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 1.0 meter above the ground level and the top of the track is 2.4 meters above the ground. What isn the distance on the ground
w.r.t. the point B (which is vertically below the end of the track as shown in (figure) where the sphere lands ?

• A. 5
• B. 7
• C. 2
• D. 3

Answer 1

The correct option is C 2

solution:

Applying conservation of energy to the top of the track (D) and point A (as shown in the diagram)

$(PE{)}_D+(KE{)}_D=(PE{)}_A+(KE{)}_A$

$\therefore$ $mg\left(2.4\right)+0=mg\left(1.0\right)+\frac{1}{2}m{v}^2+\frac{1}{2}m{\omega }^2=\frac{1}{2}mv2+\frac{1}{2}2/5M{R}^2\frac{v^2}{R^2}=\frac{7}{10}m{v}^2$

$\Rightarrow$ mg x 1.4 = $\frac{7}{10}m{v}^2$

$\frac{1.4x9.8x10}{7}=v^2$

$\Rightarrow$ $v^2$ = 19.6

$\Rightarrow$ v= 4.43 m/sec

After point A, the body takes a parabolic path. The vertical motion parameters of parabolic motion will be

$u_y=0$

$S_y=1m$

$a_y=9.8m/s^2$

ty=?

$S_y=u_y,t_y+\frac{1}{2}{a}_y{t}_y^2$

t_y = 0.45 secs

Applying this time in the horizontal motion of parabolic path 4.43 x 0.45 = 2 m

hence the correct opt: C