Answer
$0.191\;m$
Work Step by Step
To evaluate the uncertainty $\Delta p_x$ in the momentum, we must first evaluate the momentum component $p_x$. Using the classical expression for momentum, we find
$p_x=mv_x$
or, $p_x=(0.50\times 20)\;kg.m/s$
or, $p_x=10\;kg.m/s$
The uncertainty in the speed is given as $1\;m/s$, that is, $5\%$ of the measured speed $v_x$. Because $p_x$ depends directly on speed, the uncertainty $ \Delta p_x$ in the momentum must be $5\%$ of the momentum:
$ \Delta p_x=(0.05)p_x$
or, $ \Delta p_x=(0.05)(10)\;kg.m/s$
or, $ \Delta p_x=0.5\;kg.m/s$
The Planck constant is of the universe $h=0.60\;J.s$
Then the uncertainty principle gives us
$\Delta x=\frac{h}{2\pi\Delta p_x}$
or, $\Delta x=\frac{0.60}{2\times\pi\times 0.5}$
or, $\Delta x\approx 0.191\;m$
Therefore, the uncertainty in the position $0.191\;m$.
