Question

Two particles of masses 5.0 g each and opposite charges of +4.0 × 10−5 C and −4.0 × 10−5 C are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

Answer 1

The particle speed is 53.66 m/s.

Explanation:

Step 1:

In contrast to charges, we know that they attract each other. Both particles are thus going to attract eac-h other.  

Given values  

Mass (m) = 5 g. = 0.005 kg.

Charge (q) = 4 × 10⁻⁵ C.

Initial particle distance ($d_1$)= 1 m.

Final distance (d₂) = 50 cm. = 0.5 m.

Step 2:

Since there is no external force, we must use the Energy Theorem Conservation Law.  

Change in kinetic energy + potential emergence change = 0

K$E_f$ - K$E_i$ + P$E_f$ - P$E_i$ = 0.

K$E_f$ - K$E_i$ = P$E_i$ - P$E_f$.

$\left(\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}\right)-0=\frac{K(q)(-q)}{d_{1}}-\frac{K(q)(-q)}{d_{2}}$

$\begin{aligned}&\left(\frac{2 m v^{2}}{2}\right)-0=\frac{K q^{2}}{0.5}-\frac{K q^{2}}{1}\\&m v^{2}=K q^{2}\left(\frac{1}{0.5}-\frac{1}{1}\right)\\&m v^{2}=K q^{2}(2-1)\\&m v^{2}=K q^{2}\end{aligned}$

$\begin{aligned}0.005 v^{2}=&\left(9 \times 10^{9}\right) \times\left(16 \times 10^{-10}\right) \\0.005 v^{2}=& 144 \times 10^{-1} \\0.005 v^{2}=& 14.4 \\v^{2}=& \frac{14.4}{0.005} \\v^{2}=& \frac{14.4}{0.005} \\v^{2}=& 2880\end{aligned}$

$v^2$= 2880

Step 3:

Both sides of Taking square root

$\begin{aligned}&v=\sqrt{2880}\\&v=53.66 \mathrm{m} / \mathrm{s}\end{aligned}$

The particle speed is 53.66 m / s.