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Uday Tiwari Class 11 Tuition trainer in Singrauli/>

Uday Tiwari

Waidhan, Singrauli, India - 486886.

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Overview

I am in teaching line for about 4 years. Presently I am teaching in a school and I know the way that any student can understand even the tricky, challenging and tough problems.

Languages Spoken

English Basic

Hindi Basic

Education

Utttar pradesh 2014

Bachelor of Science (B.Sc.)

Address

Waidhan, Singrauli, India - 486886

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

ISC/ICSE, State, CBSE

ISC/ICSE Subjects taught

Mathematics, Physics

CBSE Subjects taught

Physics, Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Physics

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

ISC/ICSE, State and CBSE

2. Have you ever taught in any School or College?

Yes

3. Which classes do you teach?

I teach Class 11 Tuition Class.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 4 years.

Answers by Uday Tiwari (1)

Answered on 29/12/2018 Learn CBSE/Class 11/Science/Physics

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass 'm+M' move with final velocity let us say 'v' which we have to find out If the external force on system is zero then by law of conservation of momentum of system , we have , initial... ...more

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass  'm+M' move with final velocity let us say 'v'  which we have to find out 

If the external force on system is zero then by law of conservation of momentum of system , we have ,

 initial momentum of system = final momentum of system 

⇒mu +M×0 = (m+M) v 

⇒(m+M) v = mu

⇒v = mu ⁄ (m+M) 

Thus the above value of v is the required final velocity 

 

(ii)  initial kinetic energy of system =

      ½mu² +½ M×0² = ½mu²

Final kinetic energy of system  = ½(m+M) v²

Now putting the value of v= mu⁄(m+M) in final KE

       ∴final KE  =½(m+M) × m² u²⁄ (m+M) ²

                           =½m² u² ⁄ (m+M) 

∴ loss of kinetic energy = initial KE – final KE 

                     =½mu² – ½ m²u² ⁄(m+M) 

                     = {½mu²×(m+M) –½m²u²} ⁄ (m+M) 

                    ={½m²u² +½mMu² – ½m²u²}  ⁄ (m+M)

                  =½mMu² ⁄ (m+M)  

 ∴loss of kinetic energy =½mMu² ⁄ (m+M) 

Answers 3 Comments
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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

ISC/ICSE, State, CBSE

ISC/ICSE Subjects taught

Mathematics, Physics

CBSE Subjects taught

Physics, Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Physics

No Reviews yet!

Answers by Uday Tiwari (1)

Answered on 29/12/2018 Learn CBSE/Class 11/Science/Physics

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass 'm+M' move with final velocity let us say 'v' which we have to find out If the external force on system is zero then by law of conservation of momentum of system , we have , initial... ...more

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass  'm+M' move with final velocity let us say 'v'  which we have to find out 

If the external force on system is zero then by law of conservation of momentum of system , we have ,

 initial momentum of system = final momentum of system 

⇒mu +M×0 = (m+M) v 

⇒(m+M) v = mu

⇒v = mu ⁄ (m+M) 

Thus the above value of v is the required final velocity 

 

(ii)  initial kinetic energy of system =

      ½mu² +½ M×0² = ½mu²

Final kinetic energy of system  = ½(m+M) v²

Now putting the value of v= mu⁄(m+M) in final KE

       ∴final KE  =½(m+M) × m² u²⁄ (m+M) ²

                           =½m² u² ⁄ (m+M) 

∴ loss of kinetic energy = initial KE – final KE 

                     =½mu² – ½ m²u² ⁄(m+M) 

                     = {½mu²×(m+M) –½m²u²} ⁄ (m+M) 

                    ={½m²u² +½mMu² – ½m²u²}  ⁄ (m+M)

                  =½mMu² ⁄ (m+M)  

 ∴loss of kinetic energy =½mMu² ⁄ (m+M) 

Answers 3 Comments
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Uday Tiwari conducts classes in Class 11 Tuition. Uday is located in Waidhan, Singrauli. Uday takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 4 years of teaching experience . Uday has completed Bachelor of Science (B.Sc.) from Utttar pradesh in 2014. HeĀ is well versed in English and Hindi.

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