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Conservation Of Mechanical Energy

Learn conservation of linear momentum and mechanical energy, practice closest distance when one charge is fix, and distance of closest approach when both charges are free and fly away.

Conservation of Linear Momentum and Mechanical Energy

Condition for Linear Momentum Conservation (\(\vec p\))

  • Net external force on the system must be zero i.e., \(\Sigma \vec F_{ext}=0 \implies \Delta \vec p=0\)

Here, \(\vec p_i=\) Initial momentum, \(\vec p_f=\) Final momentum

\(\vec p_i=\vec p_f\)

  • Charges of the system are free.

Here, \(v_1=\) Velocity of \(q_1\) and \(v_2=\) Velocity of \(q_2\)

  • If any of the charge is fixed, momentum conservation is not applied because an external force is required to fix that charge.
  • Linear momentum conservation is not applicable.

Condition for Conservation of Mechanical Energy

  • Work done by internal non-conservative force and external force is zero.

 \(W_{{int}_{nc}}=0\) and \(W_{ext}=0\)

When all charges are free

Reason: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

When one of the charge is fixed

Reason: External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

Flowchart:

Illustration Questions

Choose the correct option out of the following for which mechanical energy and momentum both are conserved.

A

B

C

D

×

In (A), (B) and (D), one of the charge is fixed.

So, linear momentum is not conserved.

In (C), both the charges are free.

Both linear momentum and mechanical energy are conserved.

Hence, option  C  is correct.

Choose the correct option out of the following for which mechanical energy and momentum both are conserved.

A image
B image
C image
D image

Option C is Correct

Escape Velocity when a Small Charge is Released from the Surface of Spherical Charge

  • Work done by internal non-conservative force and external force is zero.

 \(W_{int}=0\) and \(W_{ext}=0\)

When all charges are free

Reason: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

When one of the charge is fixed

Reason: External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

Flowchart:

Illustration Questions

A body of mass m = 3 kg and charge q = –1 C is placed on the surface of a fixed sphere of radius r = 1 m and fixed charge \(q_0=5\,C\) . Calculate the minimum velocity \((v)\) given to a body so that it does not return to the sphere.

A \(3\times10^8 \,m/sec\)

B \(2\times10^{11}\, m/sec\)

C \(\sqrt3\times10^5\, m/sec\)

D \(1.9\times10^{10}\, m/sec\)

×

Since, non-conservative forces and external forces are absent, energy conservation can be applied.

image

Minimum velocity so that particle doesn't return.

Here, U=Potential energy, K = Kinetic energy

By mechanical energy conservation

\((U+K)_{surface}=(U+K)_{ \infty}\)

\(U_{surface}+K.E_{surface}=U_\infty+K.E_\infty\)...(1)

image

\(U_{\infty}\rightarrow\) Potential energy at infinite is zero due to infinite separation, i.e., \(r=\infty\)

For minimum velocity at surface, i.e., \(v=v_e\)

Kinetic energy at infinity = 0

image

So, from (1)

\(-\dfrac {1}{4\pi\epsilon_0}\times\dfrac {5\times1}{(1)}+\dfrac {1}{2}mv^2=0+0\)

\(v^2=3\times10^{10}\)

\(v=\sqrt3\times10^5 \,m/sec\)

image

A body of mass m = 3 kg and charge q = –1 C is placed on the surface of a fixed sphere of radius r = 1 m and fixed charge \(q_0=5\,C\) . Calculate the minimum velocity \((v)\) given to a body so that it does not return to the sphere.

A

\(3\times10^8 \,m/sec\)

.

B

\(2\times10^{11}\, m/sec\)

C

\(\sqrt3\times10^5\, m/sec\)

D

\(1.9\times10^{10}\, m/sec\)

Option C is Correct

Closest Distance when One Charge is Fix

  • Work done by internal non-conservative force and external force is zero.

                          \(W_{int}=0\) and \(W_{ext}=0\)

  • When all charges are free

Reason: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

 

  • When one of the charges is fixed

Reason: External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

Flowchart :

Illustration Questions

A positively charged particle of mass= m and charge = +q is fired with speed = \(v\) from a point far away towards a fixed charge +Q, as shown in figure. Choose the correct expression for closest distance(\(\ell_{min}\)) between them.

A \(\dfrac {Q}{4\pi\epsilon_0v^2 }\)

B \(\dfrac {Qq}{4\pi\epsilon_0v^2}\)

C \(\dfrac {Qq}{2\pi\epsilon_0(mv^2)}\)

D \(\dfrac {Q}{4\pi\epsilon_0m}\)

×

Closest distance will be when particle stops, means total kinetic energy changes into potential energy of the system.

Since, \(W_{ext}=0\) and \(W_{non-conservative}=0\)

So, mechanical energy will be conserved.

image

Applying mechanical energy conservation, 

(Potential energy + Kinetic energy)initial = (Potential energy + Kinetic energy)final

Here, U=Potential energy, K.E.=Kinetic energy

\(U_i+K.E_i=U_f+K.E_f\)

image

\(U_{\infty }+K.E_{\infty}=U_{\ell_{ min}}+K.E_{\ell_ {min}}\)

\(\implies 0+\dfrac {1}{2}mv^2=\dfrac{1}{4\pi\epsilon_0}\dfrac {Qq}{\ell_{min}}+0\)

\(\ell_{min}=\dfrac{Qq}{2\pi\epsilon_0(mv^2)}\)

image

A positively charged particle of mass= m and charge = +q is fired with speed = \(v\) from a point far away towards a fixed charge +Q, as shown in figure. Choose the correct expression for closest distance(\(\ell_{min}\)) between them.

image
A

\(\dfrac {Q}{4\pi\epsilon_0v^2 }\)

.

B

\(\dfrac {Qq}{4\pi\epsilon_0v^2}\)

C

\(\dfrac {Qq}{2\pi\epsilon_0(mv^2)}\)

D

\(\dfrac {Q}{4\pi\epsilon_0m}\)

Option C is Correct

Final Velocity When all Three/Four Charges Fly Away 

Conservation of linear momentum and mechanical energy

Condition for Linear Momentum Conservation (\(\vec p\))

  • Net external force on the system must be zero i.e., \(\Sigma \vec F_{ext}=0 \Rightarrow \Delta \vec p=0\)

           Here, \(\vec p_i=\) Initial momentum, \(\vec p_f=\)Final momentum

\(\vec p_i=\vec p_f\)  

  • Charges of the system are free

Here, \(v_1=\) Velocity of \(q_1\) and \(v_2=\) Velocity of \(q_2\)

  • If any of the charge is fixed, momentum conservation is not applied because an external force is required to fix that charge.
  • Linear momentum conservation is not applicable.

Condition for Conservation of Mechanical Energy

  • Work done by internal non-conservative force and external force is zero. 
  • \(W_{int}=0\) and \(W_{ext}=0\)

 

  • When all charges are free

Reason: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

  • When one of the charges is fix

Reason: External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

Flowchart :

Illustration Questions

Four identical charged particles of mass 'm' and charge '+q' each are placed symmetrically on the vertices of a square of side 'a'. The final speed \('v'\) of each of them when they are very far from each other after setting them free will be

A \(\dfrac{q}{4\pi\epsilon_0a}\)

B \(\dfrac{q^2}{\sqrt {4\pi\epsilon_0ma}}\)

C \(\dfrac{q}{\sqrt {4\pi\epsilon_0a}}\)

D \(q\,\, \sqrt {\dfrac{4+\sqrt2} {8\pi\epsilon_0ma}}\)

×

Direction of force on free charge is shown in figure.

image

Trajectory of free charge is shown in figure.

image image

All the four charges move according to the net force acting on them as shown in figure. By applying linear momentum and symmetry at any instant it is clear that speed of all of them will be equal.

image

By mechanical energy conservation

Here, U=Potential energy, K.E. = Kinetic energy

\(U_i+K.E_i=U_f+K.E_f\)

\(\Rightarrow\)\(U_{square}+KE_{square}=U_{\infty}+K.E_{\infty}\)

\(KE_{square}=0\)  (\(\because\)Initial speed = 0)

and \(U_{\infty}=0\)  (by assumption)

 

image

\(U_{square}=K.E_{\infty}\)

\(\Rightarrow \left( \dfrac {1}{4\pi\epsilon_0} \dfrac {qq}{a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {qq}{a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {qq}{\sqrt 2a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {qq}{a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {qq}{\sqrt 2a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {qq}{a} \right) =\dfrac {1}{2} (4m)v^2\)

\(\Rightarrow\)\(\dfrac {1}{4\pi\epsilon_0} \dfrac {q^2}{a} \left( \dfrac {4}{1} + \dfrac {2}{\sqrt 2} \right)=2mv^2\)...(1)

image

\(v^2=\dfrac {q^2(4+\sqrt 2)}{8\pi\epsilon_0ma}\)

\(v=q \sqrt {\dfrac {4+\sqrt 2}{8\pi\epsilon_0ma}}\)

image

Four identical charged particles of mass 'm' and charge '+q' each are placed symmetrically on the vertices of a square of side 'a'. The final speed \('v'\) of each of them when they are very far from each other after setting them free will be

A

\(\dfrac{q}{4\pi\epsilon_0a}\)

.

B

\(\dfrac{q^2}{\sqrt {4\pi\epsilon_0ma}}\)

C

\(\dfrac{q}{\sqrt {4\pi\epsilon_0a}}\)

D

\(q\,\, \sqrt {\dfrac{4+\sqrt2} {8\pi\epsilon_0ma}}\)

Option D is Correct

Final Velocity When One of the Three/Four Charges Fly Away

Condition for Conservation of Mechanical Energy

  • Work done by internal non-conservative force and external force is zero. 
  • \(W_{int}=0\) and \(W_{ext}=0\)
  • When all charges are free

Reason: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

  • When one of the charges is fixed

Reason: External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

Flowchart :

Illustration Questions

Three identical charges of mass 'm' and charge 'q' each are placed symmetrically on the vertices of an equilateral triangle of side 'a'. If two charges are fixed and third one is set free then the final speed \('v'\)of the third charge when it is very far away from both of the fixed charges is

A \(\dfrac {q}{\sqrt {2\pi\epsilon_0ma}}\)

B \(\dfrac {q}{\sqrt {4\pi\epsilon_0ma}}\)

C \(\dfrac {q}{\sqrt {\pi\epsilon_0ma}}\)

D \(\dfrac {q}{\sqrt {8\pi\epsilon_0ma}}\)

×

Direction of force on free charge is as shown in figure.

image

Since Wext = 0 and Winternal = 0 (non conservative)

So, by applying mechanical energy conservation

Here, U= Potential energy, K.E. = Kinetic energy

\(U_i + K.E_i =U_f+K.E_f\)

or, \(U_{triangle}+K.E._{triangle}=U_{\infty} + K.E._{\infty}\)

 

 

image

Utriangle  = Potential energy of all possible pairs

Uinfinity = Potential energy of the pair which is of finite separation.

KEtriangle = 0  [ \(\because\) initial speed = 0 ]

image

\(\dfrac {1}{4\pi\epsilon_0} \left ( \dfrac {q^2}{a}+\dfrac {q^2}{a}+\dfrac {q^2}{a} \right)+0 =\dfrac {1}{4\pi\epsilon_0}\dfrac {q^2}{a}+\dfrac {1}{2}mv^2\)

or, \(\dfrac {1}{4\pi\epsilon_0} \dfrac {3q^2}{a} =\dfrac {1}{4\pi\epsilon_0}\dfrac {q^2}{a}+\dfrac {1}{2}mv^2 \)

or, \(\dfrac {1}{4\pi\epsilon_0} \dfrac {2q^2}{a} =\dfrac {1}{2}mv^2 \)

image

\(v=\sqrt {\dfrac{q^2}{\pi\epsilon_0ma}} \)

\(v=\dfrac{q}{\sqrt{\pi\epsilon_0ma}} \)

image

Three identical charges of mass 'm' and charge 'q' each are placed symmetrically on the vertices of an equilateral triangle of side 'a'. If two charges are fixed and third one is set free then the final speed \('v'\)of the third charge when it is very far away from both of the fixed charges is

A

\(\dfrac {q}{\sqrt {2\pi\epsilon_0ma}}\)

.

B

\(\dfrac {q}{\sqrt {4\pi\epsilon_0ma}}\)

C

\(\dfrac {q}{\sqrt {\pi\epsilon_0ma}}\)

D

\(\dfrac {q}{\sqrt {8\pi\epsilon_0ma}}\)

Option C is Correct

Final Velocity When Two of Three/Four Charges Fly Away

Condition for Linear Momentum Conservation (\(\vec p\))

  • Net external force on the system must be zero i.e., \(\Sigma \vec F_{ext}=0 \Rightarrow \Delta \vec p=0\)

here, \(\vec p_i=\) Initial momentum, \(\vec p_f=\) Final momentum

\(\vec p_i=\vec p_f\)

  • Charges of the system are free

Here ,  \(v_1=\) Velocity of \(q_1\) and \(v_2=\) Velocity of \(q_2\) 

  • If any of the charge is fixed, momentum conservation is not applied because an external force is required to fix that charge.
  • Linear momentum conservation is not applicable.

Condition for Conservation of Mechanical Energy

  • Work done by internal non-conservative force and external force is zero.

                     \(W_{int}=0\) and \(W_{ext}=0\)

  • When all charges are free

Reason: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

  • When one of the charges is fixed

Reason: External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

 

Flowchart:

Illustration Questions

Three identical charges of mass 'm' and charge 'q' each are placed at the vertices of an equilateral triangle of side 'a'. If one of them is fixed, then the final speed of both the charges when they are set free, is

A \(\dfrac {q}{ \sqrt {4\pi\epsilon_0am}}\)

B \(q\,\,\sqrt {\dfrac {3} {4\pi\epsilon_0am}}\)

C \(\dfrac {q}{ \sqrt {2\pi\epsilon_0am}}\)

D \(\dfrac {q}{ \sqrt {8\pi\epsilon_0am}}\)

×

Direction of force on free charge is shown in figure.

image

Trajectory of free charge is shown in figure.

image image

Since Wext = 0 and Winternal non-conservative = 0

By mechanical energy conservation, 

Ui + KEi = Uf + K.Ef

\(\Rightarrow U_{triangle}+K.E_{triangle}=U_\infty+K.E_\infty\)

image

By applying linear momentum at any instant it is clear that speed of both the charges will be equal.

Also, \(U_{\infty}=0\) because all the pairs are separated by infinite large distance

\(K.E_{initial}=0\) [ Initial velocity = 0 ]

image

\(\Rightarrow \left ( \dfrac {1}{4\pi\epsilon_0} \dfrac {q^2}{a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {q^2}{a} + \dfrac {1}{4\pi\epsilon_0} \dfrac {q^2}{a} \right) + 0 = 0+\dfrac {1}{2}(2m)v^2\)

or, \(mv^2=\dfrac {1}{4\pi\epsilon_0} \dfrac {3q^2}{a}\)

image

or, \(v=\sqrt { \dfrac {3q^2}{4\pi\epsilon_0ma} }\)

\(v=q\,\,\sqrt { \dfrac {3}{4\pi\epsilon_0ma} }\)

image

Three identical charges of mass 'm' and charge 'q' each are placed at the vertices of an equilateral triangle of side 'a'. If one of them is fixed, then the final speed of both the charges when they are set free, is

A

\(\dfrac {q}{ \sqrt {4\pi\epsilon_0am}}\)

.

B

\(q\,\,\sqrt {\dfrac {3} {4\pi\epsilon_0am}}\)

C

\(\dfrac {q}{ \sqrt {2\pi\epsilon_0am}}\)

D

\(\dfrac {q}{ \sqrt {8\pi\epsilon_0am}}\)

Option B is Correct

Distance of Closest Approach when both Charges are Free

Condition for Conservation of Mechanical Energy

  • Work done by internal non-conservative force and external force is zero. \(W_{int}=0\) and \(W_{ext}=0\)
  • When all charges are free,

REASON: Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

  • When one of the charges is fixed,

REASON: External force is present, but work done by this force is zero since charge is fixed, i.e. no displacement occurs. So, mechanical energy conservation is applicable.

Illustration Questions

Two identical point charges of mass 'm' and charge 'q' each are thrown simultaneously with speed 'v' towards each other from very far distance. The closest distance between them is

A \(\dfrac {q^2}{4\pi\epsilon_0mv^2}\)

B \(\dfrac {q}{4\pi\epsilon_0mv^2}\)

C \(\dfrac {q^2}{2\pi\epsilon_0mv^2}\)

D \(\dfrac {q}{2\pi\epsilon_0mv^2}\)

×

At the closest distance of approach, the relative velocity of particle is zero i.e., their velocity will be same.

image

Let the new velocities of particles be  \(v_1\) at their point of minimum separation.

 Considering these two particles in a system, there is no external force acting on the system. Hence, momentum of the system remain conserved.

By Momentum Conservation,

\(p_i=p_f\)

\(mv-mv=mv_1+mv_1\)

\(\Rightarrow 0=2\,mv_1\)

\(\Rightarrow\,v_1=0\)

 

image

 Since,  Wext = 0 and Winternal = 0 (non conservative)

By mechanical energy conservation

Ui + K.E.i = Uf + K.E.f

\(\Rightarrow U_\infty+K.E_\infty=U_{\ell _{min}}+K.E_{\ell _{min}}\)

\(K.E_{\ell _{min}}=0\) (particle stops) & \(U_{\infty}=0\) (due to infinite separation)

\(K.E._{\infty}=U_{\ell _{min}}\)

or, \(\dfrac {1}{2}(2m)v^2=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {q^2}{\ell_{min} }\)

or, \(mv^2=\dfrac {q^2}{4\pi\epsilon_0{\ell_{min} }}\)

\({\ell_{min}=\dfrac {q^2}{4\pi\epsilon_0mv^2}}\)

image

Two identical point charges of mass 'm' and charge 'q' each are thrown simultaneously with speed 'v' towards each other from very far distance. The closest distance between them is

A

\(\dfrac {q^2}{4\pi\epsilon_0mv^2}\)

.

B

\(\dfrac {q}{4\pi\epsilon_0mv^2}\)

C

\(\dfrac {q^2}{2\pi\epsilon_0mv^2}\)

D

\(\dfrac {q}{2\pi\epsilon_0mv^2}\)

Option A is Correct

Final Velocity of a Charge

 

  • Consider the potential difference between two points in a uniform electric field be \(\Delta V\), then difference in potential energy is given by \(\Delta U=q\Delta V\)

[Since, potential difference is potential energy per unit charge]

\(\therefore \Delta V=\dfrac {\Delta U}{q}\)

 

 

  • Let charge q has velocity v1 and v2 at point A and B, respectively.

By energy conservation between A and B,

K.EA + UA = K.EB + UB

U— UB = K.EB – K.EA 

\(-\Delta U=\Delta K.E\)

\(-q\Delta V=\Delta K.E\)

Illustration Questions

A point charge of charge Q=–1 C and mass m=10 kg enters a uniform electric field with zero initial velocity. Final velocity of the particle after accelerating through a potential difference of V=500 Volt will be

A zero

B 10 m/sec

C 20 m/sec

D 50 m/sec

×

Since, \(-q\Delta V= \Delta K.E\)

or, \(-q\Delta V= K.E_f - K.E._i\)

or, \(-(-1)(500)=\dfrac {1}{2}mv^2-0\)

or, \(500=\dfrac {1}{2}(10)v^2\)

or, \(v=10\ m/sec\)

A point charge of charge Q=–1 C and mass m=10 kg enters a uniform electric field with zero initial velocity. Final velocity of the particle after accelerating through a potential difference of V=500 Volt will be

A

zero

.

B

10 m/sec

C

20 m/sec

D

50 m/sec

Option B is Correct

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