?

• A.
  $\frac{H}{2}$
• B.
  
  $\frac{H}{3}$
• C.
  
  $\frac{2H}{3}$
• D.
  $\frac{H}{4}$

Answer 1

The correct option is A

$\frac{H}{2}$
Let the velocity of disc at point $\text{B}$ is $v$.
In order to obtain the velocity at point $\text{B}$, we apply the law of conservation of energy. So,


$\text{Loss in P.E = Gain in K.E}$

Substituting the values, from figure we get,
$mg(H-h)=\frac{1}{2}m{v}^2$

$\therefore v=\sqrt{\left[2g\right(H-h\left)\right]}$

Let the disc comes to rest after times $t$ when its passes point $\text{B}$. On applying equation of motion we get,
$h=\frac{1}{2}g{t}^2$

$\therefore t=\sqrt{\left(2h/g\right)}$

Now, distance covered by the disc in time $t$,
$s=v\times t=\sqrt{\left[2g\right(H-h\left)\right]}\times \sqrt{\left(2h/g\right)}$

$\Rightarrow s=\sqrt{\left[4h\right(H-h\left)\right]}$

For maximum value of $s$,

$\frac{ds}{dh}=0$

$\Rightarrow \frac{1}{2\sqrt{\left[4h\right(H-h\left)\right]}}\times 4(H-2h)=0$

$\therefore h=\frac{H}{2}$

Why this question? This question helps students to interpret energy conservation principle mathematically. And also help in using the maxima and minima principle in physics.