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(A) According to Heisenberg's uncertainty principle,the relation is $\Delta x \cdot \Delta v \cdot m = \frac{h}{4 \pi}$.Thus,$\Delta x = \frac{h}{4 \p

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$2$

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(A) According to Heisenberg's uncertainty principle,the relation is $\Delta x \cdot \Delta v \cdot m = \frac{h}{4 \pi}$.Thus,$\Delta x = \frac{h}{4 \pi m \cdot \Delta v}$.For particle $A$: $\Delta x_A = \frac{h}{4 \pi m_A \cdot \Delta v_A} = \frac{h}{4 \pi m_A \cdot 0.05}$.For particle $B$: $\Delta x_B = \frac{h}{4 \pi m_B \cdot \Delta v_B} = \frac{h}{4 \pi (5m_A) \cdot 0.02}$.Taking the ratio $\frac{\Delta x_A}{\Delta x_B} = \frac{h}{4 \pi m_A \cdot 0.05} 	imes \frac{4 \pi (5m_A) \cdot 0.02}{h}$.$\frac{\Delta x_A}{\Delta x_B} = \frac{5 	imes 0.02}{0.05} = \frac{0.10}{0.05} = 2$.

The uncertainties in the velocities of two particles,$A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$ respectively. The mass of $B$ is five times that of the mass of $A$. What is the ratio of uncertainties $\frac{\Delta x_A}{\Delta x_B}$ in their positions?

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