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{\bf Question}

Near the earth the gravitational force is $-mg$ and hence if $x$
measures distance from the earths surface then the gravitational
potential is $mgx$. Hence consider a ball of mass $m$ which rolls
off the edge of a table and drops from a height $L$ onto the
ground. Using Newton's second law, with only gravity acting,
determine the ball's speed when it hits the ground. Derive an
expression for the energy of the particle and hence show this
predicts the same speed at impact. What is the work done by the
ball from the top of the table to the floor.


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{\bf Answer}

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$\ds m\frac{d^2x}{dt^2}=-mg$ with $\ds x=L$ at $\ds t=0$ and $\ds
\frac{dx}{dt}=0$ at $\ds t=0$

$\ds \Rightarrow \frac{d^2x}{dt^2}=-g \Rightarrow
x=-\frac{g}{2}t^2+At+B$ hence $\ds x=-\frac{g}{2}t^2+L$

Therefore the ball hits the ground when $\ds x=0 \Rightarrow
0=-\frac{g}{2}t^2+L$

$\ds \Rightarrow t=\sqrt{\frac{2L}{g}}$

at $\ds t=\sqrt{\frac{2L}{g}}, \,\,\,\,
\frac{dx}{dt}=-gt=-\sqrt{2gL}$.

So the speed when ball hits floor is $\ds \sqrt{2gL}$.

Energy = kinetic + potential

Kinetic = $\ds \frac{1}{2}m\left(\frac{dx}{dt}\right)^2$, \,\,\,\,
Potential = $\ds mgx$

Hence Energy = $\ds \frac{1}{2}m\left(\frac{dx}{dt}\right)^2+mgx$
because the force only depends on $\ds x$ (conservative) $\ds
\Rightarrow$ Energy is constant.

at $\ds t=0, \,\,\,\, \frac{dx}{dt}=0$ and $\ds x=L \Rightarrow$
Energy =$\ds mgL$

Hence for all time $\ds
\frac{1}{2}m\left(\frac{dx}{dt}\right)^2+mgx=mgL$

Hence when $\ds x=0, \,\,\,\,
\frac{1}{2}\left(\frac{dx}{dt}\right)^2=mgL \Rightarrow
\left(\frac{dx}{dt}\right)^2=2gL$

speed at impact =$\ds \sqrt{2gL}$

Work done =$\ds \int_{x=L}^{x=0}(-mg)dx = [-mgx]_{x=L}^{x=0} =
mgL$


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