Question

A spherical ball of mass 20kg is stationary at the top of a hill of height 100m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30mand finally rolls down to a horizontal base at a height of 20m above the ground. Find the velocity possessed by the ball at its final position.Given g=9.8m/s.

Answer 1

$\sf{\underline{Here\:we\:should\:use:}}$

$\boxed{\sf{The\:law\:of\:conservation\:of\:energy.}}$

$\sf{\underline{Note:}}$

$\implies$ I. T. E. = Initial total energy

$\implies$ F. T. E. = Final total energy

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{I. T. E.\:of\:block = F. T. E.\:of\:block}}$

$\sf{\underline{Now:}}$

The block is at rest at height 100 m,

$\sf{\underline{So\:initially:}}$ It will have a potential energy component.

$\sf{\underline{Let:}}$

Initial height = $\sf{h_{1} }$

Final height = $\sf{h_{2} }$

Final velocity = $\sf{v}$

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{v = (2gh_{1} - 2gh_{2})^{ \frac{1}{2} }} }$

$\sf{\underline{Now:}}$

By substituting the values,

$\implies$ $\sf{\underline{We\:have:}}$

$\implies$ $\sf{v = (2 \times 9.81 \times 100 - 2 \times 9.81 \times 20) ^{ \frac{1}{2}} }$

$\implies$ $\sf{v = (1596) \frac{1}{2}}$

$\implies$ $\sf{v = 39.94 \: m/s}$

$\implies$ $\boxed{\sf{v = 40 \: m/s^{-1}}}$

$\sf{\underline{So:}}$

The velocity possessed by the ball, at its final position is $\sf{40 m/s^{-1}}$.