/> Waidhan, Singrauli, India - 486886.
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English Basic
Hindi Basic
Utttar pradesh 2014
Bachelor of Science (B.Sc.)
Waidhan, Singrauli, India - 486886
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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, CBSE, State
ISC/ICSE Subjects taught
Physics, Mathematics
CBSE Subjects taught
Physics, Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Physics, Mathematics
1. Which school boards of Class 12 do you teach for?
ISC/ICSE, CBSE and State
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.
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 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)
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, CBSE, State
ISC/ICSE Subjects taught
Physics, Mathematics
CBSE Subjects taught
Physics, Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Physics, Mathematics
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 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)
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