Discussion Forum

1)

A solid spherical ball rolls on a horizontal surface at 10 m/s and continuous to roll up  on an inclined surface as shown in the figure.If  the mass of balls is 11 kg and frictional losses are negligible, the value of h where the ball stop and starts rolling down the inclination is (Assume , g =10 m/s² )

 

Answer:

Option C

Explanation:

Key idea  For rolling motion , the rotational linetic energy is given by expression k

$KE= \frac{1}{2} mv^{2}\left(1+\frac{k^{2}}{R^{2}}\right)$

 gIven, rolling velocity on horizontal surface. 

v_H=  10 m/s  , mass of ball , m=11kg and g=10 m/s²

Kinetic energy of rotation,

$KE= \frac{1}{2} \times 11\times (1.0)^{2}\left(1+\frac{\frac{2}{5}R^{2}}{R^{2}}\right)$

  ( $\because$     For solid spherical ball,    $k=\sqrt{\frac{2}{5}}R$ )

 $\Rightarrow$       $=11 \times 50 \times \frac {7}{5}$J

 By applying the law  pof energy comservation,

 $KE_{i}+v_{i}= KE_{f}+v_{f}$

$\Rightarrow$      $11\times50\times\frac{7}{5}+0=0+mgh$

 $\Rightarrow$      $h= \frac{50}{10} \times \frac{7}{5}$=7 m

hence, the correct option is (c)