## Related questions with answers

A student was given the following problem: A 0.20-kg cannonball with a $+1.0 \times 10^{-4} \mathrm{C}$ charge starts at rest 0.40 m from a fixed $+2.0 \times 10^{-4} \mathrm{C}$ charge. When the cannonball is released, it flies vertically upward.How fast is it moving when 10 m above the fixed charge? The student gave this solution:

$\begin{array}{l}{\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(+2.0 \times 10^{-4} \mathrm{C}\right)\left(+1.0 \times 10^{-4} \mathrm{C}\right) /(0.40 \mathrm{m})^{2}} \\ {=(1 / 2)(10 \mathrm{kg}) v_{y}^{2}+(0.20 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(10 \mathrm{m})}\end{array}$

or $v_{y}=15 \mathrm{m} / \mathrm{s}$. Evaluate the solution to this problem and correct any errors you find.

Solution

VerifiedGivens

$\begin{gather*} m_{ball}=0.20\; kg\;\;, q_1=q_{ball}=+1.0\times{10}^{-4}\; C , \\\\ q_2=q_{fixed}=+2.0\times{10}^{-4}\;C \;\;,r_f=10\;\text{m}\;\;\; \text{and} \;\;\;r_i=0.40\;\text{m} \end{gather*}$

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