Question

A ball of radius 11 cm and mass 8 kg rolls from rest down a ramp of length 2 m. The ramp is inclined at 35° to the horizontal. When the ball reaches the bottom, its velocity is (Take, sin 35° = 0.57 and cos 35° = 0.81)

• A. $2m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $4m{s}^{-1}$
• D. $6m{s}^{-1}$

Answer 1

Answer: (C)

Kinetic energy $K=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$K=\frac{1}{2}m{v}^2+\frac{1}{2}\times \left(\frac{2}{5}m{r}^2\right){\omega }^2$
$=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2$ $(\because v-r\omega )$
According to conservation of energy, we get
$\frac{7}{10}m{v}^2=mgh$

$v=\sqrt{\frac{10}{7}gh}=\sqrt{\frac{10}{7}gl\sin \theta }$
$=\sqrt{\frac{10}{7}\times 9.8\times 2\sin 35^{\circ }}=4m{s}^{-1}$