Question

Assuming all the surfaces to be smooth, if the time period of motion of the ball is $N\times {10}^{-1}sec$ then find $N$. Neglect the small effect of bend near the bottom. $(g=10m/s^2)$.

Answer 1

Time period for ball to move from $A$ to $B$ is given by,

$0.2=\frac{1}{2}g{t}_1^2$

$\therefore t_1=0.2sec$

Velocity of ball at reaching $B$ is given by,

$v_B^2=0+2gs=2\times 10\times 0.2=4m/s$

For reaching $B$ to $C$ time taken is given by,

$0.2=4t_2+\frac{1}{2}g{t}_2^2$

$\therefore t_2=0.863$

adding $t_1$ and $t_2$ we get total time period,

$T=t_1+t_2=0.2+0.863=1.063sec=10.63\times {10}^{-1}sec$

$\therefore N=10.63$