The mass of charge Q is m and the mass of cha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) From the conservation of linear momentum,since the system is released from rest,the total momentum remains zero: $m_1 v_1 = m_2 v_2$.Given $m_1 =

Vedclass · Accepted Answer

$\frac{8KQ^2}{5r}$

Vedclass · Answer

(A) From the conservation of linear momentum,since the system is released from rest,the total momentum remains zero: $m_1 v_1 = m_2 v_2$.Given $m_1 = m$ and $m_2 = 4m$,we have $m v_Q = 4m v_{2Q}$,which implies $v_Q = 4 v_{2Q}$.The ratio of kinetic energies is $\frac{K.E._Q}{K.E._{2Q}} = \frac{\frac{1}{2} m v_Q^2}{\frac{1}{2} (4m) v_{2Q}^2} = \frac{m (4 v_{2Q})^2}{4m v_{2Q}^2} = \frac{16}{4} = 4$.Thus,$K.E._Q = 4 	imes K.E._{2Q}$.By the conservation of energy,the initial potential energy is converted into the total kinetic energy at infinite separation: $U_i = K.E._Q + K.E._{2Q}$.$U_i = \frac{K(Q)(2Q)}{r} = \frac{2KQ^2}{r}$.Substituting $K.E._{2Q} = \frac{K.E._Q}{4}$ into the energy equation: $K.E._Q + \frac{K.E._Q}{4} = \frac{2KQ^2}{r}$.$\frac{5}{4} K.E._Q = \frac{2KQ^2}{r} \Rightarrow K.E._Q = \frac{8KQ^2}{5r}$.

The mass of charge $Q$ is $m$ and the mass of charge $2Q$ is $4m$. If both are released from rest at an initial separation $r$,what will be the kinetic energy $(K.E.)$ of charge $Q$ at infinite separation?

Similar Questions

$A$ charged particle $q$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be:

Charges $+q$ and $-q$ are placed at points $A$ and $B$ separated by a distance $2L$. $C$ is the midpoint between $A$ and $B$. The work done in moving a charge $+Q$ along the semi-circular path $CRD$ is .......

$A$ charge of $2 \ \mu C$ moves from point $B$ to point $C$ along the path shown in the figure. Calculate the work done in $J$.

$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module