Question

A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has an elevated section and a horizontal part. The horizontal part is $1m$ above the ground and the top of the track is $2.4m$ above the ground. The horizontal distance traveled by the sphere from the edge of the track where it lands on the ground is nearly :

• A. $1m$
• B. $2m$
• C. $3m$
• D. $4m$

Answer 1

The correct option is B $2m$

Applying conservation of energy to the top of the track ( D)and point A ( as shown in the diagram)
$(PE{)}_D+(KE{)}_D=(PE{)}_A+(KE{)}_A$
$\therefore mg\left(2.4\right)+0=mg\left(1.0\right)+\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)\frac{v^2}{R^2}=\frac{7}{10}m{v}^2$
$\Rightarrow mg\times 1.4=\frac{7}{10}m{v}^2$

$\Rightarrow \frac{1.4\times 9.8\times 10}{7}=v^2$
$\Rightarrow v^2=19.6$
$\Rightarrow v=4.43m/sec$
After point A, the body takes a parabolic path. The vertical motion parameters of parabolic motion will be
$u_y=0$
$S_y=1m$
$a_y=9.8m/s^2$
$t_y=?$
$S_y=u_y{t}_y+\frac{1}{2}{a}_y{t}_y^2$
$1=0\times t_y+\frac{1}{2}\times 9.8\times t_y^2$
$\Rightarrow t_y=\sqrt{\frac{1}{4.9}}=0.45secs$
Applying this time in horizontal motion of parabolic path, $BC=4.43\times 0.45=2m$