Question

Calculate the mķnimum value 'v' with which q should be projected so that it just reaches the centre of the ring–

• A. v = $\sqrt{\frac{kQq}{mR}(2 - \sqrt{2})}$
• B. v = $\sqrt{\frac{kQq}{mR}(\sqrt{2} - 2)}$
• C. v = $\sqrt{\frac{kQq}{mR}}$
• D. v = $\sqrt{\frac{kQq}{\sqrt{2}mR}(1 - \sqrt{2})}$

Answer 1

Answer: (A)

v = $\sqrt{\frac{kQq}{mR}(2 - \sqrt{2})}$

Answer 2

U_i + K_i = U_f + K_f

+ $\frac { 1 } { 2 }$mv² = + O

mv² =

v = $\sqrt { \frac { 2 \mathrm { kQq } } { \mathrm { mR } } \left( 1 - \frac { 1 } { \sqrt { 2 } } \right) }$

v = $\sqrt { \frac { \mathrm { kQq } } { \mathrm { mR } } ( 2 - \sqrt { 2 } ) }$